linear-algebra

[wintry steppe]

Quick question - can we assume b is an element in the real/complex field?

[lavish jewel]

in general, no

to be fair, what i did of saying b + b = 2 b is also not allowed

you'd rather require that b + b = b

since the addition in the field F is not specified

[wintry steppe]

so the textbook should have been more specific

[lavish jewel]

no, the proof has to be more general

just replace the 2b with b + b

[wintry steppe]

The textbook however had a preface that stated that it uses F to denote either R or C

[lavish jewel]

oh, then yes

[wintry steppe]

So in this case, b + b = 2b holds

[lavish jewel]

sure

[wintry steppe]

Also, dumb question but

{x1, x2, ...xn} are just elements in F, right? Like if F = R, one possible set is {1, 3, 7, 21, ...7}
And suppose x = {x1, x2, ...xn}. Then this list x is an element in F^{n}, right? So using the previous example, x denotes the list {1, 3, 5, 21,...7} and x is an element in R4

[lavish jewel]

they did say int was F^n, yes

this has to be specified tho

you can have for example C^n over R

[wintry steppe]

Ok then....

BTW, what does $\mathbb{R}^{\mathbb{R}}$ mean?

[stoic python]

justini

[lavish jewel]

i have no idea

[teal grotto]

typically, $Y^X$ is the set of all functions from $X$ to $Y$

[stoic python]

c squared

[wintry steppe]

Is it still a vector space?

[lavish jewel]

google tells me that c squared is right

it should mean something like all f such that f: R -> R

[wintry steppe]

So $F^{S}$ is the set of all functions from field $F$ to some set $S$

[stoic python]

justini

[teal grotto]

motivation being that in the finite case, for example, if $X={1,\dots,n}$ and $Y={1,\dots,m}$, then the number of functions from $X$ to $Y$ is $m^n$

[stoic python]

c squared

[wintry steppe]

And $F^n$ is the set of functions from field $F$ to $n = \mathbb{N}$...?

[teal grotto]

with a correct definition for operations of vector addition and scalar multiplication yes

[stoic python]

justini

[wintry steppe]

or is it the other way around?

[teal grotto]

well, if $F^n=F\times F\times\dots\times F$ $n$ times, then you can find a bijection between $F^n$ and $F^{\mathbb{N}_n}$ where $\mathbb{N}_n={1,2,\dots,n}$.

[stoic python]

c squared

[teal grotto]

thats just not the definition

you have to be precise about what you mean here by F^n

like, is it the cartesian product F^n or is it the set of functions from {1,...,n} to F?

[wintry steppe]

Like, $R^{n} = {(x_1, x_2, ..., x_n) : x_{i} \in \mathbb{R})$

[teal grotto]

Like, $\mathbb{R}^{n} = {(x_{1}, x_{2}, ..., x_{n}) : x_{i} \in \mathbb{R}}$

[stoic python]

c squared

[wintry steppe]

Ok, so you can think of this as a function mapping n to R, right? (where n is the set of natural numbers)

[stoic python]

c squared

[teal grotto]

no

[stoic python]

c squared

[teal grotto]

since there is a bijection between the two sets

[wintry steppe]

This might just be me being an idiot but....what exactly does "collection of all functions from {1,2,...n} to R" mean?

[teal grotto]

so the collection of ordered pairs {(1,1), (2,pi), (3,17), (4,20), (5,7), (6,9)} is a function from {1,2,3,4,5,6} to the real numbers

and this would be an element in R^{1,2,3,4,5,6}

the associated ordered 6-tupple in R^6 would be (1, pi, 17, 20, 7, 9)

[wintry steppe]

Wait, so does f: 1 --> 1, f:2 --> pi, etc. or does {1,2,3,4,5,6} mean the index?

[teal grotto]

the former uhhh

1 gets sent to 1, 2 gets sent to pi, 3 gets sent to 17,...

[wintry steppe]

And that's a Cartesian product...

ok what?! \😕

[teal grotto]

{(1,1), (2,pi), (3,17), (4,20), (5,7), (6,9)} is a subset of {1,2,3,4,5,6} x R

which is what functions from {1,2,3,4,5,6} to R actually are

these are different from 6-tupples in the sense that they are definitionally different objects

[wintry steppe]

...then what are 6-tuples?

[teal grotto]

they are elements of R x R x R x R x R x R, which really has different elements than the set {1,2,3,4,5,6} x R

[wintry steppe]

So how do you tell which is which?

[stoic python]

c squared

c squared

[wintry steppe]

Oh...ok

[light briar]

can i ask about linear programming qns here?

[lavish jewel]

it's a fine line between this channel and multivar-calc, but yeah

[light briar]

i'm trying to do 1b

this is what i have so far, am i on the right track?

[twilit anvil]

seems to be alright (though im not going to do the question myself and check) but what are the Y_i doing in your formulation? you take care of the alpha, beta, gamma, theta already with the first 4 constraints. it seems like you can get away with just the X variables.

[sick sandal]

idk if this goes here
is hoofman\kunze linear algebra suitable for everyone or are there prerequisites to be able to access it properly ?

[odd kite]

You need to know some basic things about numbers but there's not much prerequisites it looks like. However that doesn't mean it's suitable for everyone. People will have varying levels of comfort with abstraction. Also the focus of the book may not be appropriate depending on goals or interests.

They do want you to know some basics of complex numbers.

[wintry steppe]

I realized I really have a big problem about linear algebra and derivatives of this lesson such as analytic geometry. I really don't understand a single thing about those lessons even if I listen to some videos about fundamentals many times. I have limited resources in my native language. English also makes me slow but, what do you suggest?

imma go insane jesus last year was different

[lavish jewel]

if you have already gone through the content once, then 3b1b's videos might help you

also the MIT stuff with gil strang

but none of these cover the nitty gritty in detail as well as a book does

so maybe strengthen the intuition first and then revisit the books

[wintry steppe]

is there any english book non-native friendly for math learners

[lavish jewel]

that i wouldn't know

[wintry steppe]

ok

[marble lance]

Hmm, what would make one english book more friendly to non natives than another?

[wintry steppe]

making shorter sentences, using words with their most common meanings etc.

do I realy need to say this

[lavish jewel]

arguably, though, having simpler english would probably mean it has more intricate mathematical constructions instead

[wintry steppe]

I don't mean the terminology

it's a necessity after all

whatever man I am gonna suffer

[lavish jewel]

sounds like it

[marble lance]

I mean, yeah, I can imagine a textbook that is particularly difficult for non natives

But I feel like most math textbooks don't use intentionally difficult English

[winter harbor]

What's your native language?

[wintry steppe]

it's turkish

which is very far from the nature of english

[marble lance]

That's tough. Good luck. It's unfortunate that higher math is kinda locked behind knowing some languages

[wintry steppe]

yeah

[lavish jewel]

my guess is that most people here don't speak english as a first language, but yeah, that's a large limitation

[wintry steppe]

at least I know basic math terms

anyways guys thanks

[marble lance]

Good luck!

[lavish jewel]

wdym "spans of the same length"

there is no length related to a span, so

it means nothing unless you assign it a meaning yourself

i'm trying to figure out what you want to say

cuz there is no such thing 😛

if the bases are equal, all of that is trivial, sure

i think you're mixing up a bunch of terms

sure, by definition, that is true

i still don't know what you meant by length

[marble lance]

What is the span of a vector space?

[lavish jewel]

sure, that's sensible enough

but as luna points out, i think you're having issues with many definitions

[marble lance]

If you're just saying V = span(v1, v2, ..., vn) and W = span(w1, w2, ..., wn) and those two spans are equal, then yeah V = W

But that's kinda trivial so I don't think you meant that

[lavish jewel]

yeah, that's pretty much just by definition

[sick sandal]

https://www.routledge.com/Advanced-Linear-Algebra/Cooperstein/p/book/9781482 has anyone worked with this before?

[midnight trout]

when finding A=LU, is switching rows a legal operation to find U? And then, would I switch rows in reverse to find L?

