linear-algebra

[little crater]

for these given

but idk how you got it :/

[fair plaza]

i solve this sistem

[little crater]

oh my....

wow

i guess you have 3 variables

and 3 equations

[fair plaza]

yeah

[little crater]

but this is non linear at least how it appears or am I missing something?

well I guess you factor out

v1

ill see what is upp...

[fair plaza]

actually dont do this

this is wrong

gere

here

its v1² + 1 not v1², v2²+1 not v2² ...

because you have to sum with I

[little crater]

why do I have to sum with I?

[fair plaza]

you send that (1) is this

[little crater]

yeah I see that but I just distributed through

[fair plaza]

now i see

you already did this

so its correct

[little crater]

I still don't know where to go about solving the system :/

[fair plaza]

i'll try to solve on the paper for you

[little crater]

:/

are these the ones I work with?

omh

well

1 = 3

[fair plaza]

yeah

[little crater]

idk if that gets me far

well I guses factoring it didn't help exactly

[fair plaza]

this second is wrong, in the () is not 2v1, is 2v2

[little crater]

seems they have a common factor

[fair plaza]

with v1 = v3, u get 2v1v2 = -6 and 2v2² = 2, so v2² = 1 then v2 = 1 and v2 = -1, sub in 2v1v2 = -6, the u get v1 = -3 and when v2 is -1 v1 = 3

with this u get v

u get?

[little crater]

i am still solving the system

we get v_1 = v_3 because we can divide the whole thing by the common factor, right?

wait

how do we know that the sum isn't 0?

[fair plaza]

bro look this how to solve this fastly, look what i circle, if v1 = v3, this diff is 0, so v1 2v2 = -6, and v2*2v2 = 2

[little crater]

wait

duh

it is 6

ahh

I see

[little crater]

thanks a lot @fair plaza

[tribal willow]

what is T^+(n)? its not mentioned in thm 14

[gray dust]

@tribal willow my guess is nxn upper triangular matrices but the notation shouldve been introduced before the corollary

[zinc timber]

It looks like polar decomposition where U is unitary ( orthonormal)

not sure

[thin crystal]

Algebra isn’t linear

[native rampart]

Explains why engineering is difficult

[wheat sorrel]

Algebra is the only branch of math that is strictly non linear

[wintry steppe]

I’ve gotten to creating a matrix but I don’t know where to go from here guys

[dusky epoch]

well you can check when its determinant equals zero

[wintry steppe]

Yeah I have to row reduce but i don't really understand what's going on

Or how I could row reduce that

Or can I literally just do the math for the determinant and find when that doesn't equal 0

so the determinant came out to be 45+p^2, which means the vectors are only linearly dependent when p is complex, but that's gotta be wrong

[bold hound]

I understand that U*U=I, but im having trouble as reordering matrices, to my understanding, isn't allowed.

[wintry steppe]

my matrix is wrong, its constructed from looking at the second line of my working, not from the order of components in each vector

[normal heath]

Yeah you need all your $\hat{i}$ components to be the vector of x-values, $\hat{j}$ the elements of y... etc.

[stoic python]

Deadphool

[normal heath]

So instead you would have $$x\begin{bmatrix}-1\2\3\end{bmatrix}$$

[stoic python]

Deadphool

[wintry steppe]

Yep gotcha 👍

[prisma socket]

guys, assuming you have matrices A and B nxn and rank(A)=rank(B)=rank(AB) can you say that the space of solutions Bx=0 is the same as Ax=0?

I think that's not true

[tribal willow]

oh that makes sense. thanks

[zinc timber]

consider $A=E_{12}, B=E_{21}$ then they satisfy the condition

[zinc timber]

I hope you can find vectors (look at e_j) such that Bv=0 but Av≠0

@prisma socket

[unreal linden]

U * U can be != I, but det(U * U) = 1

[zinc timber]

sol: ||Be_2 = 0 but Ae_2=e_1||

[prisma socket]

@zinc timber its not always true that's what i mean

[zinc timber]

wym?

[prisma socket]

not for every matrix A,B that rank(A)=rank(B)=rank(AB) Ax=0 and Bx=0 has the same solutions

[zinc timber]

weren't you looking for a CE?

[prisma socket]

CE?

[zinc timber]

counter ex

[prisma socket]

yes I did

just what is E12?

what is the size of this matrix?

[zinc timber]

E_ij is a square matrix with only i,j th element =1 and rest =0

size can be whatever you want as long as i,j <= size

[prisma socket]

That is something I do know. but what is the size?

oh

[zinc timber]

1¹⁰⁰⁰⁰ if that pleases you

[prisma socket]

Thanks, and another thing. the sol's of ABx=0 are the same as Bx=0 right?

[zinc timber]

No

AB=E¹¹ here

wait

ye

[prisma socket]

why not? rank(AB)=rankB

[zinc timber]

yes

[prisma socket]

Nice,thanks!

[blissful shell]

hi, something quick, why is this equal to n^2 ?

[wet stratus]

the matrices E_ij which have a 1 at the place ij and 0 everywhere else are linearly independent and span the space

[thin crystal]

algebra isnt linear

[little crater]

Back here again seeking guidance

I have done some stuff

which I believe has lead me to this

but I don't have any clue how to get M, c, or γ

I assume I cant just say

it doesn't say they have to be different from each other but I assume they want that?

<@&286206848099549185>

[little crater]

question above ^

[thin crystal]

algebra isnt linear.

[little crater]

This one happens to be

[thin crystal]

no

[little crater]

[ripe glen]

energy (E) and mining (M) is (Picture attached)
The management of these two sectors would like
to set the total output level so that the final demand
is always 40% of the total output. Discuss methods
that could be used to accomplish this objective.```

I am not sure how to solve this, would appreciate some help!

[uncut yarrow]

why do all finite dimentional vector spaces over R have a inner product space

[wintry steppe]

pick an isomorphism with euclidean space and transport the usual dot product over

explicitly: if $T\colon V \to \bR^n$ is an isomorphism, then define $\langle\cdot,\cdot\rangle\colon V \times V \to \bR$ by $$\langle v, w \rangle = (Tv) \cdot (Tw),$$ the right-hand side being the usual dot product in $\bR^n$

[stoic python]

TTerra

[wintry steppe]

remarks:

1. you'll get a new inner product on V with each choice of isomorphism T. this is the unique inner product on V making T into a linear isometry.
2. you may do the analogous thing for complex vector spaces and C^n

and more generally, any vector space over R or C (regardless of finite-dimensionality) admits an inner product. you do essentially the same thing by taking a basis (for the foundations minded person, this requires choice in infinite dimensions) and letting the inner product of two vectors be the dot product

[little crater]

Hello

I need some assistance in this last step

I believe based on my question I need to Find M, c, gamma in terms of B, u, alpha

But I really can't figure out how to rewrite this

The only thing we know is that B is invertible

[little crater]

<@&286206848099549185>

[little crater]

<@&286206848099549185>

[robust owl]

Well I guess for finding M in equation 1 you know B is invertible so maybe you can solve for M and start trying to eliminate other stuff?

