linear-algebra

[lavish jewel]

as ann mentioned, the form the matrix has is a hint regarding what it will behave like

it's indeed defective

but it's not just because it's not symmetric

it's not symmetric and the geometric multiplicity of at least one eigenvalue is larger than the dimension of the corresponding eigenspace

[hybrid ibex]

I see, I'm learning this for the first time so I'm still fuzzy in some places

That's why I thought I'd ask

Could you explain what it is?

[dusky epoch]

well since you're learning about diagonalization

you may or may not, at a later point, learn about something called jordan normal form

which is a generalization of diagonalization in a way - every matrix has a JNF (at least if you allow yourself to work over C rather than R), and if a matrix is diagonalizable then its diagonalization and JNF coincide

JNFs consist of blocks like your A, in which there are copies of an eigenvalue along the diagonal, and 1's right above it

[hybrid ibex]

Oh

[hybrid ibex]

so in my case, the teacher is asking me to diagonalize a JNF

which is not possible

[dusky epoch]

your teacher is asking you to diagonalize a matrix that consists of one nontrivial jordan block

[hybrid ibex]

But that's not possible right? I really appreciate you helping me out as my course material does not mention JNF anywhere

[dusky epoch]

yes, it is impossible. i am just fixing your wording.

[hybrid ibex]

Thank you so much! Appreciate it :)

Gonna read about JNF, cya

[wintry steppe]

Can I always find the dimension based on a basis of a subspace?

[dim epoch]

what dimension are you referring to

[wintry steppe]

of a subspace

[dim epoch]

and you have the basis of said subspace?

[wintry steppe]

yes

[dim epoch]

then your dimension should be the number of elements in that basis since the subspace is also a vector space

if we're talking about regular finite dimensional vector spaces here at least

[wintry steppe]

yes exactly just wanted to be certain

is it the same as saying the dimension of V as a finite-dimensional v.s. is the cardinality of a basis of V?

[dim epoch]

yes

since the subspace is a VS itself

[zinc timber]

@wintry steppe u asked the same question like 5times now

[wintry steppe]

I am just trying to settle in L.A. I am a beginner

[nocturne jewel]

dimension of a space is defined as the cardinality of a basis

for fin-dim V

[wintry steppe]

yh ik

[nocturne jewel]

then your question is very vague

[wintry steppe]

I am just learning these definitions rn and it was strange to me that it is that simple so I wanted to double-check I am sorry

[wintry steppe]

could someone describe to me the set W1∩W2 in two subspaces of a vector space which is finite-dimensional?

[marble lance]

If W1 and W2 are subspaces of V, then W1 cap W2 is a subspace of V, and it is the largest possible subspace contained in both W1 and in W2. W1 cap W2 is just normal set intersection, so it is the subspace of all elements that are both in W1 and in W2.

[empty ibex]

You are given a 3 dimensional vector space V ⊆ R5. Could there be a 3 × 6 matrix A with nullspace of A being V ? Explain. Could there be 6 × 5 matrix B with nullspace of B being V ? Explain. In either case, if you were given a basis for the three dimensional space V , how would you find the desired matrix assuming it exists.

someone asked for a repost of this question so here ya go

[jade karma]

if (dim(Null(A))=0 then Ax=0 has one solution right ?

because we define Null(A)={x| Ax=0}

am i mistaken ?

[empty ibex]

i guess so cus that would only be the zero vector then

dont take my word for it tho. I might be wrong

[marble lance]

Yes

[zinc timber]

I already answered it

fir 3x6 it's literally impossible because null space should be a subspace of R⁶ not R⁵, your A is a subspace of R⁵

[empty ibex]

oh then the nullspace would work for 6X5 cus u would get a subspace R5 right?

[zinc timber]

hmm yes

you can construct one easily. say {v1, v2,v3} be basis of V in R⁵. now extend it to a basis of R⁵ by adding 2 elements u1,u2

now tou define T as T(vi)= 0

and T(ui) = ei (basis of R⁶)

then you'll get nullT= V

@empty ibex

[empty ibex]

thankyou!! This example u just gave was what I needed to understand it

[empty ibex]

is it impossible for a matrix to have 0 pivot variables?

[lavish jewel]

how about the 0 matrix

[empty ibex]

well, aside from that tho

[lavish jewel]

then no

[wintry steppe]

i would recommend you Peter Lax: Linear Algebra and its Applications, it's a wonderful book. It's very theoretical, which is exactly what you're looking for.

i liked friedberg. there's also roman's book, which is at a higher level than all of the ones mentioned so far

[versed yew]

can someone help me with 5d

the answer is suppose to be sqrt(37) <= 7 <=11

but I dont know where they got 7 from

[nocturne jewel]

Just compute norm(u+v) and norm(u)+norm(v)

Ie... part c

[versed yew]

I already did the question but I am very confused where did they get 7 from

[wintry steppe]

what were |u+v| and |u| + |v|

just the numbers

[versed yew]

|u+v| = sqrt(37) <= |u| + |v| = 11

therefore my answer:

 sqrt(37)<= 11

the actual answer to the question:

sqrt(37)<= 7 <= 11

[nocturne jewel]

Probably just loose bound on what sqrt(37) is

[versed yew]

what does that mean?

[wintry steppe]

it's easier to show that sqrt(37) <= 7 than showing directly that sqrt(37) <= 11.

sqrt(37) <= sqrt(49) = 7. now the rest is obvious. would it be as immediately clear if you had just written down sqrt(37) <= 11?

[nocturne jewel]

Since sqrt(37)=6+c for some c in (0,1)

[wintry steppe]

Sorry to interject here, but I'm struggling to get past the beginning of this problem.

Defining the image as the span{e2,e3} surely implies that any S(v) is some (0,x,y)

But then ker(S) = span {e1-e2} = span {(1, -1, 0)}

And so then S(1, -1, 0) would equal 0, but (0, x, y) = (0, 1, -1)

I know I'm wrong, I just don't know where I'm wrong here. Why does the Im(S) definition not contradict the ker(S) definition in this question?

Would appreciate some help 😅

[empty ibex]

In A=MDM**-1, could I say that A and D are similar?

[quartz compass]

what's your definition of similar matrices?

[wintry steppe]

What do you guys think is better: Representing a point with a vector vs Representing a line segment with a vector?

[gleaming knot]

Neither

A vector is a difference between two points

[empty ibex]

they have the same eigenvalues

Could someone help with 5.a?

[teal grotto]

@empty ibex just like, for now, assume A has an inverse

to get some idea for what it has to be

call the inverse B

then BA^2 = A = -B

so B = -A

now see what happens (try to use the above to see what the inverse of A should be) ||multiply A by -A||

[wintry steppe]

interesting thought

[wraith monolith]

Hi, guys, is this true: F: V-->W, and F(v_1),...,F(v_n) is a basis of W, then F is linear?

[teal grotto]

try it with W taken to be the ground field

[wraith monolith]

i am sorry, i am not sure what you mean

if v_1, ..., v_n is also basis of V, can i just define F = [F(v_1) ... F(v_n)]?

[teal grotto]

no. you cant redefine F

im just going to assume V and W are real vector spaces (nothing changes if they arent)

just look at the determinant function from the vector space of real m by m matrices to R

[teal grotto]

you never said v1,...,vn was a basis of V when u first stated the problem

[wraith monolith]

oh, sry, i just kind of think if i did not add this condition, the original statement might be false?