[lavish jewel]

switching rows in what?

[midnight trout]

for A, for example, finding U first, I'd subtract R1 from R2, then switch R2 and R3, then switch R3 and R4

[lavish jewel]

i don't think that's how LU works

[midnight trout]

So LU is to find A without a permutation

Just trying to think about how to approach this question, I guess.

[lavish jewel]

there IS a special elimination-like algorithm to compute the LU factorization

but it doesn't quite work like this

since when you change rows and stuff this changes the original matrix, you're usually left with an extra factor when doing it this way

something more like PA = LU

[midnight trout]

so the lack of P here is pretty important then

[silent dune]

Would just doing a matrix like this and showing that it is linearly independent be right or is there another way to do this question?

[winter harbor]

Yeah, all you need to do is show these 4 vectors are linearly independent, which can be done by showing this matrix has full rank/is invertible/has non zero determinant.

[silent dune]

The way I added the two extra standard basis vectors is correct tho right? like as long as the two original vectors are still in the new matrix I can show it correctly

I wasn't sure if I should have like the (1 2 3 4) run like horizontal or vertical, but I guess it doesn't matter right

[winter harbor]

Oh, if you had taken the transpose of A instead It would be the same thing, really.

Because a matrix is invertible iff its transpose is.

[silent dune]

oh ya, thanks

would I have to put that matrix in RREF?, or could I just make it so that there is a pivot in each column? and not completely 1's and 0's

[winter harbor]

There's tons of ways to check if a matrix is invertible.

Yeah, you can try putting it in RREF.

Sure, that's a way to do it.

[silent dune]

So like using Guass elim here, and getting this last matrix. Can I just leave it how it is and say that it is obviously linearly independent due to the pivots, hence has a rank of 4, and hence a basis in R^4. Or do I need to make those 2's into 1's,and that 1 in the (1st row, 2nd column) disappear

Like I don't quite understood if it is required to go to RREF fully or just get all pivots

[winter harbor]

You only need to check for the pivots.

[cosmic glacier]

Can someone help me with 1.6 please

[silent dune]

okay nice thanks

[winter harbor]

Qu'avez-vous essayé jusqu'à présent?

Substituer les valeurs de x=2,y=-2,z=4 dans l'équation et résoudre a,b,c. Ce sera un nouveau système d'équations linéaires.

[cosmic glacier]

@winter harbor ah merciiiii

[reef ridge]

so I can prove the first part of the test, but I'm not sure what to do for the second

(this is a subgroup of (Z, +) by the way)

what would the inverse be under addition?

[winter harbor]

The additive inverse of an integer n is -n.

So you have to prove if k is in nZ, then -k is also in nZ.

[reef ridge]

ahh, because they sum to the identity element, which under addition would be 0?

[winter harbor]

Yup

[reef ridge]

gotcha, thank you very much

[winter harbor]

Np

[midnight trout]

For permutation matrices A, B of dimension 3: it seems like AB != BA whenever B != B^(-1) OR A != A^(-1)...

[stoic python]

CreamyBoy

[wintry steppe]

https://cdn.discordapp.com/attachments/269573202018041856/897260266847354910/unknown.png

[stoic python]

CreamyBoy

[midnight trout]

I believe these matrices A, B don't equal their inverse and I'm noticing that when they are one of A or B, then AB != BA

Is this a property of permutation matrices? Does it extend to higher dimensions?

[midnight trout]

what does the red C in C^C mean?

[wintry steppe]

It's the complement of the set C

[midnight trout]

so everything that is not in C?

[wintry steppe]

yeah

[midnight trout]

wait is this for homework?

[wintry steppe]

yeah

[midnight trout]

Do you understand the rest of the notation?

[wintry steppe]

not rly

i just started learning this stuff and im already getting these questions on the hw lol

[midnight trout]

You should know what a union of sets is.

I can't really help you without just giving you the answer otherwise.

[wintry steppe]

hm

i thought it was this but it said it was wrong

cus if it's not c

a U b is that which is wrong

[midnight trout]

what?

[wintry steppe]

idk

[empty egret]

Can someone explain this step by step?

[midnight trout]

can you show what you've tried so far?

[empty egret]

I multiplied g first

giving me g(m1-m2)/(m1+m2)

I then multiply by m1

which gives me gm1^2 - m1m2 / (m1+m2)

multiply m1g * (m1 + m2)/(m1 + m2)

I add this with the previous answer and I get

[midnight trout]

it'd probably b e way less confusing if you just posted your work

[empty egret]

I don't have access right now so I am simply restating my work

[midnight trout]

I'm not familiar with that character above the P.

[nocturne jewel]

It's just "I want to still use P but cant use just P"

Ie just another name

[nocturne jewel]

this also isnt LinAl, this is just algebra

[midnight trout]

so P^3 = I when P!=P^(-1)

so weird

and when P=I obv

I wonder if that holds true for higher dimensions

seems like it does... wild

for permutation matrix P, dimension n, P^n = I when P != P^(-1)

or when P = I

[winter harbor]

Take
\begin{bmatrix}
0 & 1 & 0 \
0 & 0 & 1 \
1 & 0 & 0
\end{bmatrix}

[stoic python]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter harbor]

This matrix cubes to the identity

[midnight trout]

it seems like any permutation matrix with** zeros along the main diagonal works

[winter harbor]

You can also take
\begin{bmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1 \
1 & 0 & 0 & 0
\end{bmatrix}

[stoic python]

MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winter harbor]

I suppose you are not familiar with group theory.

[midnight trout]

no this is all very new to me

[winter harbor]

But this is related to the fact that elements of a finite group all have finite order, and the order of such an element is less than or equal to the order of the group.

[midnight trout]

hmmm

[winter harbor]

Well

To put in simple terms

[nocturne jewel]

Yeah Permutation Matrices sometimes get thrown into 1st year LinAl, I suffered too

[winter harbor]

We could basically take any permutation matrix as an example

And it would work

[midnight trout]

so eventually any permutation matrix P^n will arrive at the identity?

depending on the n

[winter harbor]

yeah, if P is an nxn permutation matrix.

[midnight trout]

that's so cool

[winter harbor]

Oh

[exotic wedge]

hey if u remember the problem we discussed before, how does jordan blocks relate to it? I could actually solve the problem for 2 x 2 matrix it turned out to be like 2 x2 jordan block

[winter harbor]

We also need P to be non degenerate ofc.

I forgot this detail.

[winter harbor]

Oh, yeah. The thing is that a matrix is diagonalizable iff its jordan canonical form is a diagonal matrix.

If it is not diagonalizable

We can still express such a matrix as follows

Matrices of this form are called jordan blocks

And the jordan canonical form theorem guarantess that we can always write a matrix in such a way that it is the direct sum of jordan blocks.

If if its jordan canonical form is not diagonal, we have that the matrix is not diagonalizable.

That's how these things are related.

Ofc, I am talking about complex vector spaces here.

[exotic wedge]

Is this statement true : "A 2 x 2 matrix is non-diagonalizable iff it is similar to the matrix
a, 1
0, a"

[winter harbor]

Yeah, that is correct.

[prisma sail]

im a bit confused by this question not sure how i start it

[winter harbor]

That is a consequence of the jordan canonical form theorem.

https://math.stackexchange.com/questions/2972696/how-many-non-diagonalizable-2-times-2-matrices-are-there-with-all-entries-sing

I found a math stack exchange post discussing this.

[exotic wedge]

Okay, this can be extended to a 3 x 3 matrix right ?

[winter harbor]

Yup.

[winter harbor]

Take the diagonal matrix whose all entries are equal to 8.

[prisma sail]

alright

[exotic wedge]

So a 3x3 matrix is non-diagonalizable iff it is similar to 3 x 3 jordan block ?

[prisma sail]

thank you btw!

[winter harbor]

No, it doesn't generalize like that. You can take direct sum of jordan blocks tho.

[exotic wedge]

How should the matrix look like ?

