linear-algebra

[warm harbor]

i get it now

thank you @hollow finch !!

[arctic hazel]

hey nix do you think you could help me understand my problem

it's right above

[last geyser]

Well, can anyone help with my final test?

[cunning arch]

1. Requesting help during an exam is a bannable offense. #rules

[wintry steppe]

[cunning arch]

(Looking at the second paragraph). $X = \begin{bmatrix}1&-2&4\1&-1&1\1&1&1\1&2&4\end{bmatrix}\ and y=\begin{bmatrix}-2\2\-1\0\end{bmatrix}$

[stoic python]

strawberrypocky

[last geyser]

How do you even know I;m taking the test right now lol

[cunning arch]

I-????

[hollow finch]

you have y in terms of the columns right?

[cunning arch]

Yeah, $y=-\frac{1}{4}b_1+\frac{1}{10}b_2-\frac{1}{2}b_3-8b_4$

[stoic python]

strawberrypocky

[cunning arch]

There's some kind of connection between that linear combination and the projection right?

I just don't know what the connection is

[hollow finch]

wait what is B

[cunning arch]

Orthogonal basis made up of $b_1=\begin{pmatrix}1&1&1&1\end{pmatrix},\ \ b_2=\begin{pmatrix}-2&-1&1&2\end{pmatrix}\ ,\ b_3=\begin{pmatrix}\frac{3}{2}&-\frac{3}{2}&-\frac{3}{2}&\frac{3}{2}\end{pmatrix},\vec{b}_4=\begin{pmatrix}\frac{1}{10}&-\frac{1}{5}&\frac{1}{5}&-\frac{1}{10}\end{pmatrix}$

[stoic python]

strawberrypocky

[cunning arch]

Basically all the parts before this part gave us X and y, I had to build the orthogonal basis B with gram schimt or whatever the name was, then I wrote y as a linear combination of the orthogonal basis vectors

[hollow finch]

i think the first 3 vectors of B should have been the columns of X

[cunning arch]

why?

X isn't an orthogonal basis, so that's why we had to build orth. basis B

yeah no wha?

wait i don't think they're orthogonal

yeah no first dot third column != 0

[hollow finch]

my b

[cunning arch]

lol no worries

[hollow finch]

my fault for trying to do linear algebra while im tutoring differential equations

[cunning arch]

wait is this an orthogonal projection?

wait there's a formula for this

[hollow finch]

@cunning arch did you figure it out?

[cunning arch]

well i did something, not sure if it's right but at least it's something lol

i just used orthogonal decomp formula

so it's the same, y'=y i think

[hollow finch]

oh did you delete your picture of the problem?

[cunning arch]

oops yeah

did you figure out something?

[hollow finch]

not yet at least i just wanted to take a closer look since im done tutoring

[cunning arch]

@hollow finch

that would actually help a lot if you could, thank you :)

[arctic hazel]

i don't want to take away from this problem being solved, but i still want to type out my problem here, if that's okay:

[hollow finch]

just to make sure would you be able to post the whole problem (the stuff that led you to B and y and whatnot)?

[arctic hazel]

Consider the space $\mathbb{R}^2$ endowed with the standard dot product $\varphi(x, y) = x \cdot y$. Let $\theta$ be any real number. I have already proven that $\mathcal{T}_A$ and $\mathcal{T}_B$ are isometries, where $A = \begin{bmatrix}\cos \theta & - \sin \theta \ \sin \theta & \cos \theta\end{bmatrix}$ and $B = \begin{bmatrix}\cos \theta & \sin \theta \ \sin \theta & - \cos \theta\end{bmatrix}$. \ Prove that every isometry $\mathcal{T}: \mathbb{R}^2 \to \mathbb{R}^2$ is of one of the two forms above.

[cunning arch]

it's kinda a long problem and i feel bad if i spam this channel with pictures. are you ok with it if I dm it to you?

[hollow finch]

yeah thatd be great

[stoic python]

Snodlop

[arctic hazel]

finally

sorry to embarrass myself like that lol

[frigid otter]

I'm trying to do an Ax=b style problem, like I have matrix A and I have to find the values the variable matrix to equal b (also a vector). But I think my matrix is singular. If I have a singular matrix can I do that?

[hollow finch]

yes

as long as b is in the column space of A the system has a solution

@arctic hazel oh yeah thats a neat problem

the professor i was a tutor for had that on a worksheet

so lets start with what we know about the matrices of isometries. what is the definition of an orthogonal matrix?

[frigid otter]

Could I use Cramer's rule for that Ax=b question?

[hollow finch]

cramer's rule only works on invertible matrices unfortunately

[wintry steppe]

cramer emote

[hollow finch]

hey i really like cramers rule

[frigid otter]

So what other way is there to solve it? Both ways I think I know require invertible matrices, A^-1*B = x, and Cramer's rule

[hollow finch]

well theres the most fundamental basic way

row reduction

[frigid otter]

with an augmented matrix?

[hollow finch]

that would be useful yeah

[frigid otter]

Hm, alright I'll give that a go

That's called Gauss-Jordian elimination, right?

[hollow finch]

ye

[frigid otter]

thank you 🙂

[steep hearth]

Can I get help with this problem please?

[wintry steppe]

@steep hearth read the pinned message

[steep hearth]

What channel is it in?

I read the how to get help channel I don’t see anything @wintry steppe

Ohhh I see

My bad I think I posted this in the wrong channel

[arctic hazel]

an orthogonal matrix has A * At = I

At is the transpose of A

[hollow finch]

great so then calculate AA^T for an arbitrary 2x2 matrix and set it equal to I

[arctic hazel]

by arbitrary you just mean [a b \ c d]?

[hollow finch]

yeah sure thats fine

thats what id use

[arctic hazel]

aight give me a second

ok i think i see where to go from here, i got sums of squares on the diagonal

so i can set sin theta and cos theta and they equal the identity matrix

[hollow finch]

you might want to also do A^T A=I too

[arctic hazel]

i haven't computed it yet but my guess is it'll come out to be the other isometry described?

[hollow finch]

well lets start not with the diagonal but with the other entries

ac+bd=0

it would be really nice if we could reduce the number of variables we're considering

[arctic hazel]

yes i agree

i was just about to set all 4 of them to equal trig functions but i think you have a more elegant trick up your sleeve

[hollow finch]

darn it would have been nicer if we did A^T A but its fine

[arctic hazel]

i can do that instead it's fine i got time

[hollow finch]

okay cool

so then its ab+cd=0

this is just (a,c) dot (b,d) right?

[arctic hazel]

true yeah

which means two orthogonal vectors

[hollow finch]

yeah exactly

[arctic hazel]

why is it harder to just do (a,b) dot (c,d) instead?

[hollow finch]

so if (b,d) is orthogonal to (a,c) then its some scalar multiple of the orthogonal complement (-c,a)

its just nicer because columns are almost always more important than rows

does it make sense that (-c,a) would be orthogonal to (a,c)?

or at least that it is

[arctic hazel]

ok cool yeah

and yes that does make sense

[hollow finch]

great

so now our matrix is
a -kc
c ka

which is starting to look familiar so we're on the right track

so now we can bring in our equations we got from the diagonals

[arctic hazel]

a^2+b^2, c^2+d^2?

er

i guess now we made it with fewer variables

[stoic python]

nix

[hollow finch]

since we're doing A^T A

now we can solve for k

[arctic hazel]

k is plus or minus 1

right?

[hollow finch]

there we go

thats great

[stoic python]

nix

[hollow finch]

and these look really close to what we want

[hollow finch]

and we can use your idea from before

to get expressions for the possible values of a and c such that a^2+c^2=1

[arctic hazel]

bada bing bada boom then

so that gives me both of the isometries i described!