[little crater]

but how?

I have shuffled these things around and can't seem to eliminate the terms

[robust owl]

Yeah I just did that one too.

[little crater]

problem is I can't get it in terms of just M = (B,u, alpha)

etc

c = (B, u, alpha),

[robust owl]

I'm seeing that too

[little crater]

this is bugging me because I have been sitting on this for hous

hours*

[robust owl]

Yeah some problems are just hard like that lol

Sometimes you just get stuck and have to keep plodding along, thinking about it, talking to people etc.

[little crater]

idk if I screwed up

[quartz compass]

since they just say "find a ..." I would just make M=B^-1 and gamma = 1/alpha

then see what conditions are forced onto c from there

[little crater]

but that is so dumb ....

[robust owl]

Ooh that seems nice lol

[little crater]

I wrote some example just assmung they were all identities intially

but that isn't even "solving" it

[little crater]

Sure but where does that even lead into this?

[quartz compass]

well it might not be doable like I describe I didn't work it out

just a suggestion lol

but worth trying if you haven't

[little crater]

Well it could work out but it just is already pre assigning a value

my example works also

but that is just dumb :/

[quartz compass]

not necessarily

I'm not saying what B, u or alpha are

[little crater]

i see

still

idk how I was to even get that

[quartz compass]

it might not work out

[little crater]

if M = B^-1

then

uc^T = 0

u could be 0 or c^t could be 0

or both

[quartz compass]

I read that as really saying they're orthogonal

[little crater]

wdym orthogonal

[nova yoke]

both satisfies

[quartz compass]

err I read that backwards

I was thinking you were writing the other thing

or you edited it

[little crater]

oh I just accidently wrote U instead of u

[robust owl]

What if they are both nonzero?

Are you trying to use zero product property?

I forget if it works here

[little crater]

they could be but they would have to be (0, n,0,n,0,n...)

like this

or one of them is 0

or they are both

[robust owl]

Isn't c a column vector?

[little crater]

it doesn't even say

[quartz compass]

you wrote uc^T=0 did you mean to write u^Tc=0

[little crater]

just says

"vector c"

but I think we assume it is a column based on the dimensions

[robust owl]

Yeah it seems like a column which would make c^t a row.

[little crater]

i got uc^t

BM + u c^t = I _(nxn)

[quartz compass]

yeah that's fine it's just what you're writing here is not that

[little crater]

oh :v lol

yeah that

[quartz compass]

uc^T appears in the top left term, u^Tc appears in the bottom right term

[robust owl]

Either way I'd bet zero product prop fails when you multiply a col by a row. (I'm not actually sure)

[little crater]

I believe I got the dimensions right

but not sure how important it is

[robust owl]

Makes sense to me

[nova yoke]

should work because if neither of them is 0 then there's an element a and an element b non-zero in column and row respectively and their product is in the product

[little crater]

still doesn't seem "mathy" to just say M=B^-1

[robust owl]

So, the goal is to find or construct M, c and gamma.

Often times it helps to make simplifying assumptions.

[little crater]

Sure but like :/

[quartz compass]

for what it's worth I already worked it out and it won't be true in general

[little crater]

I thought I would get some insights from these equations

[robust owl]

Maybe picking M=blah blah does the trick for a special case but there are more solutions?

[quartz compass]

there's only a unique inverse to a matrix

[robust owl]

Oh shit duh lol

[quartz compass]

😛

[little crater]

is there anything unique about matrix inverses and their inside matrices?

[quartz compass]

it turns out if we try to force this solution it will also force B to be a symmetric matrix and another property on B as well

[little crater]

which we never learned about

[quartz compass]

well if you transpose this matrix it's equal to itself along the last row and column

so they're basically saying if that's true, then its inverse has the same property

[little crater]

welp ;?

this is dumb

[robust owl]

Have you tried eliminating M and gamma?

[little crater]

wdym

[robust owl]

I'm doodling around and seems possible.

[little crater]

The problem is you can't use 2 and 3 together

[robust owl]

You can solve one of your equations for M, then sub out all the remaining M's

[little crater]

they are not the same size :/

[robust owl]

Then solve one equation for gamma and sub out all the gamma's

Then you have two equations to solve for c, which idk how to do or if it is even possible lol.

[little crater]

idk this problem is way more difficult then anything we did in the class

[quartz compass]

what happened with that problem the other day anyways

[little crater]

wdym

"that problem"

which one are we talking about

I have been trying to solve 3 problems that I couldn't solve

I got 1 of them

[quartz compass]

you had some other linear algebra problem you were working on the other day, I don't remember what it was now

did you turn in your problems or talk to your teacher or anything about them?

[robust owl]

Wait would picking c to be the zero vector work? Lol

[little crater]

These are were our exam questions and I couldn't solve the last 3 out of the 6 of them :/

so I turned them in basically blank

and it seemed to be the the same for about everyone else in the class

if you are referring to this problem

[quartz compass]

no cause it'd require B^-1 u=0 and 1/alpha * u = 0 too, which means u=0

so in that case it works

but too special of a case

[robust owl]

Ah fuck lol

[quartz compass]

good idea though

[little crater]

idk but Monday will be interesting because these problems really felt way more challenging then our HW, lecture, or in class examples

[quartz compass]

no it was something else with block matrices

[little crater]

also trying to solve this one

part (b)

[quartz compass]

like getting the matrix blocks to line up wasn't really even possible

[little crater]

oh

you mean the one were we just assumed

[quartz compass]

yeah

[little crater]

they corresponded?

[quartz compass]

exactly

[little crater]

I tried talking to him about it but he really didn't even say much

[quartz compass]

my guess is he's just bad at making up questions without working through the answer first or something lol

[little crater]

it really felt like a leap of faith because the problem didn't give any insight the initial structure of how it got divided.

like this one I assume you were reffering to

where I just said Z=C

[quartz compass]

yeah that one

we can try to dig in deeper on this problem

err, the problem we're working now

like do you know about the adjugate?

[little crater]

nope

[quartz compass]

what formulas do you know of for computing the inverse

[little crater]

I know chapter 1 and up to 2.4

[little crater]

only for 2x2

or using identity

and making augmented matrix

by doing like

[A | I ] ~ [I | A^1]

[quartz compass]

gotcha

when's this due?

[little crater]

This problem he said something about "row reduction using block matrices" but we never learned about that

[little crater]

it isn't due as we already turned in the exam

[quartz compass]

oh interesting

[little crater]

I am just working on them

[robust owl]

Is gamma zero whenever u is nonzero?