[wispy pewter]

is the det of A always 1

because it is square, invertible and in RREF

before you do the operations I mean

[gray dust]

no. 2I is invertible but det(2I)=8

[wispy pewter]

hu?

so a better way of saying that is if A is in rref and its a square invertible matrix

is it the 3x3 identity

<@&286206848099549185>

[nocturne jewel]

det(I)=1

so you can undo the EROs to get det(A), since it's known how EROs affect det

[wispy pewter]

I know that once you start doing row operations it changes

but I just wanted to verify

we are starting with det of 1

which is always the case for an invertible square matrix in rref

[nocturne jewel]

no

since A is invertible, it's RREF is I

so the operations end up with det(I) which is 1

[wispy pewter]

the operations end up with a det of 1/2

im not undrestanding

[nocturne jewel]

they dont

[wispy pewter]

you are saying an invertible matrix in rref is always 1

[nocturne jewel]

Yes

cause its RREF is I

That's one of the statements in FTIM

[wispy pewter]

oh ok so these operations are done

to get it into rref

with the final matrix after you interchange rows 3 and 1 being the identity

[nocturne jewel]

yes, you perform those 3 EROs and get A to I

so det(A) becomes det(I)=1

[wispy pewter]

and 2 det(A) = det (I) = 1

so det(A) = 1/2

nice man thanks

I clearly didnt understand it

[nocturne jewel]

oh yeah, it's 1/2

cause 3rd one negates and 1st scales by -2

so 2det(A)=1

not linear algebra

[torpid portal]

This isn't linear algebra. You want #prealg-and-algebra

[wintry steppe]

How would you show that one has listed all the subspaces of a vector space?

[desert palm]

Oh ok

[slow scroll]

By taking an arbitrary subspace and showing it has to be one of the subspaces in the list

[lavish jewel]

what's the y up top?

you made some mistakes, but yes, homogeneity won't hold here

[winter harbor]

@short magnet Sorry for the ping. That was a fun problem

I tried to solve it once I had some free time

Where was the problem proposed originally?

Let,
$$
B = \dfrac{1}{q} \sum\limits_{i=0}^{q-1} A^{i}
$$
Notice that $\forall k \in \mathbb{N}$ we have $A^{k}B$. Indeed, for $k=0$ the result is trivial and for $k=1$ we have:
$$
AB = A \left(\dfrac{1}{q} \sum\limits_{i=0}^{q-1} A^{i} \right) =\dfrac{1}{q} \sum\limits_{i=1}^{q} A^{i}
$$
But since $A^{q} = I$, we have
$$
AB = \dfrac{1}{q} \sum\limits_{i=1}^{q} A^{i} = \dfrac{1}{q} \sum\limits_{i=0}^{q-1} A^{i} = B
$$
By induction, notice that if the result holds for some $k \in \mathbb{N}$, then:
$$
A^{k+1}B = A(A^{k}B) = AB = B
$$
So the result indeed holds.
\
\
Notice that this proves $B$ is idempotent, since:
$$
B^{2} = \dfrac{1}{q} \left(\sum\limits_{i=0}^{q-1} A^{i} \right) \left(\dfrac{1}{q} \sum\limits_{j=0}^{q-1} A^{j} \right) = \dfrac{1}{q} \sum\limits_{i=0}^{q-1} A^{i} \left(\dfrac{1}{q} \sum\limits_{j=0}^{q-1} A^{j} \right)
$$
So,
$$
B^{2} = \dfrac{1}{q} \sum\limits_{i=0}^{q-1} A^{i} B
$$
But we know that $A^{i}B = B, \forall i \in {0, \cdots, q-1}$. From thus we conclude:
$$
B^{2} = \dfrac{1}{q} \sum\limits_{i=0}^{q-1} B = B
$$
Finally, it is easy to see $x-1$ and $\dfrac{1}{q} \sum\limits_{i=0}^{q-1} x^{i}$ are coprime polynomials in $\mathbb{R}[x]$ and we also have:
$$
0 = A^{q} - I = (A-I) \left(\sum\limits_{i=0}^{q-1} A^{i} \right)
$$
From which we thus conclude $(A-I)B = 0$.
\
\
So, by "Lemme des noyaux" we have:
$$
\text{ker} , (A-I) \oplus \text{ker} , B = \text{ker}(A-I)B = \text{ker} , 0 = \mathbb{R}^{n}
$$
We then have
$$
\text{dim} (\text{ker} , (A-I) )= n - \text{dim} (\text{ker} , B)
$$
But we have by rank-nullity $\text{dim}(\text{ker} , B)= n - \text{rank} , B$. And then:
$$
\text{dim} (\text{ker} , (A-I)) = \text{rank} , B
$$
Finally, since $B$ is idempotent, we know
$$
\text{rank} , B = \text{tr}(B) = \text{tr}\left(\dfrac{1}{q} \sum\limits_{k=0}^{q-1} A^{k} \right)
$$
Thus,
$$
\text{dim} (\text{ker} , (A-I)) = \dfrac{1}{q} \sum\limits_{k=0}^{q-1} \text{tr}(A^{k})
$$
As we wished to show ; $\square$

[hot kelp]

Oh, what mistakes do you see?

[stoic python]

MisterSystem

[hollow finch]

lol if its that long maybe use http://mathb.in/

[short magnet]

@short magnet Sorry for the ping. That was a fun problem
@winter harbor It was originally proposed as an oral exam question in "L'école polytechnique"

Glad you enjoyed it ^^

[still lodge]

is it accurate to say that the difference between a change of basis matrix and a transformation matrix is that matrix multiplying v by the former tells us what the coordinates of that v are in a new basis without moving it anywhere, while multiplying by a transformation matrix will tell us where v ends up after applying the transformation described by said matrix

[lavish jewel]

i'd rather say that a change of basis matrix is just a special type of transformation matrix

[still lodge]

can ye elaborate por favor

[lavish jewel]

to begin with, what do you call a transformation matrix?

[still lodge]

well cant any square matrix be interpreted as a transformation matrix

[lavish jewel]

that was exactly point

that means that a change of basis matrix is also a transformation

[still lodge]

that's what's been nagging at me

[lavish jewel]

there's no difference

all matrices represent linear transformations

it just so happens that some of them can be interpreted as a change of basis

[still lodge]

so what distinguishes those

[lavish jewel]

whenever you have something like y = Ax, you can say that x is the coordinates of y in the basis A, if you want

but you could also say that y is the result of applying a linear transformation to x

there is no difference

if no one tells you it's a change of basis, you'd never know

[still lodge]

im asking bc im trying to understand why diagonalization works

[lavish jewel]

the same equation y = Ax could also be a system of linear equations in some number of variables if you'd like, too. you can give the same thing many interpretations

that's kinda the idea behind diagonalization

[still lodge]

yeah the overlapping ideas is what gets me

[lavish jewel]

you have a transformation A

which you can choose to interpret as doing a change of basis into the eigenbasis, stretching the coordinates, and then transforming back to the original basis

or if you want, you can just say that A is the same as the composition of 3 transformations

there is no difference and you can call it what you prefer, they're both true

[still lodge]

it's gonna take me a bit to really let that marinate in my head lol

how exactly does a change of basis correspond to a transformation tho

actually wait

ig if that change of basis matrix is still being expressed in terms of the standard basis

[lavish jewel]

you said yourself that any matrix is a transformation

[wintry steppe]

how do I do this?

[lavish jewel]

pretty much by following the instructions 😛 to be a vector space, V has to follow 8 or so rules

test all of them

[wintry steppe]

It does not have zero.

(x,y) = (0,0)
0 + 0 = (0)(0)+1
0 = 1.

[lavish jewel]

what if x = 1 and y = -1?

then x + y = 0

i'm just playing devil's advocate. be careful when you check

and also don't confuse the addition there

when they say x + y, they mean for you to follow the instructions

addition is defined as xy + 1

so what you did of 0 = 1 doesn't make sense

[wintry steppe]

x+y
= 1+xy
= xy+1
= y+x

idk i suck at this

[lavish jewel]

forget everything you knew

they are literally telling you addition is now a different thing

not what you are used to

forget about the old way of adding

[stable kindle]

might help to use different symbols

[wintry steppe]

x + y : = xy+1

damn

[stable kindle]

??