[hollow finch]

yep! 🙂

[arctic hazel]

in the initial problem

holy cow thank you so much

[hollow finch]

np

[arctic hazel]

one more problem (should be fast if you're interested, if not i'm sure i can figure it out on my own):

Make a choice: either $V$ will be a complex vector space with $\varphi$ a Hermitian form on $V$, or $V$ will be a real vector space with $\varphi$ a symmetric bilinear form on $V$. In either case, suppose $\dim(V) = n$ and $W$ is a subspace of $V$ such that $\varphi(w_1, w_2) = 0$ for all $w_1 \in W$ and $w_2 \in W$.

Give an example for which $\dim(W) = \frac{n}{2}$ and $\varphi$ is nonsingular on $V$.

[stoic python]

Snodlop

[arctic hazel]

weather update: i am lost on the above problem and would like to request assistance

i know it's nonsingular for cases where the determinant of the associated matrix is nonzero, but idk how to come up with examples for that

i also don't know how it'd be zero for half of the basis but not for the full thing

[hollow finch]

@arctic hazel I would @ the helpers

It has indeed been at least 15 mins lol

[arctic hazel]

ah, then i shall do so

<@&286206848099549185>

i never know when it's appropriate to do that

[dreamy iron]

I have a vector space $\mathbb{V} $ with a basis $\eta \equiv \left{ \vec{v}_1 , \vec{v}_2 ,,\ldots, , \vec{v}_n \right}$.

Since this is a basis, then no $ \vec{v}_i$ is the zero vector.

So it’s sensible to take the span of each vector $\text{span}\left( \vec{v}_i \right )$ and we get a lot of subspaces:
$$\mathbb{U}_1 \equiv \text{span}\left( \vec{v}_1 \right ), \ \mathbb{U}_2 \equiv \text{span}\left( \vec{v}_2 \right ), \ \cdot \ \cdot \ \cdot \ \mathbb{U}_n \equiv \text{span}\left( \vec{v}_n \right )$$

I have two questions which I think the answer are yes.

1. Considered the direct sum of the several $\mathbb{U}_i $’s:

$$\bigoplus_{i=1}^{n} \mathbb{U}_i \stackrel{?}{=} \mathbb{V}$$

2. $$\bigoplus_{i=1}^{n} \mathbb{U}_i \stackrel{?}{=} \text{span}\left(\vec{v}_1 , \vec{v}_2 ,,\ldots, , \vec{v}_n \right)$$

[stoic python]

ninnymonger is a physics main.

[hollow finch]

that sounds right to me, but i would wait until someone more experienced with direct sums gives their opinion

[arctic hazel]

isn't it true that span(v1,v2,...,vn)=V? so yeah that looks right

[native rampart]

Should be correct

[hollow finch]

,rotate

[near nova]

i set up this matrix and my friend said this is the first step.

but where did she get 23/34 from!

[hollow finch]

because that would get a zero in the 2,1 entry

115/170=23/34

[near nova]

ooooo

okay thank you

[arctic hazel]

i'm gonna repeat my question if that's okay since i'm still stuck on it:

Make a choice: either $V$ will be a complex vector space with $\varphi$ a Hermitian form on $V$, or $V$ will be a real vector space with $\varphi$ a symmetric bilinear form on $V$. In either case, suppose $\dim(V) = n$ and $W$ is a subspace of $V$ such that $\varphi(w_1, w_2) = 0$ for all $w_1 \in W$ and $w_2 \in W$.

Give an example for which $\dim(W) = \frac{n}{2}$ and $\varphi$ is nonsingular on $V$.

[stoic python]

Snodlop

[pallid swallow]

#help-7｜zen1thxyz [question there about linear algebra, discuss there]

[round coral]

could anyone please tell me if I am doing the wedge product correctly
(a v_1 + b v_2) ∧ ( c v_1 + d v_2 ) = av_1 ∧ ( c v_1 + d v_2 ) + b v_2 ∧ ( c v_1 + d v_2 ) = (ad - bc) (v_1 ∧ v_2) . here v_1 and v_2 are two linearly independent vectors in a vector space V over field F and a, b, c and d are just elements of F

[wintry steppe]

looks good

[round coral]

thanks

[stray granite]

anyone ?

[native rampart]

You missed some

[stray granite]

I know but which ones ?

[native rampart]

This is not a test right?

[stray granite]

no it is review thing

for the test

[native rampart]

C also works

(The one above what you marked)

All polynomials of the form you need will be of the form p(x)(x-2)(x-5)

Since deg<=3,deg(p(x)) has to be 1 at most

[stray granite]

right

so obviously not b

[native rampart]

So,It's all polynomials of form ax(x-2)(x-5)+b(x-2)(x-5)

[stray granite]

so then only the one I marked and one above it

[native rampart]

Yes

[stray granite]

alright thank you

just wanted to make sure. is this correct?

[stray granite]

is this right for above

[native rampart]

I think it's CD

Ax=0 implies x is in null space of A and Ax is the range of A

dim(range A)+dim(null A)=dim col space=n

[stray granite]

then b works too

only A is wrong

since m is number of rows not columns

[native rampart]

Yea,BCD

mb

[stray granite]

how about this one ?

[native rampart]

BC

For C,inverse exists by construction

[stray granite]

just B and C ?

[native rampart]

Yes

[wintry steppe]

no tests here

[stray granite]

I am aware

this is a review

for the test

our prof posted questions for us to get ready

wait never mind

[hollow finch]

What have you tried? We can't just do the problem for you

[native rampart]

Just use the definition of span

[tawny tulip]

if w $\in$ SpanT then there are scalars $a_1,a_2$ so that $w=a_1 * v_1 + a_2 * v_2$, same goes for T - $w= a_3*u_1 + a_4 * u_2$.

[stoic python]

Nyrre

[tawny tulip]

compare these 2 equations

[frigid otter]

if you have the volume of the parallelepiped formed by four points, is the volume of the tetrahedron half that of the parallelepiped?

[tame mural]

lol, this is a review for the test

it's basically finals week for most colleges atm

[hollow finch]

finals are next week at my college

[steady cargo]

Can anyone tell me what this means:
Find a solution of the SLE Bx = 0 that is not contained in the solution set of Ax = 0

[hollow finch]

what is A what is B

[steady cargo]

Both matrices

I have solution set of both of em

[hollow finch]

...well yeah i assumed as much

[half ice]

What is an SLE?

[hollow finch]

the solution set of Ax=0 is the null space of A

"system of linear equations" i think

[steady cargo]

System of Linear equations yes

[hollow finch]

the solution set of Bx=0 is the null space of B

[half ice]

Okay cool. Then yeah any vector x such that Bx = 0, but Ax ≠ 0

[steady cargo]

I have the solution set of Bx=0 and Ax=0

Ohh weird

[hollow finch]

yeah so its asking "find a vector in the null space of B that isnt in the null space of A"

[steady cargo]

Is that just trial and error?

[hollow finch]

so you cant use the easy zero vector 😔

surely not

[half ice]

If you can identify both solution sets, it should be easy to just pick something that is in one, and not the other

[steady cargo]

Well one has 3 free variables and the other has 2

[hollow finch]

can you send a picture of your bases? theres a method i think may work but im not 100% on it

[steady cargo]

Is bases my work?