[quartz compass]

did he say anything else

[little crater]

I had no clue how to do them and maybe he might allow a fix up or something

[little crater]

No other than like instead of dividing you would have to be using inverses

[little crater]

in relation to this problem

but again idk how that would work because we never did anything like that

book doesn't seem to mention it that im aware of either

[quartz compass]

I say we do that, I never thought about it but I think that'll work pretty easily

[little crater]

like the whole section was like 4 pages in the book

and I am guessing X and Y in this problem can be whatever

it really isn't clear to me

He told me one of the solutions someone kinda just wrote was

and then

X = 0, Y =0

but again really isn't much insight into doing the problem and maybe more so an "oversight"

[quartz compass]

yeah I agree

[little crater]

see I never did this row reduction with "block matrices" and wasn't sure it would even work given we need to ensure the dimension all line up.

[quartz compass]

I think it will, just make sure at each step stuff works out

$$\left[\begin{array}{cc|cc}
B & u & I & 0\
u^T & \alpha & 0^T & 1
\end{array}\right]$$

[stoic python]

Merosity

[little crater]

Oh you are reffering to that problem

[quartz compass]

first thing is we want to remove the u^T with the B, so basically can we find some vector to multiply by B to make u^T?

since B is invertible we know we can find some v such that v^T B = u^T, so I'm thinking multiply the first row by v^T and then subtract it from the second row

$$\left[\begin{array}{cc|cc}
B & u & I & 0 \
0^T & \alpha-v^Tu & -v^T & 1
\end{array}\right]$$

[stoic python]

Merosity

[quartz compass]

so far so good

now let's eliminate the u in the top right

it's going to be a bit ugly but not impossible

well, I don't wanna go on without you, does what I've done so far make sense

I say you try to work out the rest, in the end whatever answer we get we can just multiply it through to see if it actually works or not, since the ends justify the means

[little crater]

I mean I get what you are saying but what vector would make u^T -> 0?

[quartz compass]

we're wanting to multiply that nasty scalar alpha-v^Tu by a vector to make u

so 1/(alpha-v^Tu) * u will work

[little crater]

also which way do you multiply

[quartz compass]

so long as alpha -v^Tu is nonzero

[little crater]

left or right

or is it just as long as you are consistent?

[quartz compass]

yeah as long as you're consistent should be fine I think

I'm gonna grab a snack and let you play around for a bit and then I'll check your work and do it myself to see what happens

[little crater]

still trying to follow this step

do I just need to introduce some varaible/vector

[quartz compass]

yeah, v is created by us and we know it exists

well, we can write it even more explicitly

v^TB = u^T means v^T=B^-1 u^T

[little crater]

don't you have to multiple on the left?

Oh you did

nvm

[robust owl]

Does this work for gamma: Bc=-gu so c=-gB^-1 u. But then u^Tc+ga=1 becomes u^t(-gB^-1 u)+ga=1. Hence g=1/(a-u^T B^-1 u)? (I guess division by zero should be ruled out here)

I hope that makes sense. I'm too lazy for latex I'm using g for gamma a for alpha.

[quartz compass]

this was wrong I should have put v^T = u^T B^-1

[robust owl]

Given gamma Bc+gu=0 can be solved easily for c I think.

Well I basically already did that.

[little crater]

so is v = n x 1 => v^T = 1 x n?

or other way around?

[robust owl]

So, now that c and gamma is determined only M is left, but BM+uc^T=I is solvable for M.

[quartz compass]

yeah that's right

I just made v to make it less cumbersome but we could directly just write in u^T B^-1, multiply this on the left of the first row

then subtract it from the second row

[little crater]

wait did you factor the gamma out in the 2nd part?

[robust owl]

Yes

I could just be failing at matrix algebra lol

[little crater]

bruh I can't believe I didn't factor

I was stuck trying to figure out how to work with the 2 gammas

still need to miss around with this row reduction idea though as i guess i need to use it for the other problem

but lets see if I can do something with this

[little crater]

what do I even do with this new gamma

back into equation 2?

[robust owl]

So, you solved for c already in terms of stuff that is all already known. (Since you now know gamma)

Now all you need is M

M can be found from the very first equation.

[little crater]

yeah

so I got

[robust owl]

I didn't actually solve for M or write c not in terms involving gamma.

I also didn't check if a-u^t B^-1 u could be zero.

You can probably check if we messed up by working out the block matrix multiplication identity from the problem stmt too.

[little crater]

yeah idk about that case

[quartz compass]

if it's 0 then that means it's not invertible because it's the lower right entry after the first step of elimination

that would make two 0s in that row

I worked it out painstakingly by hand that the result I got from row reduction was indeed the inverse

[robust owl]

You, got the same thing as us?

[quartz compass]

but we can also sorta reason out directly that each of these block row operations actually corresponds to doing regular operations too

idk I didn't see your final answer

[little crater]

^

[quartz compass]

yeah that looks about right yeah

no slightly off

some of your operations don't make sense

[little crater]

[quartz compass]

in M specifically

B^-1 u B^-1 u

[robust owl]

Was your gamma the same?

[quartz compass]

yeah

[little crater]

all I did was sub in my C

[quartz compass]

you have to be careful about what side you multiply on

in your M the B^-1 u B^-1 u multiplication doesn't line up

nxn with nx1 with nxn with nx1

[robust owl]

Okay Ixm relieved I wrote c in terms of gamma and knowns, found gamma and just said the eqn in M had a solution lol.

[quartz compass]

it turned out when I worked it out, all my row operations were left multiplication

[little crater]

wait im confused

are you saying I need to do

B^1 all on the right?

[quartz compass]

I don't think I'm saying that

not sure what you mean though

[little crater]

all I did to solve M

was replace C

and then invert the B

on the left

which unless I wrote my equation 1 wrong

[quartz compass]

this is meaningless

[little crater]

:c

that was my C

[quartz compass]

think about the vectors u B^-1 u

nx1 with nxn

you can't multiply that

[little crater]

yeah

so then I screwed up in gamma or c

[quartz compass]

just gotta go through it more carefully and make sure you're consistent with what side you're multiplying on I think to make sure stuff lines up

[little crater]

B^-1 = nxn

u = nx1

[quartz compass]

I actually wrote out my steps,
$$R_2 \to R_2-u^TB^{-1}R_1 $$
$$R_2 \to \gamma R_2$$
$$R_1 \to R_1 - uR_2$$
$$R_1 \to B^{-1}R_1$$

[stoic python]

Merosity

[quartz compass]

the side I multiplied on is also the same as how I did it, so writing uR_2 is multiplyling R_2 on the left by u

[little crater]

well maybe I need to work it out with row operations

Could it be because of of BM + ug = 0

which could have also been

MB + ug = 0

well actually idk 🤔

[quartz compass]

idk, I'm not gonna go into your garden and dig up your weeds for you 😛

[little crater]

anyways with block matrices

is it

BM + uc^T

or do you just have to see how it aligns?

[dusky epoch]

yes

assuming the blocks align

it is BM + uc^T

[little crater]

ill screw around with the row operation stuff and see how it works

[quartz compass]

yeah just requires going slow and being careful

that way you only have to do it once

[quartz compass]

I decided to check some examples, and the problem is wrong

because the answer I got was correct except that c and c^T were actually separate vectors

in general they're not going to be transposes of each other even though the u and u^T vectors are

there are likely easier examples but this is just the first random thing I tried and it broke lol

regardless, the row reduction technique works

[little crater]

well that is not very enlightening

so are you saying I can't write them in terms of the other 3 terms?