[wintry steppe]

x + y is now defined as xy+1?

[stable kindle]

ok how about this

let the two new operations be # and % for 'addition' and 'multiplication

then x#y = xy + 1

and k%x = k^2x

so the question is

to find out if there's a 0

does there exist x such that, for all y, x#y = y#x = y

so an x such that, for all y, xy + 1 = y

[wild fulcrum]

@_@

[stable kindle]

what

[lavish jewel]

and the answer would be no, since solving that for x gives you the info that an element that satisfies this property depends on what you are adding it to

this is just one property though, you have to check others and point out all the ones that fail

[wintry steppe]

thanks

[still lodge]

so just to tie my shit together

matrix multiplying a vector by some matrix A has the dual property of telling us where said vector ends up after applying the linear transformation described by the col's of A as well as changing from the basis from the basis of the cols of A to the standard basis

[lavish jewel]

sounds ok

[glad acorn]

if A is invertible does A+I necessarly need to be invertible

I was thinking of det(𝐼+𝐴)=(1+a_1)(1+a_2)...(1+a_n)

[stable kindle]

no

for example consider A = -I

[glad acorn]

i forgot to say I was considering A not equal to -I. It seems that any one of the eigenvalue of A=a_n=-1 and detA not equal to 0 can satisfy the condition

that A+I is not invertible

[lavish jewel]

sounds right

since A + I has the same eigenvectors as A, the eigenvalues of A + I should be lambda_i + 1

[torn stag]

$\det(\lambda I - (A + I)) = \det((\lambda - 1)I - A)$

[stoic python]

IlIIllIIIlllIIIIllll

[midnight trout]

"a symmetric matrix with order n" is an nxn matrix, yes?

[still lodge]

should be yeah

anyone got good resources on jordan canonical form btw

lecturer on it today confused me and 3b1b no longer has my back :(

[mossy jasper]

x+4y-z=-5,

x-y+2z=6,

2x+y+аz=3

For which number of the parametar a, the system of linear equations doesnt have a solution?

[empty latch]

Do any of you know where some good resources are to practice/learn proof-based linear algebra problems?

[winter harbor]

https://yutsumura.com/linear-algebra/

This site has some neat problems.

You can always look up book references ofc.

Axler, Serge Lang, Artin, Dummit and Foote all have some neat linear algebra problems.

Besides Axler, all these other references are not your main linear algebra introductory course books tho.

[empty latch]

@winter harbor thanks a ton

[winter harbor]

Np

[pine lion]

Quick question

Is the dimension of this 2 or 3?

Or hell I now realize I don't even know this

okay wait its 2 my b

[hollow finch]

Yeah. 2 linearly independent vectors span a 2 dimensional space. Has little to do with the dimension of the containing space

[neon holly]

hello guys how can i find the eigenvalues and vectors of the operator T f(x) = f*(x) i have that lambda = f */f

[winter harbor]

Just to make things clear, we have the operator
\begin{align*}
T : \mathcal{M}{n}(\mathbb{C})& \rightarrow \mathcal{M}{n}(\mathbb{C}) \
A \mapsto& A^{\ast}
\end{align*}
Which maps an $n \times n$ matrix with complex with complex coefficients to its hermitian transpose, right? And we are viewing such a map as map of complex vector spaces.

[stoic python]

MisterSystem

[winter harbor]

Well, the way you can approach the problem of finding the eigenvalues of such an operator is as follows.

Notice that $\mathcal{M}{n}(\mathbb{C})$ is isomorphic to $\mathbb{C}^{n^{2}}$. Via de identification which maps
$$
\begin{bmatrix}
a{11} & a_{12} & a_{13} & \dots & a_{1n} \
a_{21} & a_{22} & a_{13} & \dots & a_{2n} \
\vdots & \vdots & \vdots & \ddots & \vdots \
a_{n1} & a_{d2} & a_{13} & \dots & a_{nn}
\end{bmatrix}
$$
To the $n^{2}$-uple:
$$
(a_{11}, \cdots, a_{n1}, a_{21}, \cdots, a_{2n}, \cdots, a_{n1}, \cdots, a_{nn})
$$
Notice also that under such an identification, we can consider the canonical basis
$$
\mathcal{B} = {e_{1}, \cdots, e_{n^{2}}}
$$
Which for convinience, via the identification we can also label as
$$
\mathcal{B} = {e_{11}, \cdots, e_{ij}, \cdots ,e_{nn} }
$$
And we have $T(e_{ij}) = e_{ji}$.
\
\
Since we know how $T$ acts on the canonical basis, we can then view $T$ as an $n^{2} \times n^{2}$ matrix and study the eigenvalues of $T$ using the matrices.
\
\

I will compute things in the 2x2 case explicitly

and then maybe you can try to generalize by yourself.

[stoic python]

MisterSystem

[neon holly]

i think f(x) is just a complex valued function this is more like a 1d problem

so it is complex conjugate of the function rather than adjoint

[winter harbor]

can you phrase the full problem then?

f is a function from what set to the complex numbers?

is it f : C -> C

and f linear?

[neon holly]

it does not specify

it says T is antilinear operator that takes function and returns the conjugate

[winter harbor]

in any case

depending on what kinds of crazy functions you are considering

this approach still applies

try to find a basis for the complex vector space of all functions you are considering

and see how T acts on such a basis

If the vector space of such functions is finite dimensional

Which I am not sure because it is not specified

You can study T via its matrix representation

[neon holly]

i think is mapping from hilbert space to conjugate space

ok i understand what you mean

[winter harbor]

Yeah. But like, be sure to check what kind of spaces you are dealing with.

Because if we are not dealing with spaces of functions which are finite dimensional

We can't just apply simple linear algebra.

[neon holly]

i think simple linear algbra this is qm class

[glad acorn]

how to deduce those two statements? 2 is false 3 is true

[lavish jewel]

well, regardless of the size of the matrix, they are telling you nothing about the matrix

say i give you a 10 x 5 matrix with only zeroes

the null space is all of R^5 and it's a counterexample

thats for 2)

for 3), we now have a 4xn matrix with n > 4

remember that the maximum rank a matrix can have is max(m,n), where m is the number of rows and n is the number of columns

since n > 4, the matrix has maximum rank 4

this implies only a set of 4 columns can be linearly independent

since the columns are linearly dependent, by definition there is a linear combination of them such that the result is 0

this linear combination of vectors is identical to a product Ax = 0 for some nonzero 0

or simply look at the shape and go "more variables than equations lmao"

[glad acorn]

okay, get it

[wintry steppe]

I have the following question: Let A be a non negative matrix (whose diagonal entries are all 0 - in fact it is the adjacency matrix of some undirected simple network). By the Perron Frobenius Theorem, A has a largest eigenvalue k1 with eigenvector v1. How can I show that v1 is non negative?

i've managed to show that k1 > 0 as Tr(A) = 0 and k1 is the largest eigenvalue but I'm stuck from there

[modern palm]

Im not sure what is contained in U and W for this question

[stable kindle]

why do you need to know

[modern palm]

wait, what is U and W a subspace of ?

[stable kindle]

V

[modern palm]

So i need to first prove that U and W is a subspace of V, then I need to show it is contained in U and W

[stable kindle]

no

the question tells you U and W are subspaces of V

oh wait

you mean

'U and W' is a subspace of V

yes

[modern palm]

yes for this?

[stable kindle]

ye

[modern palm]

ok

i mean, from how they defined this set, its gonna be in U and W

theres nothing to prove?

[stable kindle]

you need to show it's a subspace

like

it's self-contained

[winter harbor]

Wait. Can't we use the fact that A is the adjacency matrix of an undirected graph here? Do you need/want to prove this result with such generality?

[modern palm]

yeah, right. So I need to show its a subspace of V, so it contains 0 vector, closed under addition and scalar multiplication

[stable kindle]

yes

[wintry steppe]

sorry for the wait, but yeah we can use that idea

[noble swan]

Hey, I've been trying to figure out how to solve this problem. Can I get some help?