[hollow finch]

bases as plural for basis. so the solution sets you got for the null space of A and B

[steady cargo]

Ohh okay yes

Solution set of A = {(-3s+2t, -s, -s-t, t, s) : s,t E R}
Solution set of B = {(-2s+t-q, 4s-3t-3q, q, t, s) : q, s, t E R}

[hollow finch]

for future ref, its fine to solve for the basis of a null space this way, but you should turn it into a set of vectors

[steady cargo]

We haven't done much on linear algebra we only did matrices and barely touched on vectors unfortunately

[hollow finch]

for example (-3s+2t, -s, -s-t, t, s)=s(-3,-1,-1,0,1)+t(2,0,-1,1,0) so a basis for the null space of A is {(-3,-1,-1,0,1),(2,0,-1,1,0)}

[steady cargo]

I only know a vector is a y x 1 matrix

[hollow finch]

ah okay

[steady cargo]

Ah I think I see what u doing, to find the solution of Bx=0 not contained in solution set of Ax=0 is there anything I can equal to or is it kind of messy?

Like no way to avoid it being messy

[hollow finch]

oh

actually i think you can do it by inspection

so for A you have {(-3,-1,-1,0,1),(2,0,-1,1,0)} and for B you have {(-2,4,0,0,1),(1,-3,0,1,0),(-1,-3,1,0,0)}

notice that for A, theres no way to have two zeros in the last two entries unless you have 0 of each vector

...but in the null space of B you have (-1,-3,1,0,0)

so that vector is not in the span of the two vectors for the null space of A (i.e. not in the solution set)

if that vector was in the solution set of B was in the solution set of A then you would be able to tell me an s and t such that

$\begin{bmatrix}-3\-1\-1\0\1\end{bmatrix}s+\begin{bmatrix}2\0\-1\1\0\end{bmatrix}t=\begin{bmatrix}-1\-3\1\0\0\end{bmatrix}$

[stoic python]

nix

[steady cargo]

Ahh okay I get you

That's smart

So we pick (-1, -3, 1, 0, 0) because that's a null space

And we make that equal to our solution set for A

[hollow finch]

i picked (1,-3,1,0,0) because it was in the solution set for B (in the null space of B) and it was clearly impossible to find an s and t from your solution set for A to equal that vector

because to get a zero in the 4th entry, t has to be zero, and to get a zero in the 5th entry, s has to be zero. but then thats just the zero vector

and not (1-3,1,0,0) like we were trying to get

so its impossible for that vector to be in the solution set of A

[thorn lichen]

@tawny tulip so these 2 are equal and thus dependent?

[steady cargo]

Ohhh I get you

Thank you very much!

[hollow finch]

what's the definition of linear dependence?

[near nova]

watching this video and im not sure why x2 is considered a free variable

why is x3 not free?

ah i think i see that x2 row is simply not there?

[hardy ocean]

Reading off the matrix, it seems as x_3 can be written in terms of x_4

[near nova]

but how would i know its specifically x2 thats missing

[spice storm]

I feel like X_2 is usually free when there is a 0. If I remember

[keen flame]

Having matrix A that has a row of 0's of nxn.
And having a matrix of B.
How do I prove or disprove that AB has a row of 0's?

[quartz compass]

matrix multiplication is basically just the dot product of the row vectors of the left matrix with the column vectors of the right matrix

so you can imagine walking through the column vectors on the right matrix and dotting them with the 0 row vector of B, and you'll get all 0s in that row because of it

[keen flame]

@quartz compass I know why that is. I'm asking how to prove it mathmatically?

[quartz compass]

to prove it, you should probably explicitly write down the summation

[cunning arch]

^ If you need something quantitive, set B as an arbitrary matrix and multiply A, another arbitrary matrix with a row of 0s, and show that AB has a row of 0s

[quartz compass]

$$(AB){ij} =\sum{k=1}^n A_{ik}B_{kj}$$

[keen flame]

Thanks

[stoic python]

merowo (◡‿◡✿)

strawberrypocky

[cunning arch]

O oops, merowo has a more classy way of doing it haha

[quartz compass]

then you have to just think about what ends up being the row of 0s in A and how that ends up getting you a row out

well your example is fine haha

it's just showing the 3x3 case is not enough to prove the nxn case

[keen flame]

Indeed

So showing that one index (rows) actually does it perfectly

[quartz compass]

yeah like confirm it is a row and not a column that kind of thing

like you tell me, if I have $A_{ij}$ and I keep i constant and look at all j entries is that a row or a column?

[stoic python]

merowo (◡‿◡✿)

[near nova]

what is that 2 symbol

[half ice]

That's actually α

Just, lazy

It represents a scalar

[near nova]

ok ty

does my matrix on the right make sense?

all 0s as the last row? which means W = 0

[mild igloo]

means infinite solutions I belive

if I have a 5x5 matrix with three pivots that means 2 rows are scalars meaning the det = 0 ?

[half ice]

Not certain I follow what you are doing

What's the question?

@near nova

[tawny tulip]

What's your transformation?

From what to what

[near nova]

its a matrix that represents L

[half ice]

You don't need that bottom row of 0s then

[tawny tulip]

How is L defined? (e.g From $R^4->R^4$)

[stoic python]

Nyrre

[near nova]

L( x y z w) = ()

[half ice]

Take (x,y,z,w) and multiply it by your matrix, you should get (x+2y, z-w, x+z)

[half ice]

Note the size of a matrix necessary to do this

Is not 4×4

[near nova]

so without the 4th row was fine

[half ice]

Exactly

It's worth actually trying the multiplication to see if you get the right result

[near nova]

awesome sometimes i doubt my answers

so i added the fourth row

[mild igloo]

is there a mathmatical process to find the solution of the second part?

I cant figure it out in my head

[tame mural]

M = [1 3 2; 1 0 1; 0 2 2]

b = [2 4 7]

Mx = b

[analog flame]

Linear algebra is easy right?

[hollow finch]

depends

easy for some hard for others

[analog flame]

moreso or less than calc bc?

[hollow finch]

thats not multivariable right?

[analog flame]

no sir

[hollow finch]

i found it easier but ive heard a lot of students say it was harder than multivariable calc/DE

it really depends on you

[analog flame]

huhh

[hollow finch]

linear algebra requires a different way of thinking

that way of thinking is natural to some and confusing to others

[analog flame]

have you ever applied it to ML?

[hollow finch]

what is ML

[analog flame]

Machine Learning

[hollow finch]

im not in machine learning so i couldnt tell you

[analog flame]

Ahh

[wintry sphinx]

is that actually true

I figure all you need for most modern techniques is a little about multiplication, inversion, a decomposition or two

least squares

[half ice]

Basic ML needs only basic lin alg

As for research, I can't know, haha

[analog flame]

Machine learning solely relies on LA yeah?

[half ice]

Lots of matrix manipulations yeah

[steady fiber]

not solely on LA

tons of optimization stuff too

[wintry steppe]

How to show that if span$(v_1,v_2,v_3,v_4) = V$ then span$(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) = V$?

[stoic python]

LINEAR_ALGEBRA_GUY

[hollow finch]

what does it mean for some vector w to be in the span of v1,...,v4

[wintry steppe]

w = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 for some a_i in F

[hollow finch]

great

so can you use that to show that w also has to be in the span of v1-v2,v2-v3 etc.?

it might be easier to go the other way around but

[wintry steppe]

What is the other way?

[hollow finch]

it would probably be easier to show that if something is in the span of v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4 then its in the span of v1,...,v4 but not by a huge amount

but thats not what the problem is asking

its the converse

[half ice]

The sum of all the vectors gives v1, so v1 is in the span,
Then v1 - (v1 - v2) = v2 so v2 is in the span
Ect

Obviously you can go the other direction as every vector in the second span is a linear combo of the first

[wintry steppe]

So what would be the best way of showing this

[hollow finch]

ah yeah thats very nice

[half ice]

In a perfect world you'd do both, show that each is a subset of the other

[spare crystal]

just try to write w in terms of v1 - v2, v2 - v3, etc.