[teal grotto]

i was reading this as, we have some vectors v1,...,vn such that F(v1),..., F(vn) is a basis for W. F does not have to be linear

[teal grotto]

whats the full statement of the problem?

[wraith monolith]

i just want to show this map is linear. F: V\tensor V --> V\tensor V, F(v\tensor w)=1/2(v\tensor w + w\tensor v)

never mind, i think i can figure it out by myself, thanks though

[teal grotto]

im not familiar enough with tensor stuff to help with that, sorry. gl

[wraith monolith]

no worries, no worries. me too, i just get touch in tensor product, very strange for me😅

[whole peak]

maybe this playlist helps:

https://www.youtube.com/watch?v=8ptMTLzV4-I&list=PLJHszsWbB6hrkmmq57lX8BV-o-YIOFsiG&index=1

[wraith monolith]

thanks!

[whole peak]

np, btw. linear maps starts at chapter 7

[wraith monolith]

very appreciate, thanks. this is very helpful

[whole peak]

^^

[random axle]

The concept of things being perpendicular in 3-D still confuses me. If we have 3 planes all travelling in different directions how can there be a fourth plane that is perpendicular to all three?

The 2nd and 3rd configurations make sense. Where the 3 planes are identical or parallel. But the others remain a mystery to me...

[whole peak]

look on the sides of the planes

[random axle]

How do you mean

[whole peak]

let us take the 4th.
how are the angel of top and bottom plane to each other in top-down direction

it seams that they are parallel to each other

now, let's look at the middle one, this intersects the other to planes

let us think the intersections would be a line on the three planes

[worldly bear]

how do you apply the gram-schmidt process to a set of 3 vectors if two of the vectors are already orthogonal

it just gives me the same vector since the dot of them is 0 since they’re orthogonal

[nocturne jewel]

pretty sure you just apply it like normal

[solid cobalt]

good day, is anyone able to explain to me how to go about doing this proof? any help will be greatly appreciated

[solemn lotus]

is <v, v> = 0 -> v = 0_v
really a necessary condition for smth to be an inner product? my course says so, but its not given as a requirement in the textbook im reading

[limber sierra]

typically yes; its required, for one, for it to induce a norm

if you relax it, you get a different sort of structure that goes by many names

"positive semidefinite hermitian form" is the most common probably

[solemn lotus]

i see

lang only defined these 3 so i was confused lol

[limber sierra]

did he?

are you sure he didnt say "positive definite" or something somewhere?

since as stated thats not even semidefinite!

[solemn lotus]

oh its given but its not given as a "basic" requirement for all i think

[limber sierra]

just a hermitian form

[solemn lotus]

hmm i see

[random axle]

Why does this method work to find an inverse matrix? I know how to implement it, but it kind of feels like magic.

[limber sierra]

row reduction corresponds to multiplication by elementary matrices

elementary matrices are invertible

so the operations to go from the A (your starting matrix) to the identity, are the operations to go from the identity to a matrix B

this implies A and B are inverses

just invert the operations.

to be a bit more explicit

we can write $E_n\cdots E_2E_1A = I$, where $I$ is the identity matrix and $E_1, E_2, \cdots, E_n$ are the elementary matrices corresponding to the $n$ row operations you apply to row reduce $A$

[stoic python]

Namington

[limber sierra]

this equation represents the entire row reduction process

but if we apply these operations to I simultaneously

we get $E_n\cdots E_2 E_1 I = B$ where $B$ is some new matrix

[stoic python]

Namington

[random axle]

Ohhhh, gotcha. Thanks

[limber sierra]

but elementary matrices are invertible, so we can multiply both sides of that by $E_1^{-1}E_2^{-1}\cdots E_n^{-1}$

[stoic python]

Namington

[limber sierra]

this gives $I = E_1^{-1}E_2^{-1}\cdots E_n^{-1}B$

[stoic python]

Namington

[random axle]

Thanks for the great explanation! I think I've got it now

[marsh hollow]

How to solve this by using matrix multiplication?

Can I use inverse?

[agile bronze]

No, the left matrix is not square so it doesn't have an inverse

If you know what a projection matrix is, you can follow that formula (which will be exact here because 3 cols vs. 2 rows)

[marsh hollow]

I dont know

[agile bronze]

https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/projections-onto-subspaces/MIT18_06SCF11_Ses2.2sum.pdf

Here, we'd call your original equation ax = b

[marsh hollow]

whoaa I haven't reached that lesson yet

[south hamlet]

can anyone help me on this

[stuck marlin]

so what's the meta for computing the characteristic polynomial by hand

[astral hare]

A possible basis could be (0 0 1) and (1 1 0)

[lime knoll]

Do u guys know how to solve this? what method should I use for part b?

[astral hare]

And then to solve part c you would just take any linear combo of the given basis

[strong shell]

I dont get it too, maybe someone can solve this

[marsh hollow]

<@&286206848099549185>

[empty ibex]

kind if stuck at c and d. Need help

of*

[marble lance]

Do you know how multiplying a row through by a constant affects the determinant of a matrix?

@empty ibex

[empty ibex]

i dont think so

[marble lance]

Multiply one row through by c also multiplies the determinant by c

So if you multiply one row of A by c then the determinant becomes c det(A)

[empty ibex]

I see

but how does that help prove that n is even

[marble lance]

Okay, now let's think about -I

It has n rows

So we are taking the identity matrix, and multiplying each of the n rows by -1

So then what will the determinant of -I be equal to?

In terms of n

[empty ibex]

I

[marble lance]

What?

[empty ibex]

-I would be equal to I right?

[marble lance]

No

-X = X only for the zero matrix

Do you know what I looks like?

[empty ibex]

yes (1,0;0,1) and so no for higher orders

on*

[marble lance]

Okay, so 1 on the diagonal, 0 elsewhere

Now -I has -1 on the diagonal and 0 elsewhere

[empty ibex]

oh the determinant would now be the negative of our previous determinant>

?

-1 then

[lime knoll]

Damn, got stucked for an hour

No idea 😦

[marble lance]

No

Wrong reply

Sorry

Multiplying the whole matrix by a constant does not multiply the det by that constant. For each row you multiply by the constant, you multiply the det by that constant. Here you have n rows, so you have to multiply by -1 n times

[empty ibex]

ohhhh

i see now, that was dumb

[marble lance]

Nah, it's okay

What is the determinant of -I then in terms of n?

[empty ibex]

(1)(-1n)

[marble lance]

-1n?

[empty ibex]

-n

[marble lance]

No

(-1)^n

[empty ibex]

-1**n

yep sorry typo

[marble lance]

Okay

Now det(AB) = det(A)det(B), happy w that?

[empty ibex]

yes

[marble lance]

So

(-1)^n = det(A)^2

[empty ibex]

=det(A)det(A)

[marble lance]

Yeah

Now can you see why n is even?

[empty ibex]

ohhhhhh

damn

[marble lance]

(btw, to see why det(-I) = (-1)^n, you can also note that the eigenvalues of -I are just -1 with multiplicity n and use det = product of eigenvalues)

[empty ibex]

thats very neat, would not have crossed my mind. Tysm for the help!