[quartz compass]

no that's not what I'm saying at all

I'm saying row reduction works, you can write them

[little crater]

oh

[quartz compass]

it's just the c and c^T vectors they put are really independent vectors altogether

one is not the transpose of the other generally speaking

[little crater]

Oh

okay

[quartz compass]

because this has become something of a nightmare I want to go ahead and just put the answer so I can forget about it lol

$$\gamma = \frac{1}{\alpha-u^TB^{-1}u}$$
$$
\begin{bmatrix}
B^{-1}+\gamma B^{-1}uu^TB^{-1} & -\gamma B^{-1}u \
-\gamma u^T B^{-1} & \gamma
\end{bmatrix}$$

[stoic python]

Merosity

[quartz compass]

good thing you mentioned off hand that your teacher gave that hint to do row reduction lmao, what other hints are you not telling us 🤣 maybe I'll be interested to dig into those more tomorrow too

[little crater]

oh he didn't until our interview after the exam

[quartz compass]

ahh, damn

It seems technically possible to do without row reduction if you go back to the original 4 equations you set up, since we're effectively just doing algebra steps

[little crater]

but he said to do it for another problem , not really this one. He still seemed to work with the system of equations and do a bunch of back subbing

[quartz compass]

but the row reduction algorithm really helps guide it

[little crater]

This is the one he mentioned using it for.

[quartz compass]

oh I used the wrong hint

well either way it worked lol

[little crater]

well I mean yeah as you said all this back subbing and what not can be viewed as row reductions steps

be interesting to see how (if) he explains these few to the class.

[quartz compass]

yeah

the fact that c and c^T were actually separate vectors actually dooms you if you try to use that in your solution

because it forces a false condition

[little crater]

wait how did you say this was wrong again?

u = nx1

B^-1

= nxn

(B^-1*u) would be defined though?

or is it the bottom?

Oh

[quartz compass]

the numerator

[little crater]

oh the bottom is a scalar?

[quartz compass]

yeah

bottom is fine yup

[quartz compass]

this is defined yes

[little crater]

weird because I had the same gamma

[quartz compass]

u is nx1 and B^-1 is nxn, the indices don't match just that simple

[little crater]

all is fine until I go to solve M

[quartz compass]

gamma is the first thing that pops up in doing row reduction

something with the way you're multiplying is wrong

[little crater]

was this your C

[quartz compass]

I'd just follow along with my steps here

[quartz compass]

@little crater this was my answer

yeah looks the same

[little crater]

well thanks. I'll look at how to do this row reduction thing tomorrow

[quartz compass]

yeah you're welcome

[leaden seal]

The eigenvalues of a matrix, A, are just the values that turn det(A-I\lambda) = 0

Am I right to say this?

[dusky epoch]

yes, you have stated the definition correctly

[hallow vector]

how can i find the complete solution for this?

I got it into reduced row echelon form (on my paper), but I dont understand how people are creating the parameterized vectors for the complete solution

[crystal wind]

The C + Dt form right?

[hallow vector]

yeah i think my professor would accept a particular solution + a linear combination

[crystal wind]

So where are you stuck?

[hallow vector]

I dont understand how to do it. I put the matrix into reduced row echelon form

but I just do not understand how to write the solution set

I watched like 4 youtube videos and I dont understand where they're getting the numbers from

[hallow vector]

I think my best question is, what do I do after I put the matrix in RREF in order to get the complete solution set

[crystal wind]

Wait a min

First find the null space of A

[hallow vector]

isn't that what I'm already trying to find?

[crystal wind]

Yeah null space + particular solution

Is the complete solution

There are two pivots and two free variables

[hallow vector]

i think that this matrix may not be solvable

when i put it into reduced row echelon i got

1 3 0 0 | 4
0 0 1 2 | -2
0 0 0 0 | 3

how can an all zero row equal 3?

[crystal wind]

Yes

W = 0

Back substitution

[hallow vector]

how did you know W = 0?

[crystal wind]

Multiply the row echelon form to the vector

[hallow vector]

but what about the last equation saying that 0x + 0y + 0w + 0z = 3?

i feel like that just isnt even possible

[crystal wind]

You're right

[hallow vector]

ok, so no solution?

[crystal wind]

I think so

I'm sorry, I screwed up

[hallow vector]

np, thanks for the help

[torpid moat]

How does one show that for matrix $P\in{M_n}$ we have $\innerproduct{h}{Ph}\geq0$ for all $h\in\mathbb{C}^n$, if and only if $P$ has an orthonormal basis consisting of non-negative eigenvalues?

[stoic python]

thestonethatrolled

[zinc timber]

that is not true in general if you are not told before hand that P is normal/self adj

[torpid moat]

Oh, right, I forgot to say that it was normal

[little crater]

Does this seem okay for my answer

My prof mentioned doing row reduction on block matrices

but it seems to assume we have a inverse to A11

I did confirm and it seems though all the sizes of the multiplication will work out

Also the "rho" = p and some reason my prof uses that notation instead of just saying R1, R2, R3, etc

so

yeah

and R1: is just denoting the "first" row operation, not acting on R1

it acts on R2

so it could read

R2 <-- R2 - (A_21)(A_11^-1)R1

[small dome]

hey dudes

quick question

how would gaussien elimination work if we were to switch columns instead of rows

For eg gaussien elimination on A but by swapping columns

[wet stratus]

what do you mean with "how". same way but with columns

you can think of it as doing the normal elimination on A^T

[small dome]

but GE on A^T wont yield the same results as GE on A

swapping a row is mathematically valid but swapping a column changes the matrix

[wet stratus]

well it depends on what kind of result you are after

row operations change the column space but don't change the row space

for column operations it's the other way around

both swapping row or column changes the matrix

importantly, neither change the dimension of the row or column space

[waxen gulch]

Hey can someone explain me what i have to do to find T(1) , T(T) , T(T^2),etc... i dont know where i have to plug those value.

T(1)=(1,1,1,1) for example from the solution on my book but i dont know how

[wintry steppe]

T(1), T(t), T(t^2), and T(t^3) mean the images of the polynomials 1, t, t^2, t^3 under T

[waxen gulch]

Yes but how do u find them in this example

[wintry steppe]

you use the definition of T

the problem tells you what it is

T sends a polynomial p(t) to the vector (p(-3), p(-1), p(1), p(3))

[waxen gulch]

P(-3) , P(-1) , P(1) , P(3) how its equal to (1,1,1,1) for T(1) im missing something

[wintry steppe]

p(-3) meaning p(t) evaluated at t = -3, for example

so you substitute -3 for t in p(t). if p(t) is t^2, then p(-3) = 9, for example

can you do the rest?