[half ice]

But there's probably a theorem to summon here

[spare crystal]

should just involve rearranging what you already have around

the problem looked familiar and i realized the exact same problem was on my a few months ago

wait jk these are plus signs

almost the same

[half ice]

It's a similar question

[wintry steppe]

@spare crystal what class?

[spare crystal]

linear algebra

[wintry steppe]

first year?

[half ice]

But note "basis" does change it a little bit

[spare crystal]

nah its the second linear algebra class im taking

[wintry steppe]

you're a sophomore?

[spare crystal]

first one was normal linear alg this ones proof based

yea

[half ice]

I'll call that other set W to keep things straight.

"Let v ∈ V. Blah blah blah. Therefore v ∈ W, and V ⊆ W.
Let w ∈ W. Blah blah blah. Therefore w ∈ V, and W ⊆ V.
Ergo V = W"

Would be the ideal form to go for

That make sense?

[wintry steppe]

Seems kind of difficult?

Or maybe wordy

[spare crystal]

well thats the main way to show sets are equal

[wintry steppe]

I know that

[half ice]

You don't need to explicitly state some of it I suppose haha

[spare crystal]

if it's too wordy just use arrows => => =>

:))

[half ice]

But yeah we should break it into the two proofs. So let's start. Assume something is in Span(v1,v2,v3,v4). Can we show it's in the other Span?

[wintry steppe]

Can someone check this proof? A list v of one vector v in V is linearly independent if and only if v is not 0.

Suppose v is not 0. If that is the case, then av = 0 if a = 0. The list containing only v is linearly independent then.
Suppose the list containing only v is linearly independent. If that is the case, then by definition, the only way for av = 0 is if a = 0.

You deleted your response, is something wrong @half ice ?

[half ice]

The list containing only v is linearly independent means av = 0 only if a = 0.

Why does that mean v ≠ 0?

[wintry steppe]

Because we assume that v is not 0

That is the case that we're doing

[half ice]

Nuu not in the second part

[wintry steppe]

What do you mean not in the second part?

[half ice]

The second part is you trying to show
v linearly independent → v ≠ 0

[wintry steppe]

Because it wouldn't linearly independent then, since a could be anything and the linear combination would still equal 0, not matching the definition of linearly independent

[half ice]

Nice, that's right

And you noticed that a contradiction is the easier way to express it

Or, actually you're going for the contrapositive

v = 0 → v is not linearly independent

[wintry steppe]

Huh

[half ice]

It's a proof method. Rather than prove
A → B
you can prove
~B → ~A
Which is logically the same thing

"If v = 0, then a can be anything, showing that the set is not linearly independent" works as your second part

[wintry steppe]

I didn't do that on purpose

idk

Never learnt that

[half ice]

Fair enough haha. Well, you figured it out on your own

[wintry steppe]

Where do I learn proof methods like that?

[half ice]

There's a pdf on google of discrete mathematics by Oscar Levin, I really like it's chapter on mathematical logic. Easy, quick read too.

[wintry steppe]

But it wouldn't cover this? Maybe

[tame mural]

Analysis 1

is where you'd learn this stuff rigorously

because there's no way for students to continue forward without

[half ice]

Many entry courses have this stuff

[spare crystal]

is it normal for the contrapositive to not feel intuitive

i need to use it more :((

[wintry steppe]

@tame mural I mean I do it subconsciously

[half ice]

Whenever I'm proving something I take a quick thought of what the contrapositive would be

[wintry steppe]

Like above I knew how to do it/had an idea how

But I don't know the "formal" thing that I did or whatever

Like Kaynex explained

[half ice]

Or, sometimes working with the opposites of stuff is just easier

Letting v = 0 is easier than finding conditions for v ≠ 0, in the above case

[spare crystal]

is this like a conscious effort

like for me, i can be completely stuck on something

but when i think of the contrapositive

im like oh thats easy

[half ice]

Hah not always, but sometimes I'm like "oh yeah check that"

@wintry steppe
The common example is prove:
"If x² is even, then x is even"

Contrapositive:
Rather than prove
A → B
you can prove
~B → ~A

[wintry steppe]

Anyway can someone confirm this proof: A list of two vectors in V is linearly independent if and only if neither vector is a scalar multiple of the other.

Suppose v_1,v_2 is linearly independent. Then by definition, the only way to write a_1v_1 + a_2v_2 = 0 is if both a_i = 0. If they were scalar multiples, that wouldn't be the only way to get 0.
Suppose that v_1 and v_2 are scalar multiples of each other. That is, there exists λ in F such that λv_1 = v_2. Then suppose that a_1v_1 + a_2v_2 ≠ 0 which we can write as a_1v_1 + a_2λv_1 ≠ 0. But now notice that this can equal 0 if a_1 = -a_2λ even if none of the three scalars are zero. This means that there does exist a nontrivial combination such that it equals 0. This implies they are linearly dependent, and therefore by negation, if they're not scalar multiples then they're linearly independent.

[tame mural]

imo you don't need to prove this

[wintry steppe]

@half ice if x is odd, then x^2 is odd?

[tame mural]

this is a definition

[wintry steppe]

@tame mural it is a iff proof

[half ice]

@wintry steppe
Bam you got the correct statement. And, it should be clear this is much easier to prove.

[spare crystal]

i mean its intuitively true but ive never seen linear independence defined like that

[umbral bay]

Hello. Is anyone here familiar with Babylonian Math?

[wintry steppe]

Is my proof correct?

[half ice]

Your first part is weak. Why wouldn't that be the only way?

[tame mural]

No, that's one of the more popular definitions of linear independence

[wintry steppe]

@half ice basically the same reason I had in part 2 lol

[hollow finch]

ye if something seems really difficult or you get stuck, then checking the contrapositive is a good thing to do

[umbral bay]

?

I guess not 😦

[half ice]

Let's say
a1v1 + a2v2 = 0
If one is the scalar multiple of the other, then what is that scalar multiple?
Why is this a problem if v1, v2 are independent?

[wintry steppe]

yeah but that's what I had in part 2

[half ice]

Second part is good, I think you meant to finish with "they are linearly dependent"

Typo

[wintry steppe]

I edited it

[spare crystal]

If $$T : V \to W$$ is a linear transformation over finite dim. inner product spaces, show that $$\ker(T)^\perp \subset \text{im}(T^*)$$

[stoic python]

Frank

[spare crystal]

does anyone know how i might approach this?

[native rampart]

Just note that (v,T*w)=0 if v is in kernel of T for all w

(That is because (v,T*w)=(Tv,w)=(0,w)=0)

Ok,This shows Im(T*) is in ker(T) perp

mb

[spare crystal]

oh yeah thats how i did the other way

[sleek veldt]

can someone tell me how the hermitian star is being pushed in/ what rules its using? A = UDU* at the start since its unitarily diagonalizable

[wintry steppe]

How do I show that $1,z,\dots,z^m$ is linearly independent in $\mathcal{P}(\mathbf{F})$ for each nonnegative integer $m$?

[stoic python]

LINEAR_ALGEBRA_GUY

[spare crystal]

@sleek veldt (TS)* = S* T*

same as transposes

[sleek veldt]

@spare crystal thank you so much :D

[spare crystal]

mhm, literally right there in the notes that were already open in front of me XD

@wintry steppe are those polynomials

[wintry steppe]

Yeah

[spare crystal]

i think you just argue by saying that a linear combination of the vectors is something like

a0 + a1t + a2t^2 + ... + amt^m

and this is only 0 if all the coefficients are 0

im sure someone can make that more rigorous though

[fossil rampart]

Is there anyone willing to help me with Linear Algebra conceptual questions

[spare crystal]

i think thats kinda what this channel is for lol

[fossil rampart]

A basis for a vector space can only consist of the zero vector ? (true right?)