[marble lance]

👍

[winged prairie]

yo im having trouble wrapping my head around this theorem

so the dimension of every linear map going from V to W, is equal to the product of the dimension

because the size of hte matrix is m * n where m is the dimV , i.e the number of basis vectors in V and n is the number of basis vectors in W

so if i have a map going from R to R

T = 5x

then the dimension of the linear map is 2?

[dusky epoch]

"the dimension of every linear map going from V to W" no

the dimension of the space of all linear maps from V to W

a linear map on its own has no dimension

[strong shell]

Can someone solve this? I also want to know

[winged prairie]

ye thats why im confused

because a linear map can also be represented by a matrix

but the matrix has a dimension

[dusky epoch]

the collection of all linear maps from V to W forms a vector space in its own right

[winged prairie]

ye

[dusky epoch]

the dimension of this vector space, in your notation, is mn

[winged prairie]

what about this

[dusky epoch]

that is in some sense the reason behind my claim

L(V,W) is isomorphic to the space of all m by n matrices

[winged prairie]

doesn't it contradict this

[dusky epoch]

why would it

a matrix also does not have dimension in the same sense as a vector space does

[winged prairie]

why not

cuz it doesn't have basis vectors?

[dusky epoch]

it's not a vector space

a matrix is not a vector space

a matrix is not a vector space

[winged prairie]

ok

but the set of all m*n matrices is a vector space?

[dusky epoch]

yes

[winged prairie]

so a matrix is techinically a vector

[dusky epoch]

sure if you want

[winged prairie]

which u cannot just define a dimension for

[gray dust]

what we call a vector depends on the context of vector space

[winged prairie]

that makes sense

ye

ty

[random axle]

just started reading about vector spaces a few mins ago. didn't know that polynomials are also vectors!

[strong shell]

Im stucked 😂

[winged prairie]

ye its pretty mind blowing

[wintry steppe]

it shouldnt be that mind blowing. polynomials are just finite sequences of numbers

[winged prairie]

hello, for this exercise i prove that ST is bijective them proved the idenity by just rearanging the LHS to equal the right one. is that correct?

[winged prairie]

Ye fair but i hadn't done linear algebra before so i just thought vectors where just objects with magnitude and direction

[wintry steppe]

vectors dont have magnitude in general

only if youre in a normed space

[wintry steppe]

no need to prove ST is bijective

[wintry steppe]

do you know what it means for a linear operator to be bijective

[wintry steppe]

also, polynomials can have a magnitude and direction btw

[winged prairie]

if its bijective its invertible

so i prove that first

then just prove the idenity

[winged prairie]

oh

[strong shell]

.

[wintry steppe]

no need to show its bijective tho. you can just show that it has an inverse

[zinc timber]

don't think that's linear algebra

[winged prairie]

ah not sure how to do that

[zinc timber]

it has matrices doesn't mean it's linear algebra

[winged prairie]

do i have to use ST = I and TS = I?

[wintry steppe]

thats not true in general

[strong shell]

I dont know about the post, Im just asking for solver

[winged prairie]

so how would u do it

[wintry steppe]

i ask again

Do you know what it means for ST to be invertible?

[winged prairie]

lemme think

[wintry steppe]

before trying to solve an exercise make sure you know what everything in it means

[winged prairie]

isn't it what i said before

like for a given linear map

there exists a map

[winged prairie]

such that this is true

and that it is equivlanet to being bijective

[wintry steppe]

no

[winged prairie]

then idk

[wintry steppe]

im asking what it means for ST to be invertible

what you said meas S is invertible with inverse T

[winged prairie]

oh ok

[wintry steppe]

and thats not necessarily true in your question

[winged prairie]

ST(T^-1S^-1) = (T^-1S^-1)ST

?

[wintry steppe]

they want you to show that those two are equal to I

[winged prairie]

oh k makes sense

thanks!

[wintry steppe]

🚬

[winged prairie]

This is a contradiction right

Cuz that cannot be the case if it’s not invertibile

[dusky epoch]

what is this string of symbols meant to be?

[marble lance]

What did you do? I don't understand

[winged prairie]

ill re-write it

[dusky epoch]

what's theta?

[winged prairie]

0 vector

Sorry that’s how we did it in class forgot to mention

To distinguish from 0 scalar

[dusky epoch]

you're confusing "this function is not the identity" to "this function has no fixed points"

just because ST(0) = 0 (which of course it is, ST is linear!) does not in any way mean that ST = I

otherwise all linear maps would be the identity

[winged prairie]

oh shit haha

ty

wait but the whole point is that its not linear because its not a subspace

[bold ether]

how

[dusky epoch]

you're using the word "it" without clear indication of what it refers to

and that is a problem

i think you don't understand what you yourself are saying

[winged prairie]

sorry i meant The set of non invertible operators

like if its not a subspace it either doesn't have the 0 vector or its not closed under scalar multiplication or addition

[dusky epoch]

yes, so at least one of the following must be true:

• the zero map is not linear
• there exist two operators S and T which are not invertible individually, but S+T is
• there exists an operator T and a scalar c such that T is not invertible but cT is

[winged prairie]

makes sense

ty

[lavish jewel]

do notice, drunkgamer, that linearity is a property of a single operator, whereas for a set of linear operators to form a subspace, they must satisfy the definition of a subspace, which relates several linear operators to each other. you can have a set of linear operators that don't form a subspace

[winged prairie]

ok 👍

ty

[marsh hollow]

If the value of | A | = - B, then the value of | B | is.....

[wintry steppe]

are you meaning determinant with "|A|"

?

or something else

[lime knoll]

Yes, determinant A

Im sure he wants to say that

[wintry steppe]

because the determinant usually returns a number and not a matrix (I think)

[lime knoll]

so it must be

if the value of det A = - B, then the value of det B is

[wintry steppe]

ok

thx

[silk scroll]

12B

[wintry steppe]

is this supposed to be the answer?

[lime knoll]

I dont think so

[wintry steppe]

me neither

but I stick with my oppinion, this question makes no sense if the entries to A are all numbers

[marsh hollow]

Sorry, Revision

[marsh hollow]

If the value of | A | = - 8, then the value of | B | is.....

[silk scroll]

96

[bold ether]

how?

[marsh hollow]

Neg?? or positive?

[wintry steppe]

ok, that makes more sense

[lime knoll]

So , the answer is??

[teal grotto]

multiply A on the left by the matrix
3 0 0
0 -1 0
0 0 4

[vast iron]

Then -12×-8=96

[teal grotto]

right

[vast iron]

Big brain moment

[strong shell]

how to solve this?

[bold ether]

diagonal multiply?

[teal grotto]

just matrix multiply

[vast iron]

If (I+3A)^-1 = B then (I+3A) = B^-1, after inverting the matrix, it should just becomes 4 independent algebraic equations.

[strong shell]

so, we move the inverse??

to the right one?

[zinc timber]

yes

It's simple algebra

[stone siren]

Help.....

[wintry steppe]

have you heard of gauss elimination

[stone siren]

yes , I have

[wintry steppe]

have you tried applying it to that matrix

[strong shell]

how about using property of inverse?