[waxen gulch]

Ok for T(T^2) it will be (9,1,1,3) for respectevly P(-3) , P(-1) , P(1) , P(3), but why its not (-3,-1,1,3) for T(1)

[wet stratus]

1 here is the constant polynomial p(t)=1

[wintry steppe]

T(t^2) is (9, 1, 1, 9)

[waxen gulch]

Yes sorry misstyping

[wintry steppe]

please distinguish between capital T and lowercase t

[waxen gulch]

i find this very confusing im having struggle to find images each time on this kind of exercice

well thx for ur answers i will try to work on this

[waxen gulch]

I tried to think with an example for P(t) if it is 1 how I get a component vector (1,1,1,1) I think I don't understand very well those if someone could explain that would be nice

I think it's because I don't realize very well what the 4 components in the vector (1,1,1,1) are

[wet stratus]

if p(t)=1 is the constant polynomial that is 1 everywhere, then p(-3)=1 and p(-1)=1 and p(1)=1 and p(3)=1

[waxen gulch]

Okay I see and now let's suppose that the basis of R4 is no longer the standard basis but this one, does it change anything to find the vectors T(1), T(T), ect.. of M?

Is T(1)=(1,1,1,1) there?

Well T(1) will remain the same because thats the same vector here

but for T(T) for example

with standart basis of R4 T(t) is (-3,-1,1,3) what about this one?

[gray dust]

please distinguish between capital T and lowercase t

recall how the matrix of T, wrt given bases of the domain & codomain, is constructed

recall how the matrix of T, wrt given bases of the domain & codomain, is constructed
this hints to why not much effort was needed when the given basis of R^4 is the standard one

[wild forge]

Can anyone explain this

[wintry steppe]

expand both sides...

this isn't linear algebra

[wild forge]

Yes it is

[wild forge]

Both sides of what

Line ad looks pretty linear

[tranquil steeple]

no

[wild forge]

Do you even know what you're looking at

[tranquil steeple]

goto #prealg-and-algebra

[wild forge]

You don't know

[tranquil steeple]

If you have a problem understanding what you posted, I would say you are the one that does not know...

people do that in middle school

[wild forge]

Do what

[tranquil steeple]

understand. why the two expressions are equal

you don't obviously since you asked.

[wild forge]

You don't realize how it relates to the figure

[ripe lance]

#help-4 ???

[wild forge]

You don't know what a square is even when looking at one

A literal square

[wintry steppe]

i'll phrase it as explicitly as possible

what you posted is euclidean geometry, which is not linear algebra

therefore it does not belong in this channel. you would have a better shot in #geometry-and-trigonometry or one of the others help channels

[wild forge]

Naw

It's literally labeled linear algebra

[wintry steppe]

what is?

[wild forge]

Book two

[tranquil steeple]

lol

[wintry steppe]

i do not know what book two is, but just because it's titled "linear algebra" doesn't make all of its content linear algebra

[tranquil steeple]

kids....

[wild forge]

You don't know Euclid

Kids .....

[wintry steppe]

lmao

i don't know euclid, but i can pretty confidently tell you that euclidean geometry is not linear algebra

[wild forge]

But muh lines

[wintry steppe]

Linear algebra is the branch of mathematics concerning linear equations such as:

{\displaystyle a_{1}x_{1}+\cdots +a_{n}x_{n}=b,}{\displaystyle a_{1}x_{1}+\cdots +a_{n}x_{n}=b,}
linear maps such as:

{\displaystyle (x_{1},\ldots ,x_{n})\mapsto a_{1}x_{1}+\cdots +a_{n}x_{n},}{\displaystyle (x_{1},\ldots ,x_{n})\mapsto a_{1}x_{1}+\cdots +a_{n}x_{n},}
and their representations in vector spaces and through matrices.[1][2][3]

Linear algebra is the study of vectors, matrices, vector spaces (including inner product spaces, the geometry of vectors, etc.), and linear transformations.

[wild forge]

We can consider pairs (A,B) of points, and define an equivalence on them which makes (A,B) equivalent to (C,D) under the conditions when we would now say that the vector from A to B is the same as the vector from C to D. (Either ABDC is a parallelogram or they are collinear and….) As far as I know Euclid never developed such an idea, but the step of abstraction involved in taking pairs of points under an equivalence is only a very modest extension of classical Euclidean geometry. I think you can count the idea as belonging to the field.

[tranquil steeple]

stop

[fallen karma]

I haven't encountered much in the way of dual spaces/dual maps yet. When do those start becoming useful?

I'm thinking about row space which I suppose spans the range of the dual map

[wintry steppe]

the dual space and its exterior powers are used to describe covectors and more generally differential forms on manifolds

[little crater]

Does this check out

[wintry steppe]

another related example where duals come up is in the correspondence between bilinear forms V x W -> k and linear maps V -> W^* (or W^* -> V if you'd like). this is a rather easy, but important, observation

chances are, if you were to name any given field of mathematics, you could find some instance of a dual space in it

(... one that makes any use of linear algebra)

[fallen karma]

Thanks!

[fallen karma]

Did you make use of the transpose at all?

[little crater]

yep

I said

Q^TQ = I

and since Q is nxn

we know it inverts

that is about it

not sure if I was suppose to approach it like that or not.

[fallen karma]

Right but I'm just curious why Q need be orthogonal rather than just invertible

[little crater]

Idk anything about A = QR or what is special about it

[little crater]

well in this case we have shown it is in fact invertible

idk about in general

[fallen karma]

Your proof looks ok to me

[little crater]

Not really sure the actually solution

[fallen karma]

Oh wait, why is upper triangular significant?

[little crater]

Im not really sure why half the information is in there

we just learned about A = LU

not even how to factor it (yet)

just how to use it to solve Ax=b

this problem set just has a few of these weird A = ....

not sure the significance but I just roll with it.

like this one

I feel like I am just going to use IMT

given U and V are nxn

and they have stated inverses which are their transpose

I really do feel like I am not doing this right though

well D is going to screw this one up....

but seeing it is diagonal and none zero it must be row equilvant to the I and thus have an inverse

are diagonal matrices alway square?'

[fallen karma]

ye

[little crater]

Does this idea extend

to like

(ABCD.....)^-1 = ....D^-1C^-1B^-1A^-1

basically

are the inverse of a product of matrices

the inverse of those matrices but in the opposite order

nvm I think I got it

[fallen karma]

Rx=Q^t•b has a unique solution, then QR=b has a unique solution

[little crater]

I think we are doing leastt squares like the last week of class

so idk if this will show back up or not

I really do feel like I approached those 2 wrong though

[fallen karma]

Still don't know where upper triangular comes in

[little crater]

Not sure either

okay

well I guess I was okay

that is the answer in the back of the book for my question

the other question is even so I don't have a solution in the back

[little crater]

I think this chegg thing answers something about upper triangular

I guess when a matrix is triangular it is faster to compute which makes sense but whatever

[jaunty knot]

Consider solving Rx = b where R is upper triangular, try writing out an explicit example, maybe 3x3 or 4x4

Can you see perhaps why it is easy to compute the solution? @little crater

[frigid fiber]

Linear Operator has been given as a matrix. Find the basis and defect.