😅🙈

[tame mural]

NU.

[native rampart]

Sure

Span{0} is a valid vector space

[spare crystal]

not over the real numbers

[tame mural]

he said can only consist

[spare crystal]

wait is this even true in any case? isnt span{} = span{0} for any field

[native rampart]

I don't know what span{} means but prob that

[tame mural]

span of empty set

[native rampart]

Yea,Makes sense the empty set is contained in every vector space and the smallest vector space is span{0}(wrt inclusion)

[spare crystal]

span of empty set by convention is just {0}

i think

[tame mural]

one can also think of it as the value of an empty vector sum

[spare crystal]

yeah

[wintry steppe]

Can someone confirm this? Show that if we think of $\mathbf{C}$ as a vector space of $\mathbf{R}$ then the list $(1 + i, 1 - i)$ is linearly independent.

Proof: We want to show that if $a(1 + i) + b(1 - i) = 0$ then $a = b = 0,$ which will prove that it is linearly independent.
Suppose $a=b=0.$ If that is the case then it's obviously true.

Suppose $a(1 + i) + b(1 - i) = 0.$ If that's the case then we get the system of equations $a + b = 0, a - b = 0.$ Solving this system, we see that $a = b = 0$ which means that they are linearly independent.

[spare crystal]

so i think NU is correct

[stoic python]

LINEAR_ALGEBRA_GUY

[tame mural]

plus

you want your basis to be linearly independent

anything with the 0 vector is never independent

[fossil rampart]

Thanks peeps 🙌

[wintry steppe]

Is my proof correct?

[native rampart]

Span{(1,0),(2,0)} is perfectly valid

Nvm,misread the question

[tame mural]

Why work so hard to show that [1, 1] is independent from [1, -1]

[sleek veldt]

anyone know why this is a valid use of the quadratic formula lol im not sure which corresponds to which term

[gray dust]

it's a quadratic eqn. QF applies

[sleek veldt]

okay that was a lot more obvious than i thought lol thank you :)

[blissful pagoda]

I wanna make sure, the cross product between two vectors gives a vector that is orthagonal to the two vectors right

[native rampart]

Yes

[blissful pagoda]

Say vector v is the result of the cross product, then v/||v|| would give you the unit vector that is orthagonal to both the original vectors used in the cross product?

[slow scroll]

yep

[wintry steppe]

how to calculate the complex roots of this polynomial

im having a brainfart

is it just discriminant

[native rampart]

Use Quadratic formula

[wintry steppe]

ah yeah i already see what i did wrong

squared c for some reason 🤦

[wintry steppe]

is there an easy way to get the quadratic form of a 3variable function

this is the answerbut no idea how to compute it except the diagonals

[half vector]

what set is this? n is natural numbers, x = n-1, thus x is natural numbers including 0, but then it says that x is natural again

[tropic trail]

It is the set of all natural numbers x that can be written als n-1 with n a natural number, so indead it is the set of all natural numbers

[half vector]

so what's the point of this writing? why not just say A = N

[acoustic path]

x is natural numbers and n-1

[fringe lance]

help

[acoustic path]

#prealg-and-algebra

[idle current]

Can anyone help me with a matrix problem?

[tawny tulip]

send it

[idle current]

Are you able to help with that @tawny tulip

[acoustic path]

write out the equations

[tawny tulip]

$\alpha = \alpha_1 e_1 + \alpha_2 e_2$ and $\alpha = \alpha_3 (Ae_1) + \alpha_4 (Ae_2)$ compare these 2 equations (they are both equal to $\alpha$)

[stoic python]

Nyrre

[tawny tulip]

@idle current

[idle current]

yup

[idle current]

I am not sure what you mean

[tawny tulip]

if you have alpha = x and alpha = y then x=y

[idle current]

How do i answer the question going from that?

[tawny tulip]

You know that $e_1 = (1,0)$ and that $e_2=(0,1)$. transfer your equation into vectors

[stoic python]

Nyrre

[idle current]

How do i do that?

[tawny tulip]

for example: $\lambda(1,0) = (\lambda,0)$

[stoic python]

Nyrre

[idle current]

Can you show me the entire process and i can pick out the part i do not get. This is not material i have covered yet, i am reading ahead. I just want to see processes. @tawny tulip

[tawny tulip]

just keep going from what i helped you, its not far from the solution

[idle current]

I do not know how to get from vectors though.

[tawny tulip]

do you understand why $a_1e_1+a_2e_2=a_3(Ae_1)+a_4(Ae_2)$?

[stoic python]

Nyrre

[idle current]

No

[tawny tulip]

a3 = a'1 a4 =a'2 (too lazy to write that in latex)

[tawny tulip]

read this again

[acoustic path]

Bro its cause a=a

[idle current]

Okay

[acoustic path]

That is vector a

[idle current]

What is the final solution to the question then @acoustic path

[acoustic path]

Use factorization

[stoic python]

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

[frigid otter]

Is this an example of how the multiplication of two nxn elementary matrices is not always an elementary matrix? Or is the result elementary?

[hollow finch]

product of elementary matrices is definitely not always elementary

[frigid otter]

that's gotta at least be like 3 row operations

[hollow finch]

every invertible matrix can be written as the product of elementary matrices, so if the product was always elementary that would imply that every invertible matrix is elementary

[frigid otter]

wow. that's a good way of explaining it

conceptually that makes a lot more sense than how he explained it

[frigid otter]

If Av = w, then w is in the column space of A. Is this true because Av is a linear combination of the columns of A?

[half storm]

@frigid otter 👌

[frigid otter]

but this relationship does not extend to the row space by virtue of how matrix multiplication works?

[wintry sphinx]

basically

the row space is the set of all vectors that come out of vA

where v is a row vector

[frigid otter]

so vA = w, then w is in the row space of A
but Av = w, then w is in the col space of A

[wintry sphinx]

yes

an easy way to remember this is that the row space is the column space of A^T

[frigid otter]

ah yes, I knew that. that makes sense

If A is nxn with a nullity n-1, and A has two eigenvalues, is it diagonalizable?

[analog flame]

|

Linear Algebra

Call me by your name and I shall do the same

Flamentix...

Flamentix..

Flamentix.

[hollow finch]

i prefer to think of the row space as being the orthogonal complement of the null space, but thats just me

[tawdry yacht]

hey

amyone know product?

[hollow finch]

this isnt linear algebra. check the pin

[stoic python]

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

[frigid otter]

I end up with an erroenous k on the constant in the x, which means that it does not satisfy the original solution set, right?

[hollow finch]

wdym?

that looks like a fine start to proving its closed under scalar mult

[frigid otter]

but say I sub kt and kr for t and r respectively, I end up with a first variable 3k + 2t - r, which isn't of the same form

[hollow finch]

oh i see

i didnt see the three in the first entry

[frigid otter]

that means it is not closed under scalar multiplication, right?

since for k = 2 or something, it's 6 + ....

I end up with a similar situation for proving closed under addition

[hollow finch]

the system you started with was Ax=b right?

[frigid otter]

x in R^3 | Ax = v

[hollow finch]

and not Ax=0

[frigid otter]

with a specific A and v

[hollow finch]

right v b doesnt make a difference

but v was nonzero right?