[stone siren]

no

[wintry steppe]

do it then

[stone siren]

how

[lime knoll]

Consider a triangle, the smallest angle measures 10o less than one-half of the largest angle. The middle angle measures 12o more than the smallest angle. Let x, y, and z represent the measures of the smallest, the middle, and the largest angle, respectively. The linear equation system for this problem can be represented as

kinda confused

[zinc timber]

remember sum of the interior angles of a triangle is pi

also this is more of a #help-i kind of question rather than LA

[lime knoll]

btw , Is this correct?

[zinc timber]

what do you think?

think about the relation they have

[winged prairie]

does anybody know how to do the first part

i.e given that nullT1 = nullT2 prove that there exists an invertible...

the other way round is simple but i have no idea how to do this first part

[zinc timber]

$V/\ker T_1 \simeq $ to a subspace of $W$, same for $V/\ker T_2$

[stoic python]

Ryu

[zinc timber]

Also from rank nullity you can conclude that dim range T_1 = dim range T_2

just find a isomorphism between them and extend it to a map from W to W

[winged prairie]

ok ty

[shut orbit]

does anybody know how to answer this question?

[wintry steppe]

My first step would be to write down what the implications of being non invertible means

[shut orbit]

can you show me an example?

[marble lance]

Give an example of two non invertible matrices who sum is invertible, then the set is not closed under addition

[charred dagger]

can anyone help me understand this question

[marble lance]

It wants you to explain in your own words the terms elliptic geometry and hyperbolic geometry, and why/why not they are called that

[shut orbit]

does anybody know how to answer this question?

[marble lance]

Did you find one?

[shut orbit]

yeah

[marble lance]

Okay

[shut orbit]

i got it

[shut orbit]

but i think my example is too short can you give me one?

[marble lance]

Too short?

What do you mean?

What is your example?

[proven oracle]

would it be too greedy if i asked help about constructing a linear equation system ?

[marble lance]

I will help

x = 0

There you go

I constructed one

[proven oracle]

brilliant

[dim epoch]

[marble lance]

[gray dust]

luna on fire

[proven oracle]

seriously can i post the problem lol

[dim epoch]

spittin fr

[shut orbit]

a is
0 0
1 1
and b is
1 0
0 0 so a+b is invertible

[marble lance]

Yeah, that's fine

[shut orbit]

okay

[marble lance]

Of course, if no one wants to help, you will find out lol

[shut orbit]

can you help me with this question?

[dim epoch]

$\subseteq$ denotes subspace for you i suppose?

[stoic python]

Timo2

[proven oracle]

G's are presents and R's are raw materials. There are still 20 R1, 30 R2 and 10 R3's. x1 amount of Gi(i = 1,2,3,4) should be produced so that all raw materials are used.

Do you guys have an idea how to construct a linear equation system to solve for x1 ?

or is it simply just multiplying this matrix with x1 = (a1 a2 a3) ?

if any of you speaks german i can send the full question so that you can understand it better

[dim epoch]

i do

but im not promising anything

[marble lance]

So you are producing G_1, G_2, G_3, and G_4 using materials R1, R2, R3 and the table tells you how much material is needed to produce each thing? And you want to know how much of each thing you should produce to use up all the materials?

[proven oracle]

here is the full question

[proven oracle]

yeah it seems so

[marble lance]

Then I would think you actually need to have four variables, not 3. So more like x = (a1, a2, a3, a4), and the matrix would be the table basically

So if A is the matrix in the table, your system is Ax = (20,30,10)

[proven oracle]

then should the matrix be 4x3 instead of 3x4 ?

[marble lance]

No, 3 x 4 is correct

Since your x is 4 * 1

[proven oracle]

oh wait right

then i will solve the system normally buy making the 1x2, 1x3 and 2x3 equal to 0 right ?

[marble lance]

I don't know what you mean by that

[proven oracle]

okay i think i got it nevermind, thanks a lot for helping 🙂

[marble lance]

👍

You weren't greedy at all

[proven oracle]

since i posted a whole problem, it feels like i am asking for a lot from you guys lol

[marble lance]

Nah, it was fine

[proven oracle]

glad it was, we shall meet again in the future then

[dim epoch]

i did not do anything but i think it was fine as well

[marble lance]

[dim epoch]

i cant even read it

[jade karma]

sec

e) from here then

[marble lance]

Don't ask for help with tests and quizzes

[wintry steppe]

if you don't immediately see a counter-example, the best strategy is to just go try to prove it

[jade karma]

i already took the test

it's just i didn't know how to do it

and i wanna understand

[wintry steppe]

so just try all the requirements

[marble lance]

It says time remaining 34:39

[jade karma]

bro it's 8:53pm

[wintry steppe]

do you know the requirements for a subspace

[marble lance]

Idk what time zone you are in, but it's weird for it to say there is time remaining if there is not.

[jade karma]

ight lemme find another pic of it

[marble lance]

Or come back in 35 minutes

[jade karma]

I was thinking of using

but idk again

i'll wait 35 mins

just to prove my case

[wintry steppe]

well to show something is a subspace you have to check if it is closed for addition and multiplication by a scalar

[jade karma]

so i just text the axioms ?

[wintry steppe]

you don't need to check all the axioms, if you know that you are working with a subset of a known vector space

you have to only verify that this subset is closed for addition and multiplication by a scalar

everything else is automatically true

here is a simple example

[jade karma]

ight ty

[wintry steppe]

https://math.stackexchange.com/questions/871145/proving-a-subset-is-a-subspace-of-a-vector-space

[jade karma]

ty

[shut orbit]

can you?

[dim epoch]

can i what

[shut orbit]

what do you mean about this?

[proven oracle]

okay another quick question, vectors x1 and x2 are solutions for a linear equation system, is (x1) X (x2) or x1 + x2 also a solution for the system?

[marble lance]

What is the product of two vectors?

And let's think about it. If Ax1 = b and Ax2 = b, then what is A(x1 + x2)?

[proven oracle]

it just says, x1 and x2 are the solutions of the system; Cx = d, C is a 4x4 matrix and d = (0,0,2,1) transposed

[proven oracle]

i think it wouldnt work but im not sure if i think correctly

[marble lance]

Matrix multiplication distributes over addition

So C(x1+x2) = Cx1 + Cx2 = what?

[proven oracle]

so it works then, right ?

[marble lance]

No

= d + d = 2d ≠ d

It only works if d = 0, which it is not for your question

[proven oracle]

then neither the multiplication nor the sum is also a solution ?

okay got it

[wintry steppe]

in highlighted letter v, would be any difference if it was u for the proof stated?

[marble lance]

I don't know what is meant by the multiplication

So I can't say

[proven oracle]

okay, thanks a lot again

[marble lance]

👍

[wintry steppe]

what do you mean? if there was u instead of v, you would have two u's defined?

I thought I could do it like this

[slow scroll]

no, <u,v> = 0 if and only if <v,u> = 0. If you're just dealing with real inner product spaces then you can say <u,v>=<v,u> in general

[forest quiver]

I assume you are done

I am really bad at this planes unit

But I know that (a) is a plane equation

I just don't know how to get it in that form

If I have the plane equation in the form ax+bx+cx+d=0, then i could get it into parametric form

But Idk how to go from system of equations to matrix equation

[wintry steppe]

so I can follow the proof I screenshot?