This is the work I've done so far, but it didn't give me a helpful answer

[lavish jewel]

they presumably want you to show this in general, not just for 2x2 matrices

maybe A^-1 A = A A^-1 = I and det(AB) = det(A)det(B) help you out

[noble swan]

Hmm, I don't get how the last identity you used would help me

Oh wait

Nvm

[noble swan]

Ty!

[winter harbor]

I think we can mimick the proof of Perron-Frobenius for the adjacency matrix of a connected graph here.

https://people.orie.cornell.edu/dpw/orie6334/Fall2016/lecture3.pdf

There's a nice proof here.

The idea is basically to apply the min-max theorem, a.k.a variational theorem.

[wintry steppe]

Thanks!

[forest quiver]

Yo when am I allowed to do row operations on a determinant?

I missed last class and my teacher is doign stuff I don'

t understand

He does row operations sometimes, but also has to make the determinant negative for row swapping sometimes?

[buoyant radish]

I’m confused about the R² at the end of this problem:

What sense does it make to ask whether a set of three vectors in R^3 are linearly independent in R²? And if does make sense, how do you go about determining that?

[stable kindle]

typo probs

any three vectors are trivially lin-dependent in R^2, right?

[winter harbor]

Abstractly, the determinant can be seen as an alternating (or skew-symmetric if you prefer) n-multilinear form on $\mathbb{R}^{n}$ (or any field really).

[stoic python]

MisterSystem

[forest quiver]

i don't know what multilinear and field means

im in hs

[winter harbor]

Yeah

I will explain what all of this means

[forest quiver]

Ok

ty

[winter harbor]

But just to give you some intuition

Are you familiar with the interpreation of the determinant as giving you the signed volume of a parallelepiped ?

[forest quiver]

Somewhat from 3b13

3b1b

[winter harbor]

Nice

[forest quiver]

But I really suck at that

[winter harbor]

No, that's the intuition I want you to have.

[forest quiver]

I think it would be better for you to just send me article/textbook pg. /video or something

I legit started this today

i don't have my classes textbook yet so Its a struggle

[winter harbor]

Ok... I want you to think about this in the case of a plane then. Just to develop some intuition.

Notice that a parallelogram in the plane is specified by two vectors.

Which correspond to its sides.

Right?

[forest quiver]

let me @ you when my class is over in 40 mins

Thank you 🥺

[winter harbor]

Aight then

https://www.youtube.com/watch?v=AVMym1KfVXc

I found a video which prolly goes over everything I wanted to say lol.

And is prolly way better explained.

[forest quiver]

Thank you very muchg

It would be great to undersand how the operations defined for determinant actually get the visual / scalar area thing

[winter harbor]

Yeah, that's what I was trying to explain.

How this characterization the the determinant as the unique multilinear alternating map such that on the canonical basis (identity matrix if you prefer) is equal to 1 is related to the fact that determinants measure signed volume.

[buoyant radish]

I’m thinking typo. Assuming it’s R^3 the answer is simple and what I get is consistent with the book.

[stable kindle]

yeah

that's what i said

[buoyant radish]

Im agreeing with you.

I appreciate your input on it. It’s frustrating to spend an hour trying to sort something like this out on for it to turn out to be a typo.

But, in fairness, I also miss typed the numbers into the matrix calculator, so I had two things going on that didn’t make sense

[stable kindle]

f

[raven badger]

How do I model this in a matrix?

a3+-2b = 0
sqrt(a^2+b^2) = 1

The first equation I could easily model, the second one is the one causing me trouble.

Do I just go

[3,2|0
sqrt(1^2),sqrt(1^2)|1]?

please tag me with an answer

?

Someone just wrote something Dx

[wintry steppe]

what is the first equation supposed to be

3a + 2b = 0?

[raven badger]

The dot product between unknown vector v = (a,b) and known vector u = (3,-2)

Actually, it was u = (3,-2) not u = (3,2)

second is the norm of vector V which has to be equal to 1

[wintry steppe]

you're not going to be able to write it in the form M(a, b) = (0, 1) since the second equation isn't linear

[raven badger]

oh...

yup

But, couldn't I manipulate

sqrt(a^2+b^2) = 1

<=> a^2+b^2 = 1^2 = 1
<=> a + b = sqrt(1)

And then I could just compute that, right?

[wintry steppe]

uh, that last <=> is definitely not true

[stable kindle]

(a + b)^2 != a^2 + b^2

[wintry steppe]

they're working in Z/2Z give them a break

[raven badger]

What are you trying to say?

[wintry steppe]

a^2+b^2 = 1^2 = 1
<=> a + b = sqrt(1) 

why

how did you go from the first line to the second

[stable kindle]

by your logic

since (1+2)^2 = 9

1^2 + 2^2 = 9

[raven badger]

sqrt(x) = x^1/2

(sqrt(x))^2 = x^(1/2*2) = x

->

sqrt(a^2+b^2) = nvm.

Well, then can I manipulate it into a linear system?

My current guess, would be no. But then I am lost 🙂

[wintry steppe]

i can't follow your reasoning

anyways, no, you can't

[raven badger]

I see no problem here.

[wintry steppe]

but you can still solve your system

[raven badger]

?

a hint

please

[wintry steppe]

why do you need it to be a linear system?

[raven badger]

Because I am doing linear algebra

so I felt like, I would be too dumb, if it turned out to be a linear system, and I couldn't solve it

• but now I feel dumb for not realizing, it couldn't be soled like that

[wintry steppe]

anyways if you want to solve it, start by solving 3a - 2b = 0 for one of a or b and sub that into a^2 + b^2 = 1

[raven badger]

I did do that, twice

I will try again :/

[raven badger]

See,

I did it again, and got

a = 2/sqrt(5),
b = 3/sqrt(5).

2 a-3b = 0

But

sqrt(a^2+b^2) != 1

[wintry steppe]

idk check your computations

also 2a - 3b is not zero here

[raven badger]

I am legitamately going crazy right now.

[wintry steppe]

have you tried solving it by hand

[raven badger]

I did 3a -2b

Yeah, that's what I did, and then I checked on the PC.

Look at line 2, where I confirm 3*sqrt(5)/5 = 3/sqrt(5)

Alright, did it all on PC. A calculation mistake.

THanks for the help 🙂

[still lodge]

does a non square matrix still describe a linear transformation

i know we can obv squish something like R^3 down to a plane in R^2

[nocturne jewel]

yes

a non-square matrix just means the linear transformation maps 1 space into a different one (as in different dimension)

[still lodge]

ok fair enough

im trying to prove injectivity and surjectivity of a linear transformation from V to Rn where V is just an n-dimensional vector space w basis {v1, v2, ... , vn}

my idea was to say that by virtue of being square, it would be invertible and then use that for injective but idk if that's valid

[nocturne jewel]

being a square matrix doesn't guarantee invertibility

example the 0 matrix

[still lodge]

yeah that's what i mean

but i have no matrix to describe the transformation so idk how to go about injectivity

[nocturne jewel]

what can you say about an injective mapping's kernel/nullspace?

[still lodge]

must be 0

[nocturne jewel]

$\ker(T)={0}$

[stoic python]

Mosh

[nocturne jewel]

yes

ie you show that the only vector which maps to 0 is the 0 vector of V

[still lodge]

🤦‍♂️

i forgor

thanks

[nocturne jewel]

and then surjectivity has a similar characterization, but with the im(T)

[still lodge]

but wait how do i prove it's the only one

[nocturne jewel]

depends on T

[still lodge]

this is what it looks like

i already proved 1

[nocturne jewel]

Well, not a trick question what's the 0 vector of R^n

[still lodge]

{0, ... , 0}

[nocturne jewel]

Right, so under T, you need the vector 0v1+...+0vn

Ignore the shit notation, on phone

Alternatively you can just find the matrix of T and check invertibility

Since that then implies bijectivity of T

[still lodge]

that was my first idea but idk how to find the matrix for that

oh wait

so can i just show that the only linear combination of V that goes to 0 vector is all 0's

or something along those lines, not V i think

[wintry steppe]

any1 tryna help me understand this

[still lodge]

look up inclusion exclusion

[wintry steppe]

hm

[still lodge]

$A \cup B = A + B - A\cap B$

[stoic python]

nitezba

[wintry steppe]

ive been tryin to look up the difference between ∪ and ∩ but google cant put it in simple terms lol

[still lodge]

the difference between an intersection and a union?