[frigid otter]

correct

[hollow finch]

then yeah the solution set will indeed not be a subspace

you already seemed to have showed it but the idea is basically

if $Ax_1=v$ and $Ax_2=v$

then $A(x_1+x_2)=Ax_1+Ax_2=v+v=2v$

similarly $A(kx_1)=kAx_1=kv$

[stoic python]

nix

[hollow finch]

and kv is only equal to v if v is zero

however if v=0 then as we see above, then the solution set would be closed under addition and scalar mult

[stoic python]

satazero

[frigid otter]

I just did the addition of solution sets

[hollow finch]

thats just an algebraic proof

showing A(x1+x2) is not equal to v is sufficient

as well as A(kx1) like you did which would be easier

[frigid otter]

Thanks, I was loathing this problem lol

For $\begin{bmatrix} a & a+b\a-b&b\end{bmatrix}$ where a and b are real numbers, is that basis just $a\begin{bmatrix}1&1\1&0\end{bmatrix}+b\begin{bmatrix}0&1\-1&1\end{bmatrix}$

[stoic python]

satazero

[frigid otter]

hey I'm getting better at LaTeX lol

[hollow finch]

yep

and thats great. its super useful

[frigid otter]

or is the basis just the two matrices, and any element of W is going to be of the form above

[hollow finch]

yeah the matrices form the basis

[frigid otter]

since there are two matrices, is the dimension of the basis 2?

[frigid otter]

I have one last question on my practice final

and it seems ridiculous, but

[stoic python]

satazero

[wintry sphinx]

what about it?

are you asked to prove it?

[frigid otter]

no, just whether that's the case, and I'd kind of like to understand why if it is

[hollow finch]

try multiplying by A and see what happens

[frigid otter]

I don't have an example A, it's just in general

[hollow finch]

i mean algebraically

you have enough information to do it

just simplify A(2v-w) and see what you get

[frigid otter]

A(2v-w) = A(2v) - A(w)

[hollow finch]

good start

[frigid otter]

uhhh.. 2A(v) - A(w)?

[hollow finch]

thats great

[frigid otter]

I'm at a loss after that lol

[hollow finch]

take a look back at what youre given

you usually have to use all the pieces of information you are given in a problem

[frigid otter]

is it the fact they're from the same eigenvalue?

[hollow finch]

thats absolutely going to come into play but theres a more basic assumption we're making

[frigid otter]

that 2v-w != 0

[hollow finch]

not that

yet

[frigid otter]

I'm not sure what other information was given

that they're eigenvectors?

[hollow finch]

yeah!

so what do we know about Av and Aw?

[frigid otter]

hm.

Av = eigenvalue * v?

[hollow finch]

exactly right

[frigid otter]

So since they're the same eigenvalue, when one is subtracted from the other, it should give you the eigenvector?

[hollow finch]

well lets just start with simplifying 2Av-Aw using what you just told me

[frigid otter]

2(eigenvalue*v)-(eigenvalue*w)

[stoic python]

nix

[hollow finch]

yeah

[frigid otter]

right

oh

but the lambda is the same

[hollow finch]

mmhhmmmm

[frigid otter]

so wait, don't you just come back to where you started

2v-w

[hollow finch]

well we have a new term/factor now

[frigid otter]

$\lambda(2v - w)$

[stoic python]

satazero

[hollow finch]

yeah!

that sure looks like an eigenvector to me

since we started with A(2v-w)

and got lambda (2v-w)

[frigid otter]

ahhhh, okay that makes a lot more sense

[hollow finch]

the only caveat is that if 2v-w=0 then its technically not an eigenvector since we dont call the zero vector an eigenvector
but we're given the it isnt equal to zero in the problem so we gucci

you also said you wanted to know why this fact is true in the first place

and thats because the span of the eigenvectors of a single eigenvalue forms a subspace

[stoic python]

nix

[hollow finch]

you would prove this exactly the same way we did your proof

you were just given the specific case where c1=2 and c2=-1

[frigid otter]

I think I get it 🙂

[opaque plover]

Can somebody help me please. I need to find a base of the kernel of the following linear maps.

[hollow finch]

how do you normally find the kernel of a linear map?

[opaque plover]

i usually try to find a matrix, that does the same as the linear map, then find the free variables of the matrix with A*b = 0 -> then build the base

[hollow finch]

well that is one way i suppose

but i mean all youre trying to do is to find which vectors from your vector space get sent to zero

so transform an arbitrary vector

set it to zero

and see what the coefficients have to be

[opaque plover]

yeah, but e.g. in a) the p(0) results in the term p0, so to find the kernel I need looks for vectors where this p0 = 0 and that can be any vector with the condition that p0 is already 0, right?

[hollow finch]

yep

that seems to be the only condition for an element to be in the kernel

[opaque plover]

ahhhhh yeah sure, and for b) only the 0-vector comes into question, because the linear map results in p0+p1+p2+p3

no wait, that could be anything

[hollow finch]

what about 1-x-x^2+x^3?

so clearly theres something in the kernel but we just have to figure out what

its nothing sneaky though

its kind of right in front of us

[opaque plover]

the polynomial itself?

[hollow finch]

well sure but you already did most of the work

if the result of a map is p0+p1+p2+p3 then to find the kernel we just set it to zero

p0+p1+p2+p3=0

simple as that

finding a basis is a little more work though

[opaque plover]

but aren't p0, ...., p3 are scalar?

the kernel consists of vectors

[hollow finch]

ah i see what youre saying

[stoic python]

nix

[opaque plover]

yep

sry for my bad notations

[hollow finch]

alright so then the coordinate vector with respect to the standard basis is (p0.p1,.p2,p3)

bad notation? i like it

[opaque plover]

yeah

[hollow finch]

now we've turned this into a vector/matrix problem rather than a polynomial problem

p0+p1+p2+p3 is just (p0.p1,.p2,p3) dotted with (1,1,1,1)

[opaque plover]

so e.g. A * b -> with A = (1 1 1 1 ) and b = (p0 p1 p2 p3 ) (vertically)

[hollow finch]

yeah exactly

that is actually the matrix of your transformation

(with respect to the standard basis)

[opaque plover]

if I go now my usual way to find the basis, I see that this matrix A got 3 free variables, which can be used to get a basis

[hollow finch]

absolutely right

[opaque plover]

thank you very much, I got that
but an other question how would it look like if my linear map is the formal derivation?
then a polynomial of grade 3 would look like p1 + 2* p2 x + 3 p3 *x^2

then I got a matrix with A = (1 2x 3x^2)

right?

[hollow finch]

we dont include the x in our vectors/matrices. rather we use matrices with coordinate vectors (bringing vectors from other spaces to euclidean space since its easier to work with)

so basically what we have it D((p0,p1,p2,p3))=(p1,2p2,3p3,0)

that is if its D: P3->P3 sometimes they would do D:P3->P2

which one are they asking for?

[opaque plover]

1 sec picture incoming

[frigid otter]

I have some other T/F questions if you don't mind after his question 🙂

[opaque plover]

i wrote this down for general polynomial

[hollow finch]

hm okay

the difference is negligible, you can fix it pretty easily

so right we have $D((p_0,p_1,p_2,p_3))=(p_1,2p_2,3p_3,0)$

[opaque plover]

yes

[stoic python]

nix

[opaque plover]

so the right one is my new coordination vector

[hollow finch]

yeah. so we can break it up by each coordinate to see where each basis vector from our original space goes

[opaque plover]

may I ask, why you put the 0 on the right side of the = to the back? because p0 results into 0

[hollow finch]

so the first entry is the constant term, the second is the linear term, the third is the quadratic term, and the fourth is the cubic term right?