[slow scroll]

that is a correct proof, yea

[forest quiver]

For the rref o fthis matrix I got x=-1/2 , y=1/2, z=free variable

So can I add these up or something to get a plane equation?

x+y+z=t

if z=t

Oh wait it's a line equation I think

Ok wtf why is this wrong

[marble lance]

It shouldn't be wrong

Possibly the answer needs to be entered in a different form

Oh wait

I believe you swapped x and y

x should be 1/2 t and y = -1/2 t

[untold panther]

hi hi, by chance is it okay if i post a question asking for help to solve it in this chat ? I would really appreciate any guidance on it bc i really am totally lost on how to move forward

[charred dagger]

would this adequately answer it?

hyperbolic geometry has infinitely many lines through a point that diverge, while in elliptical geometry any two lines no matter what intersect. Elliptical geometry doesn't have to do with an ellipse- it's also known as spherical geometry because the positive curvature of the plane gives it a spherical model. Meanwhile, hyperbolic geometry's curvature is based on the function it is modeled on, not a hyperbola itself.

[marble lance]

I got the feeling you were just supposed to restate the content of that paragraph

By basically saying it's not about ellipses/hyperbolas, but it's called that because of the analogy and then explain the analogy

[charred dagger]

What's the analogy there?

I guess it says a line in elliptical geometry is just a circle so it can't actually approach infinity, while a line in hyperbolic space goes to infinity in both directions

but I don't get what that has to do with hyperbolas/ellipses

[marble lance]

The passage explains it. No asymptote vs no point at infty, and two asymptotes vs two points at infty.

[slow scroll]

yea, that's what this channel is for

[hot swallow]

quick question kinda stuck here

https://i.imgur.com/zTOhVUf.png would this be diagnolizable?

[steel moon]

do we know if a not equal b/

[hot swallow]

no

that the statment

[steel moon]

are u trying to prove this

[hot swallow]

ye

trying lol

[steel moon]

tbh im not sure i was going to ask a diagonalizable question too

maybe they want you to show there are there 2 unique solutions to $(a - \lambda)(b - \lambda) = 0$

[stoic python]

rupiee

[hot swallow]

thats what i was thinking as well

[steel moon]

well

if you assume a not equal b, then those are unique solutions

so you have proved one side

then when proving the => side maybe try contradiction

assume its diagonlaizable and that a = b

[steel moon]

then you arrive to this equation

and say it is a contradiction

so a not equal b

and thats the entire proof done i guess?

[hot swallow]

hmm it makes sense

gonna give it a shot with what you siad thanks

[steel moon]

np

ive been stuck on this for a while, i computed det(M_0 - lambda I ) = 0
and got
$(\cos(\theta) - \lambda)^2 + \sin^2{(\theta)} = 0$

[stoic python]

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

[steel moon]

i tried using the fact that (cosx)^2 + (sinx)^2 = 1 after expanding the bracket

but not sure how to go from there

[nocturne jewel]

isolate for lambda

[steel moon]

So i have $\lambda^2 - 2\lambda \cos{(\theta)} + 1 = 0 $

$\lambda^2 - 2\lambda \cos{(\theta)} + 1 = 0 $

$is this working$

[stoic python]

rupiee

[steel moon]

$\lambda^2 - 2\lambda \cos{(\theta)} + 1 = 0$

[stoic python]

rupiee

[steel moon]

i dont quite see how to do it with this

[nocturne jewel]

$(\cos(t)-\lambda)^2=-\sin^2(t)$

[stoic python]

Mosh

[nocturne jewel]

as a hint

[steel moon]

hmm..

square root both sides?

since its complex number.. that would mean its a valid eigen value?

but there should be 2 distinct eigenvalues since its a 2x2 matrix

[nocturne jewel]

yeah, you get $\cos(t)-\lambda=\pm i\sin(t)$

[stoic python]

Mosh

[steel moon]

oh whoops yeah

plus or minus

that gives two

[nocturne jewel]

and so lambda is your typical unit circle complex number and its conjugate

[steel moon]

doesnt this simultaneously prove (a) and (b)?

[nocturne jewel]

I can't recall how to prove diagonalizability as in what's sufficient and what's necessary

[steel moon]

oh i see

i saw somewhere that for an nxn matrix if there are n distinct eigenvalues then its diagonalizable

[nocturne jewel]

Yeah, sum of geometric multiplicities is the size of the matrix implies diagonalizability

[steel moon]

i vaguely recall but i believe conjugate is unique

so that should do it

not sure how to relate this to c) though

c) looks like the lambda expression

[nocturne jewel]

if t is an eigenvalue of A, then what is an eigenvalue of A^n?

[steel moon]

does the exponentiate on the matrix denote each entry raised to that power?

[nocturne jewel]

no

[steel moon]

oh n matrix multiplications?

[nocturne jewel]

it means A*A*...*A

[steel moon]

i would say that the value is lambda^n then

[nocturne jewel]

Yeah

so you should be able to verify De Moivre for part c now

[steel moon]

i dont think im seeing it, all i can see is $\lambda^n = (\cos(t) \pm i\sin(t) )^n$

[stoic python]

rupiee

[nocturne jewel]

yeah, and that's an eigenvalue of the rotation matrix^n

so find the eigenvalues of R_{nt}

[steel moon]

is that substituting theta for 2theta?

[nocturne jewel]

wdym.

rotating by theta n times is rotating once by n*theta

[random axle]

Isn't the vector space V (the coordinates of all points on the plane) a three dimensional vector space? The textbook states that it's two dimensional

[nocturne jewel]

the plane is spanned by 2 vectors

so it's a 2 dimensional space

[steel moon]

sorry i meant n*theta substituting into the matrix

[nocturne jewel]

yes

[random axle]

So the dimensions of a vector space are not the number of elements that make up a general coordinate in that vector space?

[nocturne jewel]

dimension of a space is, for fin-dim spaces, the cardinality of a basis

[random axle]

oki thanks

[onyx jasper]

hii, can you help me aswell

[nocturne jewel]

dont ask to ask.

also dont ping random people

[glad acorn]

you should check the free variables, think of the degree of freedom of the vector space

[worldly bear]

how do you find the transition matrix from B to B’ for example if B isn’t the standard basis

like it’s so trivial when it is the standard basis but is it just as simple as doing k1 * b1 + k2 * b2 = b1’ if i’m in R^2 for example

[slow scroll]

Call S the standard basis. Find a transition matrix from S to B, call it X. Then find a transition matrix from S to B', call it Y. Then the transition matrix from B to B' is YX^-1

Since X is the transition matrix from S to B, X^-1 is the transition matrix from B to S, so if i have a vector v, then
YX^-1([v]_B) = Y([v]_S) = [v]_B'

[slow scroll]

I answered exactly the question you asked, but to be clear, it should actually be easier in practice to make a transition matrix from B to S than from S to B.

[worldly bear]

this is because X^-1([v]_B) is ([v]_S) right

[slow scroll]

yep

[worldly bear]

so i need to find a coordinate vector X right

[slow scroll]

X and Y are matrices. not quite sure what u mean

[worldly bear]

let me just post a problem with numbers, i’m not sure how to apply it with just the definition

[slow scroll]

aight

[worldly bear]

so what i started with was [1,3] = k_1[2,2] + k_2[4,-1]

and then did [-1,-1] = k_1[2,2] + k_2[4,-1]

solved them both, and put them into a 2x2 matrix

[onyx jasper]

can someone check if this is right

[slow scroll]

[1,3] = k_1[2,2] + k_2[4,-1]
[-1,-1] = n_1[2,2] + n_2[4,-1]
so you solved something like this and your transition matrix became:
k1 n1
k2 n2
?