[wintry steppe]

yeah

[still lodge]

u sure this is linear

[wintry steppe]

prob not lol

idk what it was

[still lodge]

also use a venn diagram to see it

what class is it for

[wintry steppe]

Mathematics for Business and Economics

so idk what to put it under

lol

[still lodge]

#math-discussion

either way what ive said above should be enough to get you goin

[nocturne jewel]

Just for reference, this wasn't LinAl

More #discrete-math

[hazy forum]

Linear algebra isn't my forte and I'm blanking on how to prove this theorem using the relevant definitions. Could someone please help? So far I've got that for the forward direction, uv=(1/4)((u+v)^2-(u-v)^2), but I'm not sure how to incorporate the fact that T is a distance preserving map

[hazy forum]

ahh thank you!

[hazy forum]

How does this look?

I know it doesn't yet prove the statement since I still have to show T is a bijection to complete the second direction

[scenic fulcrum]

T(u)=0 iff u = 0 so T is injective

[hazy forum]

right. I was gonna do "Let T(u)=T(v). Then 0=|0|=|T(u)-T(v)|=|u-v|=u-v, so u=v and hence T is injective."

I'm not sure how I wanna do the surjective part tho.

the hard part is I can't define a rule for T.

since it's just some arbitrary linear transformation with T(u)T(v)=uv

[scenic fulcrum]

No need to show surjective part

[hazy forum]

how come?

don't I need to show that as well to show it's bijective?

[scenic fulcrum]

T is linear from R^n to R^n
Bijective <=> Injective <=> Surjectivee

[hazy forum]

ah right

the domain R^n is the same as the codomain

thanks!

[fossil void]

Can someone please explain this one too me? I thought it would be true but I guess I'm not understanding something here.

[lavish jewel]

if you row reduce the augmented matrix, you will get 2 rows with zeroes

the other 3 will specify some relationship among the 4 variables

you only need to parameterize one of the variables to solve for the other 3, since you have 3 linearly independent equations

it means one of the equations is extra and unneeded, as it's just a linear combination of the others

[fossil void]

Thank you

That gives me a foothold to dive in further it's really helpful

[sick sandal]

is this a typo ? or am i missing something

[lavish jewel]

yeah, looks like it

[glad acorn]

Are there any good ways to proof this statement? I did find a proof but it's too redundant to write down it. I need to divide into two case that those coefficient equals to 0 or not. Then applying row operations

[lavish jewel]

show that A1 = A2 means the set of columns is linearly dependent. this immediately means the matrix is rank deficient and not invertible

[crystal hound]

What’s the most efficient way to find the upper triangular form of a matrix A, assuming for an n dimensional vector space we don’t have n eigenvectors, ie it can’t be diagonalised?

[half ice]

Upper triangular form of a matrix?

Like, A = PUP^(-1) or something?

[crystal hound]

But like how do you form the P

Because we don’t have a basis of eigenvectors to do it

[half ice]

Such a P wouldn't be unique

Trying to think of a way to find ANY P, haha

You really sure you want this form? An LU decomp is likely to behave much better

[dusky epoch]

jnf?

[lavish jewel]

sounds like a jordan normal form indeed, LU i think is only for invertible mats

nvm i think i was thinking of something else when i said that about LU

but anyway you can just do gaussian elimination

unless you really want a similarity transformation

[crystal hound]

Yeah that’s what my lecturer wants, they’ve done an example but it’s really confusing

[crystal hound]

What’s jnf

[lavish jewel]

jordan normal or canonical form

[dusky epoch]

yeah, jordan normal form

[crystal hound]

What’s jordan normal form, havent studied that yet

[dusky epoch]

diagonalization on steroids

[crystal hound]

Ok, so I’m guessing it involves eigenvectors?

[lavish jewel]

not really

generalized eigenvectors

[balmy wolf]

How would you guys go about doing this? (By contradiction? Is this way valid?)

[lavish jewel]

if you know the fundamental subspaces related to a matrix, you can use that

[wintry steppe]

When there say find all vector v of F^n satisfying some equation involving it, what does all "find all vectors" mean? am I looking for 1 or a collection?. I was only able to find one when solving for v in the equation.

[balmy wolf]

Okay! Thank you!

[nocturne jewel]

find all means.. find all

if there's only 1 vector, then there's 1 vector

if there's 17, then there's 17

[wintry steppe]

oh ok thank you.

[lavish jewel]

it presumably means there are infinitely many solutions and you have to find the general form of all of those vectors

[glad acorn]

a solution set can contains even only one element

if there's no element in it, it's the null set

which means there's no such a solution satisfied the required conditions

[exotic wedge]

Just to make sure I understand. Let v_1, v_2 be the result of orthogonalization grom GS. and let E = span{v_1, v_2}, then the matrix of projection is on the from [Pe_1, Pe_2, Pe_3] where e_k is just the standard basis in R^3. Is that correct?

[gray dust]

yes

[exotic wedge]

One more question, is there a simpler way to find the distance from a vector to subspace spanned by two orthogonal vectors other than finding the projection ?

[gray dust]

not w/o other info. but considering the projection is just dot products, it’s already quite neat

[exotic wedge]

I can do this with the projection but not sure how to do it without it

Tried to find perpendiculars or parallels or if the given vector is actually spanned by the two ortho. vectors but i got nothing

[fresh nimbus]

I have a question with regards to simplexes and the orthocenter of simplexes

Is there a general formula for it?

I am trying to prove that the intersection of all the altitudinal hyperplanes is this said orthocenter but is having trouble

[fresh nimbus]

does that give something that'll solve my question?

or are u posting a new question lol

[fresh nimbus]

If a, b, c are linearly independent then there exists no alpha beta, gamma that will make that true since r will be a 3d space and not a plane

[lavish jewel]

it can be a plane in 3d

try using the definition of a plane passing through a point

[fresh nimbus]

how do u mean?

isn't a plane 2d?

[lavish jewel]

sure, but it can be embedded in a higher dimensional space

[fresh nimbus]

not sure I understand

are u talking abt hyperplanes?

[lavish jewel]

here's an image with several planes in 3D space

in this case they count as hyperplanes, sure

[fresh nimbus]

oh yeah ofc, but can that be represented using a single linear combination?

[lavish jewel]

well a hyperplane passing through a point p0 is defined as all points p such that n dot (p - p0) = 0, for a normal vector n

[fresh nimbus]

hmm ok, suppose I find the relationship of alpha beta and gamma, I still don't see how that helps me find the intersection of all the hyperplanes of unknown dimneion

[fresh nimbus]

or is the question asked not an answer to my question lol

[lavish jewel]

oh yeah, i was just answering coueee, idk about your question lol

[fresh nimbus]

lol yeah I thought his question was suppose to answer mine lmao

[lavish jewel]

nah

[nimble wedge]

how can i prove that ({R{1},+ ) is not a group?

[marble lance]

Do you mean R\{1}? @nimble wedge

[nimble wedge]

yeah

[marble lance]

Closure under addition is not met

Inverses

Look at one of these conditions

For an issue

[nimble wedge]

-1 does not have an inverse

correct?

[winter harbor]

Yeah, the additive inverse of -1 would be 1, which is not not in R \ {1}

[elfin meteor]

this isn't a math problem question but

what tools can i use to calculate e^(i27)?

google is able to calculate e^i but when i type anything more, it doesnt give result

most online calculators dont allow me to input i

[winter harbor]

Wolframalpha

[elfin meteor]

thank you!