[opaque plover]

yes

[hollow finch]

so when we apply the transformation we get p1 as our constant term and so on but we no longer have a cubic term

[opaque plover]

yeah

so the dimension is also decreased by 1

[hollow finch]

that is true

and also why a lot of the time we have the derivative transformation be from $$D: P_n\to P_{n-1}$$

[stoic python]

nix

[opaque plover]

yeah

[hollow finch]

okay so where does the transformation take p0?

[opaque plover]

well it is not here anymore, so it is 0 now?

[hollow finch]

yep

p0 is sent to zero

which has a coordinate vector of (0,0,0,0) right?

[opaque plover]

yeah

that's the only way

[hollow finch]

very true haha

so since p0 corresponds with our first basis vector (1 or the constant term) then the first column of our transformation matrix will be (0,0,0,0)

[opaque plover]

yeah

[hollow finch]

great

and where does the transformation take p1x?

i.e. what is the derivative of p1x

[opaque plover]

1* p1 * x^0, so just p1

a constant term

[hollow finch]

yeah

so what would the coordinate vector of that be?

[opaque plover]

(0, 1, 0, 0)

[hollow finch]

well theres the second column of your matrix

[opaque plover]

ahh and the next one would be (0,0,2,0) and (0,0,0,3)

no wait I omitted the x

[hollow finch]

oh wait hold on my brain just broke for a second

the coordinate vector of p1*1 is (p1,0,0,0) meaning the second column would be (1,0,0,0)

[opaque plover]

but why is it now in the first place?

[hollow finch]

because its the constant term

[hollow finch]

also you dont want to put x or anything in your coordinate vector since its just the coordinates/numbers associated with the basis vectors

[opaque plover]

ah yeah, you told me that already

so for p2* 2 *x it is (0,2,0,0)

[hollow finch]

exactly right

[opaque plover]

alright I got that execpt one thing

why is the 0 now to the right?

it looks like it is shifted to the left

[hollow finch]

since after we take the derivative we no longer have a cubic term

yeah a shift left should make sense since the degree of each term is lowered by one

[opaque plover]

ohhhhhhhh yeah god I am stupid, that's so logical

[hollow finch]

nah youre not stupid

definitely not

you caught on faster than some of the students ive tutored where english is their first language haha

[brittle orchid]

Why does the eigenspace have dimension 1? Isn't dimension equal to the number of vectors in the basis?

[opaque plover]

thanks for the help and the motivation @hollow finch wish you a great day

[hollow finch]

you as well. best of luck 🙂

[hollow finch]

yes it is! and how many vectors are in the basis?

[brittle orchid]

isn't the column vector {1,1} (idk how to type it) made up of 2 vectors?

{1,0} and {0,1}

[hollow finch]

thats true that it can be written that way, but if it was those two vectors in the basis then i could do 2(1,0)+6(0,1)

which is definitely not a scalar multiple of (1,1)

(1,1) is indeed its coordinate vector with respect to the standard basis since its 1(1,0)+1(0,1) so i think that might be the connection youre making

just like the coordinate vector of (6,9) is (6,9) because (6,9)=6(1,0)+9(0,1)

[brittle orchid]

do you not express the basis with respect to the standard basis, if that makes sense?

[tame mural]

there's no need to

[stoic python]

satazero

[frigid otter]

this question has something to do with multiplicity of eigenvalues doesn't it

[gray dust]

by definitions nullity=dim(0-eigenspace)

so nullity=4 gives 4 lin indep eigenvectors

1 being the other eigenvalue guarantees a 5th eigenvector needed to make an eigenbasis

[frigid otter]

so since it has 5 linear independent eigenvectors, it is diagonalizable

[gray dust]

yes that’s from “A is diagonalizable iff A has n LI eigenvectors”

[frigid otter]

gotcha, thanks, I forgot how nullity gives information about the number of linear independent eigenvectors

[gray dust]

nullity(A)=dim(ker(A)), dim(0-eigenspace)=dim(ker(A-0I))=dim(ker(A))

[near nova]

why is this matrix representing a 0,1 vector and 1,0 vector? shouldnt it be 1 0
0 1

[wintry steppe]

probably they meant to make the second column of the matrix (0,1)

[near nova]

something about them being basis vectors

[tame mural]

is the matrix real?

[tall thunder]

Can anyone explain this ?

It says “use properties of determinants”

[hollow finch]

row operations affect the value of the determinant

[tall thunder]

I did matrix steps but I’m not coming out with the right answer even though I got matrix 1 to be matrix 2

[hollow finch]

if you can find what row operations were applied, you can adjust the value accordingly

[tall thunder]

Ok so for example if I multiply r3 by 3 then + r1

Then it becomes 16x3

+?

[hollow finch]

which row operation is that?

as in what kind out of the 3 different types of row operations

[tall thunder]

Uhhh multiplication ??

[hollow finch]

well multiplication is multiplying a row by a constant, theres no adding involved

and we're certainly not simply switching rows

so that leaves one option

[tall thunder]

I’m lost lol

It’s a combination

Of row 3 with row 1

That’s a thing right ?

[hollow finch]

whats the third row operation

[tall thunder]

Adding one row to the other ??

Haha idk

[hollow finch]

very close

adding a multiple of one row to another

[tall thunder]

😂 so how does that effect the answer ?

I already have that down

So is it multiply 16 by 3? But then what do I add ?

[hollow finch]

you tell me. how does the row operation of adding a multiple of one row to another affect the value of the determinant?

check your notes/textbook if necessary

[tall thunder]

Idk haha if I knew that I’d prob have the answer already xD

And idk math writing is too complicated 🤣

In textbooks

I know the answer is 48 so they multiplied by 3 somewhere .. but in order to get them the same you have to multiply twice

So that’s what I’m really asking

And you also have to multiply by -1 at some point

So it should be -144

But the proper answer is 48

[hollow finch]

there are 3 row operations required to get from the first matrix to the second

you identified adding 3 times row 3 to row 1 as being one of them

[stoic python]

nix

[hollow finch]

there are 2 more

[tall thunder]

But the 16 part is wrong

I thought that whatever you do to one of the rows you have to do to the answer, but that’s clearly not the case

[hollow finch]

no definitely not

[tall thunder]

And I already know the answer is 48 so I only need help with the 16 part

[nocturne jewel]

If you scale up a row, you scale the det
If you swap 2 rows, you change the sign of the det
if you add a scaled up row to another row, you keep the det the same

[hollow finch]

the type of row operation affects the value of the determinant

[tall thunder]

Ah thanks

[nocturne jewel]

(scale up being multiply the row by a scalar)

so the 1st operation doesnt affect the determinant, since it's adding a scalar multiple of a row

[tall thunder]

Ohhhh

So only if I multiply a row but then don’t add it

Like step 2

Would make 16x3

Got it

So step 1 nothing step 2 multiply by 3 step 3 nothing

So 16x3

So this is correct ?? I just wanna make sure because my final is Monday

[nocturne jewel]

@tall thunder yes

(that's why you want to do R_i + kR_j operations as much as possible)

[tall thunder]

Awesome thanks

[near nova]

why does this not affect the det value?

[hollow finch]

the transformation is a shear which preserves area

[near nova]

is it because the row that was scaled up doesnt change

[nocturne jewel]

You can show that the determinant of an elementary matrix corresponding to the operation is 1

[hollow finch]

both of these have the same area

[nocturne jewel]

So if I have some matrix A which is equivalent to the elemenyary matrix (E) times some matrix B
A = EB

[hollow finch]

and thats what adding one row to another does

[nocturne jewel]

|A| = |EB| = |E||B| = |B|

so the row operation doesnt change det

[near nova]

thank u i understand now

what would be a real life example of a shear transformation happening to a matrix?