[worldly bear]

yes exactly

so that matrix is my transition matrix, but from B to B' or the other way around

[slow scroll]

this channel is occupied, but what you have is a better fit for #proofs-and-logic anyway. try asking there

[worldly bear]

i think it would be from B' to B but im not certain

[slow scroll]

yep this is correct. because [(1,3)]_B' = (1,0) and if you plug in (1,0) into that matrix you get (k1, k2) = [(1,3)]_B according to the equations you calculated. Same thing with [-1,-1]

[worldly bear]

where do you get the (1,0) from

[slow scroll]

well the idea is that the first basis vector looks like (1,0) relative to its own basis and and the second basis vector would look like (0,1) relative to its own basis

[worldly bear]

thats the standard basis right

[slow scroll]

(1,3) is a vector written in terms of the standard basis (unless stated otherwise), and (1,0) is (1,3) in the basis u1', u2'

[worldly bear]

and since i found B' to B, i can easily find the other direction by the inverse

[slow scroll]

yep

[worldly bear]

what about part c?

i take my transition matric from B to B' and multiply it by W?

[slow scroll]

you need to get w in the basis of B first

so you could set up something like
(3, -5) = x1(2,2) + x2(4,-1)
like with the other problems

then [w]_B = (x1, x2)

and you can find [w]_B' from there

[worldly bear]

oh so im doing the same thing

and then the [w]_B' is just (3, -5) = x1(1,3) + x2(-1,-1)

the coordinate vector

[random axle]

Are spanning sets the same as a basis? If not, then what is the difference?

[slow scroll]

nope, when you have [w]_B, you want to use your transition matrix from B to B' to compute B'

actually, i take that back, it doesn't matter, but its less work to use the transition matrix you already have

[worldly bear]

oh i see that, [v]_B' = P B to B' * [v]_B

[slow scroll]

here you would get [w]_B' = (x1, x2), that's right, but you've already computed a matrix that does this for you

[worldly bear]

so i could do it that way, but matrix multiplication is another way to get it

rather than resolving a system

[slow scroll]

yea, once you've computed a transition matrix B --> B', you can toss in any vector [v]_B and get out [v]_B'. If you already have [w]_B, its just easier to use the transition matrix you computed from part b

[worldly bear]

okay, and in part d when thye say computing directly they mean by solving the system right\

rather than using the transition matrix

[slow scroll]

yep

[worldly bear]

[1,3] = k_1[2,2] + k_2[4,-1]
[-1,-1] = n_1[2,2] + n_2[4,-1]

and why is this B' to B and nit the other way around again

i need to understand this so i dont do it backwards lol

[slow scroll]

yea so [v]_B for example would mean "v as a linear combination of (2,2) and (4,-1)" so in particular
[(1,3)]_B means "(1,3) as a linear combination of (2,2) and (4,-1)"
So in the equation:
[1,3] = k_1[2,2] + k_2[4,-1]
you are solving for [1,3] as a linear combination of (2,2) and (4, -1), so this is [(1,3)]_B by definition.

Now, once you've don this for both vectors and get the matrix
A = k1 n1
k2 n2

you can verify that A([(1,3)]_B') = A((1,0)) = (k1, k2) = [(1,3)]_B and A([(-1,-1)]_B') = A((0,1)) = (n1, n2) = [(-1,-1)]_B

[worldly bear]

Okay so when im writing B' as a linear combiniation of B, i am finding my transition matrix from B to B'

since the x1 x2 n1 n2 i get are what transitions B to B'

[slow scroll]

if im following what you're saying it should be the other way around

[worldly bear]

[1,3] = k_1[2,2] + k_2[4,-1]
[-1,-1] = n_1[2,2] + n_2[4,-1]

this is writing B' as a linear combination of B, no?

[slow scroll]

okay yea i agree this is, yea

[worldly bear]

so this is the right thought process?

[slow scroll]

but this gives you a transition matrix from B' to B, not B to B'
you plug in, say, [(1,3)]_B' and get [(1,3)]_B = (k1, k2)

[worldly bear]

that seems so backwards

but yeah, thats the same thing you said earlier... i misread

[slow scroll]

i understand this sentiment

3blue1brown has a video on change of basis. It might help internalize some of these things

[worldly bear]

So an easy example... transition matrix from B' to B would be 2, 1 as the first column and -3, 4 for the other

oops didnt mean to include question 3

[slow scroll]

correct

[slow scroll]

btw this is what i meant by this earlier. To go from B' to S you just write the vectors in B' as columns of your transition matrix

[worldly bear]

yes i see that

and then for [w]_B, im doing [3,-5] = x1[2,1] + x2[-3,4]

[slow scroll]

yep

or well

do you mean B or B'?

because i was thinking B = S for prob 2

[worldly bear]

B but you make me think thats not right

[slow scroll]

well aren't we saying B' = (2,1), (-3, 4)

so [3,-5] = x1[2,1] + x2[-3,4] computes [w]_B', not [w]_B

[worldly bear]

B' to B, yeah

[worldly bear]

and it gives [w]_B' since im using u'_1 and u'_2 right

[slow scroll]

right

[worldly bear]

i guess since it wants me to find [w]_B first, i should be doing [3,-5] = x1[1,0] + x2[0,1]

[slow scroll]

yep

[worldly bear]

and then use [v]_B' = P B to B' * [v]_B

okay i think i get it now

[slow scroll]

epic

[worldly bear]

im gonna try this one in R^3 now, thank you for your help

[slow scroll]

npnp

[slow scroll]

a spanning set is a set of vectors that spans your space. so like if you have a vector space V, and a basis v1, v2, .., vn for V, then {v1, v2, ..., vn} would be a spanning set, but so would any set containing {v1, v2, ..., vn} for example

even though something like {v1, v2, ..., vn, w} would not be a basis

if u have a matrix in the complex numbers, then its diagonalizable iff the roots of the minimal polynomial all have multiplicity 1

(diagonalizability is a special case of jordan canonical form)

[worldly bear]

update, got the correct answers for problem 3 which was in R^3... granted i used my calculator for solving the systems but ive done row reductions so much im not worried about that part of this problem. Just understanding what im solving for when

[agile bronze]

Well, what did you get as the determinant

[hollow void]

I'm still finding that

Wait

[lime knoll]

Can someone explain me for part b?

[zinc timber]

notice you can make the 3rd column almost zero

[hollow void]

How?

Operation?

[zinc timber]

think

[hollow void]

I Don't have any Idea

Shit

[random axle]

How comes in 2-D we consider rotations about a point (the orgin), but in 3-D it's always about an axis?

Can't we form a matrix for rotation about the origin in 3-D?

[zinc timber]

try imagining rotation without an axis.

because dimension of R³ is 3, you'll always get a real solution for the characteristics equation giving you the axis of rotation,

[dusky epoch]

rotation happens not around a point or axis, but in a plane regardless of the ambient space's dimension

[random axle]

Gotcha, thanks!