[stable kindle]

i mean you could just apply euler's identity and compute the real/imaginary parts separately

[elfin meteor]

ooo

TYTY

https://media.discordapp.net/attachments/852655248652763156/898251925546483712/image0.jpg?width=620&height=465

im trying to use the advice you gave me to solve this

r2 should come out as 786.0514 but im stuck on second to last step, and it's coming out as negative

[winter harbor]

This isn't really linear algebra tho.

[lavish jewel]

yeah i'd suggest moving over to #real-complex-analysis

[elfin meteor]

thank you

[lavish jewel]

or a questions channel

[sick sandal]

i dont see why this isn't related to the power series or how its not an infinite linear combination

and i dont want to stick to finite dimensions as they suggested
can someone explain this example to me

[wintry steppe]

linear combinations are by definition finite

what part do you need explained? you posted an entire page

[lavish jewel]

presumably the last paragraph

[sick sandal]

ok bare with me
if we have an infinite basis
why would it still remain finite?

what makes linear combinations "finite" by definition
or do we just agree they are

[wintry steppe]

can you pull up the definition of linear combination

infinite sums don't make sense in arbitrary vector spaces

[main hawk]

Suppose, you are given a set S, then when you define the span of S, you only take a linear combinations of finitely many elements of S. (This is by the definition as given in Hoffman Kunze and otherwise).

Thus, an infinite power series is not considered to lie in the span of the infinite set {1,x,x²....}. Only finite degree polynomials can lie in this set

[wintry steppe]

finite sum

[sick sandal]

i guess i supposed infinite polynomials can lie in it
i can see the issue with my vision now ty all

[half ice]

You can have an infinite basis, but you need to talk about convergence in that case. That's beyond a linear algebra course

You'll always say "polynomials of degree ≤ n" to make it finite

[lavish jewel]

which book is this btw

lots of lip from someone with a typo on the same page

[main hawk]

Hoffman and Kunze Linear Algebra

[sick sandal]

hoffman kunze

ye i have found 5 typos by now

but everything else is good

[main hawk]

I agree. It's a pretty nice book for abstract lin al.

[main hawk]

Hey, I have a question here,
Even if we have an infinite base, we only take finite linear combinations, so why does convergence come into play?

Are you talking about some other algebraic structure where infinite linear combinations are permitted? (If yes, I am curious as to what they are)

[half ice]

What does it mean to "take a linear combination"?

Nothing is stopping you from summing infinite elements, but only if you know they converge

Again, you'll never do this in a 1st lin alg course. But, you will do this in a functional analysis course

[main hawk]

Ah I see now. (Yes sorry, we assumed throughout our lin al course that lin combinations are finite). Thanks

[nocturne jewel]

The Harmonic linear combination

[dawn copper]

Hello, does anyone have a good source to study from for linear dependent vs independent vectors?

i am studying Gaus elemination and we reached this question

I have issue with how fast the professor decided that 0 and 8 are independent and the 0 in the middle is dependent

[bold sun]

hey i need some help on this question - i dont get part a or b

basically i know for a line to be a subspace it has to satisfy va1 va0 and sm0 i did va1 by making lambda equal zero but idk how to do the rest- like prove the other 2 conditions or how to do prt b

ngl im finding subspaces n subsets bit tricky

[nocturne jewel]

subspaces are subsets of vector spaces which contain the 0 vector, and are closed under addition and scaling

no clue what va1 va0 sm0 is suppose to mean

[bold sun]

va1 is vector addition 1 there the properties basically what u said about being closed under vect addition 0 bring an element and closed under scalar mult

[nocturne jewel]

ok, just poor shorthand

[bold sun]

idk its in the lect notes lol- i didnt make it up

[nocturne jewel]

anyway.. you just take 2 vectors and add them, then show that the summed vectors are in the space

[bold sun]

but like would i make up a vecotor like y1 y2 to do it?

and how to i show it in subspace?

[bold sun]

so like lambda x1 x2 plus y1 y2? like the one thats given and a new one to add

[nocturne jewel]

$v=a[x_1,x_2]^T \ u=b[x_1,x_2]^T$

[stoic python]

Mosh

[nocturne jewel]

adding those together clearly give you something in the set

[bold sun]

whats with the t?

[nocturne jewel]

it's transpose, just means it's a column vector

[bold sun]

ah ok lemme try it see if it works

[fickle citrus]

$Ax=0$ means $(-A)x=0$ right

[stoic python]

ShatteredSunlight

[nocturne jewel]

yes

[fickle citrus]

Right I needed this transformation, as trivial as it sounds ;_;

[bold sun]

so would it be like this i still not so sure ?? plz dont laugh

[nocturne jewel]

yes, you've shown that u+v is in the space.

[bold sun]

okk and the last one with closed undr scalr multiplication is that right to?

[nocturne jewel]

you can explain why a scaled v is in the space.

[bold sun]

hm wdym?

scaled v?

[nocturne jewel]

you need to check scaling

so you need a vector in the set then check if it scaled is in the set.

[bold sun]

whatttttt

[fickle citrus]

In Euclidean spaces it just means if a vector v is inside the set, the infinite straight line containing v is also inside the set

[bold sun]

why am i always so lost?

[fickle citrus]

Think about Euclidean spaces, it will help you

Mastering R^1, R^2 is almost the same as R^n

[bold sun]

never heard of that in my life ng;

ngl

[fickle citrus]

Euclidean means nice geometry

[bold sun]

its only been 2nd week of uni 😦

[fickle citrus]

Non-euclidean is weird, you won't need it now/yet/soon

Euclidean means sum of internal angles of a triangle is 180 degrees

It's nice geometry

[bold sun]

oh ok yh so how does that help with vectors

[nocturne jewel]

Euclidean space is more generally just called R^n, n-tuples of real numbers

[fickle citrus]

The standard picture of (finite) vector (in undergraduate studies) is as positions in this idealised coordinate space

So (1, 2) is 1 step to the 'right' and 2 steps 'up'

(2,2) is 2 steps to the 'right' and 2 steps 'up'

[bold sun]

ummmm.... oh ok yh i get that thats 2d vectrs

[fickle citrus]

You should probably relate back to these kinds of simple objects when thinking about the abstraction in vector spaces

But you can challenge yourself for abstract vector spaces because there are abstract vector spaces, but to get some grounding it's good to go back to the simplest R^n

[bold sun]

yhh....my vectr knowledge aint great we skipped alot out in a levels this yr cuz lockdwn inne so only did 2d i think that makes it worse

[fickle citrus]

Even more if you are an engineer but that's a digression

Oh that's really bad, standard R^n vectors are really important

And standard euclidean geometry is standard geometry of shapes, like cuboids, cylinders and stuff

It should somewhat be useful but that's if you're doing applied math

[nocturne jewel]

You don't really cover geometry in LinAl, bar parallelograms/parallelpipeds/orthogonality

[bold sun]

currently at pure (the module is just called maths and 1st yr dont have applied which is a good thing)

[fickle citrus]

Well lines to me are essentially geometric objects

[golden reef]

For matrices, how would a you be able to rearrange this multiplication of matrices. (XY)^T(XY)

X is a 3x3 matrix, Y is 3x1

[nocturne jewel]

but yeah, U is a subspace of V if $0\in U, \forall u,v\in U, u+v\in U$ and $\forall c\in\mathbb{K}, cu\in U$

[stoic python]

Mosh

[bold sun]

so i did the first one and last one

[fickle citrus]

It's nice to see subspaces as subsets of the bigger space containing it

[bold sun]

just not 2nd- still dont really get that

[nocturne jewel]

you did the inclusion of 0 and additivity

you havent finished scaling/homogeneity

[bold sun]

this is how we learnt the 3 is the 2nd one the same thing as u saying

[nocturne jewel]

?

anyway...

showing it's closed under scaling is pretty straightforward by closure of R

[bold sun]

how? like do we go with the origional one they gave us or do we use the u and v we did ?

[nocturne jewel]

Assume $v=a[x_1,x_2]^T$ is in the space, then is $\lambda v$ in the space?