[hollow finch]

https://puu.sh/GWEdJ/aa0be1c9e9.gif

thats one application

[alpine ferry]

hey guys! im interested in learning linear algebra, what is a good resource to use?

[wintry steppe]

a book

e.g. "linear algebra done right" by axler
"linear algebra" by friedberg, insel, and spence
"linear algebra" by hoffman, kunze

those are common recommendations

[dreamy iron]

What is your background in LA? Have you gone through a computational first course yet? Or is this meant to be a second pass through LA for you?

[alpine ferry]

I know absolutely nothing about it but I want to learn

I'm a junior in high school and I learned calculus and I want to branch out

I taught myself calculus through khan academy and other resources

[dreamy iron]

Okay a first look then. So maybe don’t do Axler....nor hoffman and kunze

You’ll possibly benefit from something more computational.

There’s Hubbard and Hubbard.
Lemme just throw some books at you.

[alpine ferry]

ok :p

[dreamy iron]

https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-1-the-geometry-of-linear-equations/

[alpine ferry]

oooo ok thank you!

[frosty vapor]

ok time to do that over the winter break 😌

there's assignments too

so cool

[dreamy iron]

Strang 3rd edition. Strang 4th edition is too big for discord apparently.

@alpine ferry youve got two recommendations for David Lay (i dont have that book)

i recommend Hubbard and Hubbard, myself.

[alpine ferry]

okay thank you very much!

[arctic stratus]

( 2x + 5 ) > 7 x – 8, x>2

solve for x

[wintry steppe]

read the pinned message in this channel

[dusky epoch]

@Nyctophile❄#9458 you are not in the right channel

[blissful pagoda]

I have no idea how to do part b

Do I just use A from part a

[marble lance]

Uh no

A is your input

So it's not one matrix

[blissful pagoda]

Uhh how could I find the nullspace (kernel) of a matrix that I don't even know

[round coral]

have you studied dual maps

[blissful pagoda]

I have never heard of that so probably not lol

Just learned linear transformations

Eigenvectors and eigenspaces soon

[round coral]

A ^T is the matrix of the dual map T *

[blissful pagoda]

Definitely never learned that

[round coral]

by the way you don't need to know it

transpose is just rotate the matrix A clockwise by 90 degrees

[blissful pagoda]

That part I know

I know what transpose and inverse is, I just don't know what dual map is

[round coral]

well if you don't know about it study it , which book you follow?

[blissful pagoda]

Using Larson, R. Elementary linear algebra. 8th edition. Boston, MA: Brooks/Cole, Cengage Learning, 2017. Textbook ISBN: 978-1-305-65800-4

[round coral]

you just need to find the ker, image, nullity etc of transpose of A

you don't need to study dual maps for it

just do it the way you did part a)

I don't know about that book

maybe just do a search check if dual space and dual map is covered on it or not?

[blissful pagoda]

For part a though it was very simple to find the kernel/nullspace of A. Because I actually know the matrix I can just create a homogeneous system and solve

[round coral]

I just told you how to take the transpose of matrix

once you get A^ T , just do the same way

[native rampart]

Do you understand what ker T is?

[blissful pagoda]

[2:21 AM] gotta go fast:Do I just use A from part a
[2:22 AM] Lunasong:Uh no
[2:22 AM] Lunasong:A is your input
[2:22 AM] Lunasong:So it's not one matrix

What exactly am I transposing?

[marble lance]

An arbitrary matrix

[blissful pagoda]

"Let T: V -> W be a linear transformation. Then the set of all vectors v in V that satisfy T(v) = 0 is the kernel of T and is denoted by ker(T)"

[native rampart]

What's the 0 in your question?

[blissful pagoda]

The zero matrix

[native rampart]

Yes,Now take an arbitrary matrix and transpose

Check for what matrices,that gives you the zero matrix

[blissful pagoda]

The transpose of a m x n zero matrix would give me a zero matrix of n x m

[round coral]

you don't have to transpose 0

[wintry steppe]

hey guys, is it true that if we have a linear transformation between E and F with dim E = dim F, it doesn't necessarily imply that the transformation is surjective ?

For example if I take f: R3 -> R3 : (x,y,z) - > ( x −z,3x + 3y,x + y) we have ker(f) = (1,-1, 1) so dim(ker(f)) = 1 so dim(Im(f)) = 2 so f isn't surjective

is my reasoning correct ?

[round coral]

yes, that's right

you can have any map between two vector spaces

even when they have same dimension

[wintry steppe]

alright thanks 🙂 it's when dim(f) = dim F that the transformation is surjective

[round coral]

for the map to be surjective dim E >= dim F

ah yes you are right for it to be surjective , the dim(range) = dim F

I misread your response

@wintry steppe

[wintry steppe]

yes my notations are confusing sorry

thanks for your answers anyway

[blissful pagoda]

"The transpose of a m x n zero matrix would give me a zero matrix of n x m"
So I found the ker(T) to be just the zero matrix of my domain, which means nullity(T) = 1. How could I find the column space (range) of an abitrary matrix though

[round coral]

the columns which are linearly independent those contribute

[blissful pagoda]

I can create matrices in this domain with different numbers of linearly independent columns

Finding column space is pretty easy if it's not an arbitrary matrix where your values can be anything

[keen flame]

How would a matrice in the size of nx1 would look when transposed?

[marble lance]

It would be 1xn

[keen flame]

so all the numbers just turned sideways?

[marble lance]

Uhhhh, depends on what you mean by that lol

It sounds like you mean write $\infty$ for 8

[stoic python]

Lunasong

[keen flame]

(a)
(b)
(c)

(a,b,c)

[marble lance]

Yes

Well, not equals

The transpose of the first equals the second

[keen flame]

Right, my mistake

Thanks @marble lance

[marble lance]

Np

[sick tartan]

Is there any way for my work to be shortened

I mean if this has more values, it wouldve taken me longer.

[tropic trail]

I am not an expert, but maybe it is faster with the matrix of the linear map?

[calm hamlet]

@sick tartan what is T?

[stoic python]

xiaochenee

[calm hamlet]

It is a bit faster with matrices, but not that much

[sick tartan]

was my proof clear though?

I mean I think my prof will be confused on what I was trying to do.

[brittle orchid]

Hey, I was wondering why this matrix isn't considered to be in "Jordan Normal Form"? Along the diagonal don't we have J1(3), J2(3), J2(5), J2(5)?

[calm hamlet]

Idk how it can be clearer, it will be disgusting anyways @sick tartan

All these computations everywhere

@brittle orchid just look at Wikipedia's article about Jordan normal form

Here your 1 is surrounded by a 3 and a 5 => not a Jordan form

[brittle orchid]

Okay, thank you^

[round coral]

it isn't because as you can see, for eigenvalue 5 has 2 generalized eigenvectors but no eigenvector, that's not possible

@brittle orchid

[last gazelle]

ok quick question
to check if a set of four vectors is a basis for R^4, if the determinate of those 4 vectors is non zero can i just assume that the set is also a basis for R^4

[tawny tulip]

$det \neq 0$ then the rows are linearly independent. 4 vectors which are linearly independent in $\mathbb{R}^4$ are a base.

[stoic python]

Nyrre

[last gazelle]

ah ok

and if i didnt find the determinate, if i had to put the matrix into rref

if i get a unique solution then its also a basis for R^4?

[tawny tulip]

Yes, there are many ways to prove vectors are linearly independent.

[last gazelle]

ok

thank you

[tame mural]

Is it correct to say that two vectors are orthogonal when their spans don't have overlapping subspaces?