[zinc timber]

A must be 0

@blazing sage you can prove it by showing first that A has to be diagonalizable, then assume A is diagonalizable, then every non-zero eigen value of B has algebraic multiplicity 2 (or 4, 6,..) but the geometric multiplicity is 1 (2, 3 respectively). so A has to be zero for B to be diagonalizable

[tranquil steeple]

Since your notation is very ambiguous, then you can not say much at all. (but I assume Ryu's answer is what you intended) Here is an example where A is not 0 ```
julia> A=rand(8,2)
8×2 Matrix{Float64}:
0.74659 0.73478
0.0111299 0.1941
0.790505 0.647498
0.57515 0.236845
0.273026 0.243877
0.942947 0.702215
0.329656 0.341478
0.24785 0.224575

julia> B=hcat(A,A,zeros(8,2),A)
8×8 Matrix{Float64}:
0.74659 0.73478 0.74659 0.73478 0.0 0.0 0.74659 0.73478
0.0111299 0.1941 0.0111299 0.1941 0.0 0.0 0.0111299 0.1941
0.790505 0.647498 0.790505 0.647498 0.0 0.0 0.790505 0.647498
0.57515 0.236845 0.57515 0.236845 0.0 0.0 0.57515 0.236845
0.273026 0.243877 0.273026 0.243877 0.0 0.0 0.273026 0.243877
0.942947 0.702215 0.942947 0.702215 0.0 0.0 0.942947 0.702215
0.329656 0.341478 0.329656 0.341478 0.0 0.0 0.329656 0.341478
0.24785 0.224575 0.24785 0.224575 0.0 0.0 0.24785 0.224575

julia> e=eigen(B);norm(B-real.(e.vectors*diagm(e.values)*inv(e.vectors)))
3.1932002353824708e-15

[zinc timber]

I think he meant $\mqty[A & A \ 0 & A]$

[stoic python]

Ryu

[wintry steppe]

rotacione

[stable urchin]

hey I was just wondering how to ig start the second part of the question

[slow scroll]

think cayley-hamilton tricks

[stable urchin]

right so this is an introductory lin alg course I’m taking

[lavish jewel]

by second part, you mean I-A and A^-1? or before that?

[stable urchin]

the I-A part

[winged prairie]

hello, i was wondering why you cannot do this for this exercise, given that S and T are invertible

[lavish jewel]

have you seen how det(AB) = det(A)det(B)?

[slow scroll]

are you implying you can't use cayley-hamilton?

[stable urchin]

yeah I don’t particularly know what that is unfortunately

[slow scroll]

ah okay i was probably on the wrong track or something anyway

[winged prairie]

Sorry I mean = x for both lines

[stable urchin]

mhm I was tryna think that out, I can’t really see how I can factor |(I-A)-xI|

[lavish jewel]

where did that x come from

i was thinking that A - xI can be evaluated at x=1

but we do some trickery before

i.e. xI - A = -I(A-xI)

so you can use det(-I)det(A-xI), and then evaluate the second one by looking at the poly they gave you, and substituting 1

[stable urchin]

ahh I see

[lavish jewel]

for the first term, det(-I), this will be something like (-1)^n, where n is the number of rows

something like that

[stable urchin]

right

I’ll see what I can do with that

thanks

[winged prairie]

any ideas for this?

[slow scroll]

You said "suppose ST = I", how are you getting T(S(x)) = I?

first of all, i think you meant T(S(x)) = x

[winged prairie]

ye

cuz since they are both operators

they are both invertible

[slow scroll]

oh. anyway yea its not clear how you go from S(T(x)) = x to T(S(x)) = x

[winged prairie]

thus ST = I is invertible too

ST i mean

[slow scroll]

Im confused. Yes the identity is invertible, but that doesn't mean S and T commute

[zinc timber]

u can also try to find the eigen values

[winged prairie]

if S is invertible and T is invertible, doesn't that not imply that TS and ST are invertible?

[zinc timber]

and noting that EV of 1+A will be 1+ev of A

[slow scroll]

we don't know that S or T are invertible

[winged prairie]

ye we do cuz they are operators

[slow scroll]

that would make this problem trivial

[winged prairie]

doesn't this imply that

[slow scroll]

the square matrix with all zeros is an operator

yea, those are equivalent conditions. Its not saying "all operators are invertible"

[winged prairie]

ohh

[zinc timber]

if S or T not invertible then ST can't be invertible

[winged prairie]

why r they allowed to do this then compared to what i did

[slow scroll]

okay, they're using facts from a previous exercise?

[winged prairie]

oh shit haha

its just this

[slow scroll]

ah okay there you go

[zinc timber]

also how many times have you asked this question?@winged prairie

[winged prairie]

once

[zinc timber]

I feel like having a dejá vu

[winged prairie]

i've asked a few questions on invertibility but not this one

thanks for the help btw

[zinc timber]

[winged prairie]

they are not the same

[zinc timber]

ig

,w roots of -x^3+x+1

[stoic python]

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Requested by Ryuzaki#5323

[zinc timber]

[stable urchin]

cant do it by hand for A

[zinc timber]

yeah I can see why

btw were u able to solve it?

[stable urchin]

nope .-.

[zinc timber]

IG what you can try next is express -x³+x+1 in terms of (x-1)

like using synthetic division or something

then replace x-1 by x, or alternatingly you can replace x with (x+1) in the original polynomial

to find the characteristic poly for I-A

this can be done because
$|I-A-xI| = |(1-x)I-A|$

[stoic python]

Ryu

[zinc timber]

now we know char poly for x, the for this one it will just be p(1-x)

[stable urchin]

got it

i figured it out on my own tho like i just figured it out

[zinc timber]

lol

[stable urchin]

its 2am my brain does not do the think sometimes

[zinc timber]

and what about A^-1

the cayley hamilton will give you the hint u need

[wintry steppe]

How do I know that U perpendicular is non empty?

[zinc timber]

because <0, x> is always zero

so U \perp contains 0 atleast

[wintry steppe]

oke ty

<0,x> means 0 dot x right?

[zinc timber]

hm

[wintry steppe]

?

[zinc timber]

means 'yes'

[lavish jewel]

i had never seen hm used as yes

save for few cases. very japanese of you

[zinc timber]

is used in my local language

[lavish jewel]

interesting

[zinc timber]

[lavish jewel]

but then again, you also nod no for yes lol

[zinc timber]

lol

[dire granite]

Let W = {f: R -> R | f’’ = f}, how can i show that W is a subspace of R^2

by using calculus

[stable urchin]

new to it so im wondering how to apply that to this

[zinc timber]

hint: ||if x is an eigen val of A then 1/x is an eigen val of A^{-1}||

[zinc timber]

because derivative is linear you can say...

[dire granite]

completely don’t understand

It’s actually new to me

[lavish jewel]

show that it satisfies the definition of a subspace 😛 there's like 3 things you have to test

and as it turns out, they should be pretty easy to check.

[wintry steppe]

its enough to check only 1: that its closed for finite linear combinations

[marble lance]

At least 2. Also that it is nonempty.

[zinc timber]

empty set is a vector space over all field

[dusky epoch]

no it isn't

what's the zero vector

[zinc timber]

yes ik, I was joking

[winged prairie]

yo quick question. Why is the composition of linear operators commutative?