[stoic python]

Mosh

[bold sun]

so how do we show that we just say if lamda is equal to a then it is an elemnt of the subspace?

[nocturne jewel]

??

a scalar cant equal a vector

[bold sun]

i meant a

[nocturne jewel]

I never said that either

what's $\lambda v$?

[stoic python]

Mosh

[bold sun]

lambda (A) (x1 x2)

[nocturne jewel]

yes

which is (real number)[x_1,x_2]^T

so it's in the space

[bold sun]

oh a is element of real no?

[nocturne jewel]

yes.

since the scalar field is R

[bold sun]

so we just say as a is elemnt of real no its in the subspace?

[nocturne jewel]

???

no

lambda*a is a real number

[bold sun]

woops

[midnight trout]

for some matrix u, the transpose of u times the transpose of the inverse of u:

u' * (u^(-1))' = I

correct?

[nocturne jewel]

yes

[bold sun]

oh ok ....ngl the concept of subspaces n substs is still tricky hard to like get your head round

[golden reef]

how would a you be able to rearrange this multiplication of matrices. (XY)^T(XY)
Would (XY)^T = Y^T X^T

[nocturne jewel]

subsets are something different, subspaces are just subsets of a space which contain 0 and are closed

[bold sun]

uko i have like 2 more qs on this like proving subspces or wtver can i try it first and get it checked to see if it right if i do it without getting stuck - or if im still stuck in between ill ask here agian

[fickle citrus]

I think it's better to phrase that subspaces are subsets which are also vector spaces

[nocturne jewel]

obviously you should attempt it first..

[fickle citrus]

Which is also more technically correct, but the point is that being a vector space is a strict condition

[bold sun]

yah--i will i always do try myself

[bold sun]

btw wdym closed?

[nocturne jewel]

the operations keep you within the space

[bold sun]

everyone uses that term i dont get it?

[golden reef]

when multiplying matrices, (AB)^t (AB), would i be able to do (AB)^t and (AB) seperatly and then multiply the results of both together?

[nocturne jewel]

yes as long as you maintain order

[golden reef]

is there any other way to rearange it? The best i had was B^t A^t (AB)

[midnight trout]

,rotate

[golden reef]

is there anyway to rewrite that equation

[half ice]

@golden reef
Note T can distribute over products. It doesn't change the order though

[nocturne jewel]

transpose does change the order

[midnight trout]

did I mess up the algebra somewhere? This is supposed to equal the identity matrix in the end.

[half ice]

Oop!

[nocturne jewel]

(AB)^T=B^TA^T

[golden reef]

would B^TA^T(AB) = B^TB(AA^T)

or is that wrong

[nocturne jewel]

yes

that's wrong

since matrix * isnt commutative

[golden reef]

would i be able to apply this property here. A(BC) = (AB)C

[nocturne jewel]

yes, matrix * is associative

[golden reef]

i mean that rule in this B^TA^T(AB)

[midnight trout]

B^TA^T(AB) = B^TA^TAB

the order of multiplication is important for matrices

you can't just re-arrange them

[golden reef]

im trying to avoid multiplying AB together, I know what A^TA and B^TB is so I was wondering if theres anyway to move them arround using the rules

[midnight trout]

(AB)'=B'A' -- that might help

[golden reef]

ok yeah i had that

[exotic wedge]

I was trying to solve problem

is this proof correct

[pine lion]

At the top I have the question and I think I have a proof that the zero vector is in the space and that the space has closure under addition but I just want to know if what I have so far is good

[wintry steppe]

your proof that 0 is in the space looks good

as is the proof that it's closed under addition

seems good

[pine lion]

Finally getting the hang of this!!

Thank you

[empty ibex]

How to prove that a 3 by 3 matrix of 1's is diagonalizable?

[hollow finch]

find a basis of eigenvectors

[exotic wedge]

can someone check my proof ?

[gleaming knot]

Meanint that A^(k-1) is not zero but A^k is zero, right?

Since A^(k-1) isn't zero try picking v to be a vector such that A^(k-1)v =/= 0

that might work

curious to see your solution

Ah very nice

[cinder lotus]

hi can anyone explain this

[oak obsidian]

Multiply by inverses of matrices on both sides to get an equation that looks like C = some shit

[glad acorn]

does randomly drawn vectors tends to be all mutually orthogonal in R^{infinity}?

[lavish jewel]

i don't know that that means what you think it does

one doesn't say R^infty

[gleaming knot]

One does actually, it means all sequences approaching 0

Actually hmm

It means the set of all sequences finitely many terms of which are nonzero

1st is what I thought I remembered from Munkres

2nd was obtained from a perusal of stackexchange

[lavish jewel]

but you can show this statistically. say you have a random process {X_n} of independent and identically distributed variables. say they follow a distribution with 0 mean. then any E{X_i * X_j} is 0

ah, i wasn't aware of that, icy

[gleaming knot]

After much searching without Ctrl+F capability I have found the relevant Munkres excerpt

Turns out my memory was wrong and in both references, $\bR^\infty$ means the set of all sequences with finitely many nonzero terms

[stoic python]

Icy001

[flint hinge]

Is a scalar a vector with only one entry?

[wintry steppe]

pretty much

[empty ibex]

How do I do this? i was able to get as far as showing that det((A-2I)(A-3I)(A-4I))=0 but dont know what to do after

[dusky epoch]

(D-2I)(D-3I)(D-4I) = 0

conjugate both sides by M to get M(D-2I)(D-3I)(D-4I)M^-1 = 0

insert a couple more MM^-1 pairs in the right places to eventually arrive at (MDM^-1 - 2I)(MDM^-1 - 3I)(MDM^-1 - 4I) = 0

[undone hamlet]

Hello is there a site where I can see the linear algebra proofs completed?

[teal grotto]

the linear algebra proofs

[undone hamlet]

??

[lavish jewel]

many of the basic ones are on wikipedia

[sick sandal]

can someone explain the intuition behind this theorem ?
i can see how its applicable but i find it hard to understand it

[lavish jewel]

looks like a change of basis transformation

as an example, say we have a vector [1,2,3] in the canonical basis

but we want to represent it in a different basis

we can put the vectors of the new basis as columns of a matrix P

then there is some coordinate vector v such that Av = [1,2,3]^T

A has all lin indep columns, so it's invertible, and what it's doing is taking coordinates in some basis, and converting them to the canonical basis

and if you want to know what v is, the coordinates of the vector [1,2,3]^T in this new basis, then you get that v = A^-1 [1,2,3]^T

so A can take coords in one basis and convert them to the canonical one, and A^-1 does the opposite. it takes the vector in the canonical basis and gives the coords in the other basis

idk if that helps you @sick sandal

[sick sandal]

im gonna need to look more about it to get my head around it but yeah that helps paint the picture

[eternal jetty]

im stuggling with part 3

i know how to test if bectors are a basis, its just linear indpendance and span

but unsure of how to do it in the case because idk what the dimensions of H are

[lavish jewel]

do you know how the number of basis vectors relates to the dimension of a vector space?

[eternal jetty]

do you mean like R^3 having 3 basis vectors?

or R^2 having 2

oh so it cant be because H is a 3 dimensional vector space so because we only have 2 vecotors they cannot span H

[lavish jewel]

sounds about right

[modern palm]

Is it possible to have a linear transformation T with dim ker T = 0?

As in no vectors map to the 0 vector?

[lavish jewel]

(other than the 0 vector)

yes

all invertible matrices behave this way, for example

[modern palm]

ok, ill think about it

thanks\

[glad acorn]

(A+aI)B=-bA, A,B are square matrices

How do I proof whether (A+aI) is invertible

a, b are real scalars, I is the identity matrix, all matrices are same sizes

[lavish jewel]

is there any other context?

[glad acorn]

a, b are both non-zero

AB+bA+aB=O, where O is the zero matrix

[sick sandal]

as far as i know if a matrix A is invertible then there is some B such that AB = BA = I right?

[lavish jewel]

i don't see any useful relationship to exploit