[frigid fiber]

can anyone please explain this. answer is given but i need explanation

[quasi vale]

@frigid fiber It would be better if you watched a video on yt or somewhere else about how to find nullspace of a matrix

[quiet salmon]

Let $V = \mathbb{R}$, the set of all real numbers, and let $x + y$ and $ax$ be ordinary addition and multiplication of real numbers.

[stoic python]

texaspb

[quiet salmon]

how to check if this satisfies the axioms for a linear space?

[wet stratus]

well what is your definition of the real numbers?

this is either trivial or pretty hard

can you use that the real numbers are a field already?

[quiet salmon]

no not really, I'm on the first chapter of Apostol Calculus Vol 2.

he hasn't mentioned anything regarding fields yet

[wet stratus]

how did he define vector spaces if he hasn't mentioned fields

[quiet salmon]

yeah one moment let me type down the axioms

AXIOM 1. CLOSURE UNDER ADDITION. For every pair of elements $x$ and $y$ in $V$ there corresponds a unique element in $V$ called the sum of x and y, denoted by $x + y$ \
AXIOM 2. CLOSURE UNDER MULTIPLICATION BY REAL NUMBERS. For every $x$ in $V$ and every real number $a$ there corresponds an element in V called the product of $a$ and $x$, denoted by $ax$. \
AXIOM 3. COMMUTATIVE LAW. For all $x$ and $y$ in $V$, we have $x + y = y + x$.\
AXIOM 4. ASSOCIATIVE LAW. For all $x, y,$ and $z$ in $V$, we have $(x+y) + z =x +(y+z)$.\
AXIOM 5. EXISTENCE OF ZERO ELEMENT. There is an element in $V$, denoted by $0$, such that $x + 0 = x$ for all $x$ in $V$ \
AXIOM 6. EXISTENCE OF NEGATIVES. For every $x$ in $V$, the element $(- 1)x$ has the property $x+(-1)x= 0$. \
AXIOM 7. ASSOCIATIVE LAW. For every $x$ in $V$ and all real numbers $a$ and $b$, we have $a(bx) = (ab)x$ \
AXIOM 8. DISTRIBUTIVE LAW FOR ADDITION IN V. For all $x$ and $y$ in $V$ and all real $a$, we have $a(x + y) = ax + ay$ \
AXIOM 9. DISTRIBUTIVE LAW FOR ADDITION OF NUMBERS. For all $x$ in $V$ and all real $a$ and $b$, we have $(a + b)x = ax + bx$ \
AXIOM 10. EXISTENCE OF IDENTITY. For every $x$ in $V$, we have $1x = x$

[stoic python]

texaspb

[quiet salmon]

like

I just don't know if I need anything else

[wet stratus]

ok so he just defined vector spaces over R

[quiet salmon]

to show that this satisfies

[quiet salmon]

yep

[wet stratus]

how did he define R?

[quiet salmon]

not mentioned in this book

[wet stratus]

just handwavy "the set of real numbers" ?

[quiet salmon]

yeah kinda confusing ngl

[wet stratus]

ok so some backstory.

$\bR$ is what is called a field. we have two operations on $\bR$, + and $\cdot$ which satisfy some rules,

$\bR$ is closed under both operations ($x+y,xy\in\bR$ for all $x,y\in \bR$),

commutativity and associativity for both,

distributive laws,

existence of 0 and 1 such that $0+x=x+0=x$ and $1x=x1=x$ for all $x\in\bR$

and additive/multiplicative inverses which means there exist $-x$ such that $x+(-x)=0$ for all $x\in\bR$ and $x^{-1}=\frac{1}{x}$ such that $x\cdot x^{-1}=1$ for all $x\in\bR\setminus{0}$

[stoic python]

Denascite

[quiet salmon]

so I just need these functions defined

$+$ and $\cdot$

[stoic python]

texaspb

[quiet salmon]

?

[wet stratus]

yeah. here they are just the usual addition and multiplication

in general if you have any set F with those two operations satisfying all those axioms, then F is a field and you can define a vector space over it. this means that you use the same axioms as the one you posted, just with x,y in F instead of x,y in R everytime

[quiet salmon]

are they defined like this or something? $+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$

[stoic python]

texaspb

[wet stratus]

yes

[quiet salmon]

perfect

then because R is a field, the axioms automatically satisfy?

do I still need to check thatt?

[wet stratus]

you mean the vector space axioms?

[quiet salmon]

yeah

[wet stratus]

yeah basically immediately satisfied

[quiet salmon]

alright

thanks!

uuhhh

just another question

given that $+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, taking a pair $(x ,y) \in \mathbb{R} \times \mathbb{R}$, so that $x + y = y + x$ gives me another real number, I can choose a $z \in \mathbb{R}$ satisfying the associative law?

[stoic python]

texaspb

[quiet salmon]

$(x + y) + z = x + (y + z)$

[stoic python]

texaspb

[quiet salmon]

because like

(y + z) is a pair

and a real number

[wet stratus]

on the left you take the pair (x+y,z) and on the right you take the pair (x, y+z) if that is what you mean

[quiet salmon]

yes

exactly

[wet stratus]

let's write + as $f:\bR\times\bR\to\bR$. Then associativity means $f(f(x,y),z)=f(x,f(y,z))$

[stoic python]

Denascite

[wet stratus]

but that is just inconvenient

[quiet salmon]

ye ye

I got it

man

that's so cool

like actually interesting

lol

[wet stratus]

yeah it is

[limber forum]

Excerpt from the chapter about history of linear algebra 'It is partly the move from analog to digital, in which functions are replaced by vectors'

Does that mean that a lot of problems of continuity can be mapped to linear algebra ?

[wet stratus]

I think that is just supposed to mean that previously we had functions f(x), now we often store/work with them digitally in the form of lists/vectors (f(x1),...,f(xn))

technically, functions are still elements of infinite dimensional vector spaces

[limber forum]

Solid answer. Thanks.

[hard verge]

anyone could show me how go to the first form to the second knowing that r = p^kcos(kθ)

[wet stratus]

if $z=\rho e^{i\theta} = \rho(\cos(\theta)+i\sin(\theta))$, then what is $z^k$ and $\Re(z^k)$ ?

[stoic python]

Denascite

[hard verge]

Moivre formula so $z^k =\rho^k (\cos(k\theta)+i\sin(k\theta))$

[wet stratus]

rho^k

[stoic python]

suzuya.eth

[hard verge]

yes you right.

then ?

[wet stratus]

good and now Re(z^k) ?

[tranquil steeple]

fun fact: the second line is the generating symbol of the Kac-Murdock-Szegő matrix https://nhigham.com/2021/07/06/what-is-the-kac-murdock-szego-matrix/

[hard verge]

$Re(z^k) = \rho^k (\cos(k\theta)$

[stoic python]

suzuya.eth

[wet stratus]

and finally 1/(1-z) ? and then Re(1/(1-z))

[hard verge]

$1/(1-z) = 1/1-\rho (\cos(\theta)+i\sin(\theta))$

[wet stratus]

that is 1-z^k ?