[dusky epoch]

it isn't

[winged prairie]

that are invertible

[wintry steppe]

The zero vector is the empty linear combination

[winged prairie]

going from V to V

is that true?

[wintry steppe]

its not commutative

[dusky epoch]

it's not commutative even if you restrict to invertible maps

[zinc timber]

∅ is a VS over ∅

[wintry steppe]

Ø is not a field

its not closed for 0 and for 1

[wintry steppe]

existence of empty linear combination guarantees that its nonempty

[zinc timber]

@dire granite share what u have tried, someone will help

[dire granite]

ive not even started to write

have no idea if i should integrate so i'll get a set of solutions that depends on 2 free variables

[zinc timber]

remember you are not solving for the function here

just showing if f,g in the VS then f+g is also in the VS

[random axle]

In the third determinant, I did Column 3 - z times Column 2. Is this a valid row/column operation? Typically we only add scalar multiples of rows/columns to other rows/columns, so I just want to check if what I've done is allowed. The answer is correct, but I just want to be sure it was not by luck.

[wintry steppe]

a matrix to the power of a matrix

[random axle]

huh? are you referring to my image? sorry i know it's incorrect use of the power notation, but i am just using it to denote what row/column operation that i am doing

First one is Row 1 + Row 2

etc

[wintry steppe]

I am partly referring to your image, and this is what I thought of first

Ok, ic

ic

what do you want to acomplish with that manipulation anyways?

[random axle]

The goal of the question was to simplify the determinant. Expanding out the determinant directly would obviously lead to a lot of monotonous algebra which would be hard to factorise. So the goal is to use row/column operations to simplify the determinant. I have recently learned that we can add scalar multiples of rows/columns to another row/column without changing the determinant

Does this also apply to variables being multiplied to the rows?

[wintry steppe]

ic

[winged prairie]

sorry im still have trouble wrapping me head around this

i still don't fully understand why u can just assume S and T are invertible

[wintry steppe]

if S isnt invertible or T isnt invertible then ST not equal to I

[winged prairie]

ye but why

[wintry steppe]

theres a known theorem that says

if TS=I then ST=I

for finite dimension

[winged prairie]

so u can never have

TS = I and ST != I

[wintry steppe]

in finite dimensional spaces, no

[winged prairie]

ok ty

wait but

isn't that what we are typing to prove

[wintry steppe]

idk i didnt read the text before lol

[winged prairie]

this is the question

[wintry steppe]

i see

if you knew TS=I implies invertible itd be trivial so probably cant assume that

[winged prairie]

the solution does this but i don't understand why they can just come to this conclusion

[wintry steppe]

well they did exactly what i said

the exercise before that apparently proves that if ST is invertible then S and T are

[winged prairie]

ye but ST = I isn't enough to say its invertible right

[wintry steppe]

I is invertible

[winged prairie]

like u need to know that TS = I too

[zinc timber]

why don't u show that T and S have to be invertible

then ST=I means STS=S => TS =I

[winged prairie]

is this what i have to prove? STS=S => TS =I

[winged prairie]

I still don't get how that implies that TS = I

[zinc timber]

that's one way

[winged prairie]

ok

[zinc timber]

because if either S or T is not invertible than ST cannot have full rank, since RHS =I which has full rank, S,T must themselves be invertible

[winged prairie]

true

[zinc timber]

then what's your doubt?

can't you just conclude from here that TS=I

[winged prairie]

didn't realise this

[zinc timber]

hmm, said it like 3 times

[winged prairie]

ye sorry im just a bit slow haha

i really appreciate ur help

[zinc timber]

anyway, invertibility if S and T justifies this step

[winged prairie]

ayt

[granite mesa]

$r(AB)\le \min\left{r(A),r(B)\right}$

[stoic python]

RaD0N

[winged prairie]

for this question can u just prove T is linear. You then know its isomorphic because Dim(V) = Dim(F^(n,1))? Or do u have to prove its injective and surjective?

[marble lance]

You need to prove it is injective and surjective.

For example, Tv = 0 would also be a linear map but not an isomorphism.

[winged prairie]

ye but the Dim of this map is not equal to the dimension of F^(n,1)

why does this theorem not apply here?

[marble lance]

What do you mean dimension of the map?

[winged prairie]

oh nvm

[marble lance]

The theorem only tells you that there is an isomorphism between the spaces, not that this specific map is an isomorphism.

[winged prairie]

ye

[marble lance]

The 0 can be the zero vector from any space, so it could be from a space with the same dimension

[winged prairie]

But T has already been defined in the question

[marble lance]

Yes

So you need to show it is an isomorphism

This theorem doesn't tell you anything about T as it has been defined

It says the two spaces are isomorphic, so they have an isomorphism between them. That doesn't mean T is an isomorphism.

[winged prairie]

ok makes sense

ty

[marble lance]

👍

[random axle]

Apparently, it's not necessary for $P = R$ for this equality to be true. But I cannot see any other solution that will make this true if we don't use 0 matrices

[stoic python]

azeem321

[wintry steppe]

try stuff with a lot of zeroes maybe

[random axle]

Apparently null matrices not allowed

[limber sierra]

a lot of zeroes, not all zeroes

easy example.

[wintry steppe]

any help of how do I prove projU ◦ projU = projU?

[marble lance]

proj_U maps every element of R^n to an element of U, and every element of U to itself. So once v has been projected onto U, then applying the projection operator again just keeps it fixed.

So prove those two things

[wintry steppe]

do I need to use Sum of Series?

[dim epoch]

you have linear combinations which are finite sums

no no series

you could start by showing that it's linear and then just plug the definition in, use linearity etc

[wintry steppe]

where can I study these? bcs I have not been taught of projections.

[marble lance]

This is you being taught them.

You are being introduced to the concept now

Can you show that for any v in R^n, proj_U (v) is in U?

[dim epoch]

this is the introduction to them

so if you want to learn about projections do exactly that

there's no prerequisites you're missing if you've covered subspaces and linear functions in general

which i assume you have, considering the questions i saw from you

[wintry steppe]

oke I ll try thanks

[dim epoch]

work with the definitions of linearity etc

[heavy crown]

is it correct to say that every matrix that is diagonalizable over R also is diagonalizable over complex field ?

for the same D P?

[dim epoch]

R is in C so every matrix that's diagonalizable over R is per Definition diagonalizable over C

im not sure what you mean with D P tho

[heavy crown]

yea thats what I was thinking as well, thank you for affirming my thoughts 🙂

[swift minnow]

A and B are squared matrix
I need to prove that AxB = I

how can i do it?
i wrote down examples for it, its true, but how can i formally prove it (for every squared matrix)

[zinc timber]

multiply

[swift minnow]

any chance you can explain to me how

[zinc timber]

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

[stoic python]

Ryu

[swift minnow]

the solution for this should be I

[zinc timber]

try manipulating indices

[swift minnow]

aha

[zinc timber]

Notice if i=j then $\sum_k A_{ik}B_{ki}$

[stoic python]

Ryu

[zinc timber]

now look for cases where k>i or k=j+1 or something

[swift minnow]

thx

[zinc timber]

[whole peak]

When $A * B = I$ then $B = A^{-1}$ because $A * A^{-1} = I$

[stoic python]

Moriarty