[stoic python]

suzuya.eth

[wet stratus]

forgot brackets

$\frac{1}{1-z} = \frac{1\cdot \overline{(1-z)}}{(1-z)\cdot\overline{(1-z)}}$

[stoic python]

Denascite

[hard verge]

oh okay i got it

thx

[hard verge]

is there a name for this way of making appearing numbers to rearrange the operation or its just instinct ?

[wet stratus]

I don't know if there is a specific name for complex numbers. if it was for example $$\frac{1}{1-\sqrt{2}} = \frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})}$$ I've heard it being called rationalizing the denominator

but it's the general way to divide by complex numbers

[stoic python]

Denascite

[hard verge]

okay thank you.

[cunning violet]

I want to calculate the Eigenvalue of this

the solution says "the first one is obviously 8"

I dont understand how its obvious 8

[wintry steppe]

look at the first column

if A is any matrix, Ae_1 is the first column of A

here, Ae_1 = 8e_1

[cunning violet]

R says the third Eigenvector is (1,0,0). why isnt it the first one?

[wet stratus]

well how you order them is kind of arbitrary

[pastel brook]

hey, I have some quick questions about eigenvectors, eigenspaces, and symmetric matricies:

1. is a one-dimensional eigenspace an eigenspace whose span is that span of a single vector. e.g. E = Span { u } would be one-dimensional, but E = Span { u, v } would be two-dimensional, right?

2. I know that according to the Spectral Theorem, a n x n matrix A has n real eigenvalues (counting multiplicities). Does this mean that A has n one-dimensional eigenspaces?

3. If A is orthogonally diagonalizable, then A could an n x n dimensional matrix with n distinct one-dimensional eigenspaces. If A is an n x n matrix with n distinct 1D eigenspaces, is A guaranteed to be orthogonally diagonalizable? I was considering

|` 1  2 `|
|. 3  4 .|

as a counterexample, as it is not orthogonally diagonalizable (e.g. isn't symmetric) but has n = 2 distinct eigenvectors, and hence n distinct 1D eigenspaces. Is this reasoning sound?

[wintry steppe]

1. a non-zero single vector, lest the space be zero-dimensional. the latter space span{u, v} could be one or zero-dimensional as well, unless you ask that u and v are linearly independent.

2. absolutely not. the spectral theorem says that a real symmetric matrix has n real eigenvalues and is orthogonally diagonalizable. this does not mean they are distinct (in which case your conclusion would hold). the identity matrix is an easy counterexample.

3. this example is correct, though you ought to state linearly independent eigenvectors instead of distinct.

[pastel brook]

Thank you for the thoughtful reply, I really appreciate it! Let me read though it :)

Great answers, really clarified the requirements for 2 to hold.

So an n x n symmetric matrix has has n real eigenvalues, and eigenvectors corresponding to different eigenvalues are orthogonal. Is the converse true? (e.g. is a matrix with an orthogonal eigenspace symmetric?)

[winter harbor]

A matrix that admits orthogonal diagonalization is always a normal matrix (i.e, it commutes with its transpose), but not in general symmetric.

In fact

The spectral theorem for normal matrices says that a (real) matrix is normal iff it admits orthogonal diagonalization. In the complex case iff it is unitarily diagonalizable.

[pastel brook]

1  0  0
0  0 -1
0  1  0

limited to real entries, would this constitute a counterexample of the converse in my original statement (e.g. matrix has an orthogonal eigenspace, but is not symmetric?)

[pastel brook]

huh, interesting, thank you for pointing out this correspondence :)

[winter harbor]

Hmm

Ok

We have to be a bit careful with the real case though

In the complex case normal iff unitarily diagonalizable works just fine

[pastel brook]

So if A, is say, a 3 x 3 matrix with mutually orthogonal eigenspaces, then A is normal, but not necessarily symmetric? Or do I have that completely wrong?

[winter harbor]

https://math.stackexchange.com/questions/225918/a-matrix-is-symmetric-iff-its-eigenspaces-are-orthogonal

This is quite helpful

[winter harbor]

Yeah, forget what I said about the complex case tho.

Things really are different in the real case

This post on mathoverflow addresses this

[pastel brook]

Thank you for the link, I appreciate it.

Ah, because 0 is orthogonal to everything

@winter harbor @wintry steppe — I think I've got a handle on what I set out to clarify, thank you so much for the help, you've made my day :)

[errant palm]

why is this not reduced echelon form? i thought as long as theres 1 pivot in each column and they're all 1's, then it's reduced

https://cdn.discordapp.com/attachments/811392047180808192/983834879437463602/unknown.png

[hard drum]

If a column has a leading 1 then all other entries in that column must be 0

[errant palm]

ohhhhh

yeh makes sense

thank u potato

[hard drum]

Ig we want RREF to be unique and what they've done puts it in a form where you can't really do any more 'moves' to simplify

[hard drum]

np

[queen olive]

Does this transformation exist, and if not why?

condition 3 should have f(x) < y replace with f(x) > x

[little crater]

👉 👈 can I get a little bit of help with this question

So today we learned about LU decomposition but I am really not sure what we need / are suppose to generalize

I am not sure if this has to do with my book and prof showing different ways to compute L and U.

But It seems when you compute the LU of A

It doesn't seem to exist

,w lu decomposition {{2,-4,-2,3}, {6,-9,-5,8}, {2,-7,-3,9}, {4,-2,-2,-1}, {-6,3,3,4}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Brandon7716#6524

[hexed urchin]

I am given 4 matrices and want to know if a 5th matrix is in the span of the first 4. To do this, I made an augmented matrix with the left side rows equal to each of the vectors respectively as below:

the right side of the augmented matrix is just 0,5,-6,3 from top to bottom given from B. I solved this to get an inconsistent matrix. I conclude that this means that B is not in the span of the 4 A vectors and I was wondering if that sounds reasonable

[mortal stream]

sounds right

[hexed urchin]

Awesome, thanks

[grave garden]

Guys, how do I find Jordan normal form given minimal polynomial ?

My problem is $m_A(\lambda)=(\lambda-1)^3$

[stoic python]

Potato

[wintry steppe]

the multiplicity of an eigenvalue as a root of the minimal polynomial is the size of the largest jordan block corresponding to that eigenvalue, so in general you're not going to find a unique jordan form

without more information, all you can write down is the possible jordan forms

[grave garden]

A is 4×4 too

[wintry steppe]

then in that case you can find a unique jordan form

[grave garden]

Hmmm i still dunno how to find

Can you explain more ?

[wintry steppe]

what is the size of the largest jordan block for A?

[grave garden]

4×4?

[wintry steppe]

read my first message again

[grave garden]

Hmmmmm

[grave garden]

This is am right ?

[wintry steppe]

no

read more carefully

as a root of the minimal polynomial

in this case, it's 3

so the size of the largest jordan block is?

[grave garden]

3 ?

[wintry steppe]

yes

and your matrix is 4x4, so the only remaining jordan block could be what?

[grave garden]

0 ?

[wintry steppe]

i'll ask a more specific question

the size of the only remaining jordan block could be what?

you can tell from the minimal polynomial that it has to correspond to the eigenvalue 1, since all eigenvalues must be roots of the minimal polynomial. so the only problem left is determining how many and what size they are

but you already have a 3 x 3 block, and you need a 4 x 4 matrix, so the only possibility is...

[grave garden]

1×1 right ?

[wintry steppe]

yes

[grave garden]

And hmmm what's next ?

[wintry steppe]

you are done

you have the jordan form

[grave garden]

This ?

[wintry steppe]

yes

[grave garden]

I just randomly put it kwkw

Is this too ?

[wintry steppe]

do you know what a jordan block looks like?

the second picture is just a 4x4 block

the first one was right

[grave garden]

Like you have matrix in the matrix ?

[wintry steppe]

see the definition of "block matrix"

[grave garden]

Hiii I'm back

[grave garden]

Just understand it XD

Thanks mate !

[umbral spindle]

Anyone know how to start this ?

[dusky epoch]

@umbral spindle is this your first time doing things w/ linear transformations?

also do you still need help with this?

[serene solstice]

The space asked 6 is not a vector space over R because the two aforementioned objects are not in R, right?

And how do we do 2 and 5?

[zinc timber]

objects do not have to lie in lR, example lR is a vector space over Q. You argument is not right

for 2, if a=0 then done, otherwise F being a field can you multiply something with av=0 giving you v=0?

[torpid moat]

How do we find the norm of matrices?
If we define the operator norm on a matrix $A$ as $\norm{A}=max_{\norm{v}=1}\norm{Av}$, then what would the norm of, say, $\norm{\begin{pmatrix}0&0\0&1-e^{i\phi}\end{pmatrix}}$ be? or $\norm{\begin{pmatrix}1&0\0&e^{i\phi}\end{pmatrix}}$?

[stoic python]

thestonethatrolled

[zinc timber]

what norm are you using your normed space?

[torpid moat]

induced norm on a Hilbert space

[zinc timber]

there are other methods to calculate norm without using the def. for example if you are using l² norm then operator norm is the largest singular value

[zinc timber]

not what I meant

if it's l¹ norm instead then it's max absolute sum of the cols of the matrix

[torpid moat]

ohh

[zinc timber]

for ref https://www.sciencedirect.com/topics/engineering/matrix-norm

[toxic apex]

In this question, I could calculate the eigenvalues for the projection (0 or 1) but I can't prove the second part which is to show every projection has atleast one eigenvalue

[zinc timber]

hint: Look at Pv for some v

you need cases when P ≡ 0 and when it's not

@toxic apex

[toxic apex]

so if v is nonzero and Pv=0, then it is not true that Pv=1v?

[zinc timber]

remember Pv=0=0v

still an eigen vector with eigenvalue 0

[toxic apex]

got it thank you

[zinc timber]

sure np

[hollow finch]

for this problem, you could use that

$A
\begin{bmatrix}
1&1\1&3
\end{bmatrix}=
\begin{bmatrix}
0&1\1&1\1&0
\end{bmatrix}$

and solve for $A$

[stoic python]

nix

[umbral spindle]

This is my first one like this

[little crater]

do you still need help?

[umbral spindle]

@little crater just on how to show its one to one

[little crater]

Do you have your matrix reprsenstaition for T

if some T is one - to -one then a != b then T(a) != T(b)

so no 2 different input vectors, x, in R^2 map to the same R^3

given your matrix A

is

3x2 with 3 rows and 2 columns

as long as both of those column vectors have a pivot then you know you don't have a free variable and thus Ax=0 only has the trivial solution and then Ax must be one-to-one.

Essentially if all your vectors are pivot vectors => means the set of vectos of A are linearly indepedent and you have no way of of expressing the one column vector in the form of the others and thus cant have something where a!=b but T(a)=T(b)

[velvet moss]

Let V be the set of real numbers. Regard V as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finite-dimensional.

I know I need to show there exists no finite subset of independent vectors in v.

I’ve tried to prove it by induction but failed

[wet stratus]

If it is finite dimensional, then what is the cardinality?

[velvet moss]

the span of its basis

oh wait I think I get it

[little crater]

for this question

when they say "form a basis for R^m"

does that mean A has to be square mxm or just that we have at least m columns?

nvm if the columns form a basis then that means they must be linearly indepedent and span R^m.

im going with A must be a m x m square matrix

[wintry steppe]

you are correct

every basis of R^m has exactly m vectors, so you would need exactly m columns

[hollow finch]

equivalent to what @little crater said, if you can show that nullity(T)=0 then it is one-to-one. this can be done by showing A has nullity zero, or the zero null space.

[wintry steppe]

zero null space

[hollow finch]

righto. me dumb plum forgot

[wintry steppe]

happens to the best of us

[umbral spindle]

Dummy check:
V € R^2 means V has 2 rows ?

Using “€” as “element of” lol
R is real integers

[winter harbor]

Real integers

Anyways

V being an element of R^2 means that v = (x,y) for some real numbers x and y.

Or

equivalently

you could write it as a column vector with two components.

[umbral spindle]

Oh okay

So 2 by 1

[winter harbor]

Also, you probably meant "real numbers".

[umbral spindle]

I did! Didn’t even notice I did that lol

[hidden void]

Recommend me THE BEST book on linear algebra?

[wintry steppe]

every book has flaws and is written with different purposes in mind; there is no best

that said, the book by friedberg insel and spence is pretty good

it is consistent in its quality, contains many good exercises, and i honestly can't think of any glaring issues with it (as i could for some other recommendations on this subject...)

it starts out with vector spaces and linear maps, then moves on to applying it to solving systems of equations (the original purpose of linear algebra)

then a deeper look is taken at linear maps themselves, discussing determinants and normal forms (diagonal, spectral theorem in inner product spaces, jordan form)

lots of little asides here and there on other interesting and related material

very solid book

[hidden void]

Thanks! I’ll check it iut

[wintry steppe]

it doesn't really stand out in any particular way

it's all stuff you can find in other books, there really isn't anything unique to it

it just does things consistently and well

[wintry steppe]

you know p(A) = 0, try multiplying both sides of this equation by A^{-1}

[umbral spindle]

So I assume P(t) is 0 ?

[indigo vigil]

Could smb explain what is relationship bw S and V, pls?

In this example

[wintry steppe]

no, p(A) is p(t) with A plugged in. this is zero by the cayley hamilton theorem

[umbral spindle]

Oh okay!

[wintry steppe]

p(A) is A^2 + 3A + 2I. this is zero

[umbral spindle]

Then isolate 2I and multiple by A-1

[wintry steppe]

yeah that would work