linear-algebra

[stoic python]

moshill1

[brisk fractal]

well A specifically is just T, with domain Im(S)

which is why this proof works

T and S respectively bring the fundamental subspaces to the subspaces of TS

(which makes me wonder if there is a categorical way of defining matrix multiplication based on fundamental subspaces of compositions of linear maps)

[nocturne jewel]

wait how do you get dim(Im(TS)) <= dim(Im(S)) from this?

[brisk fractal]

null(T) term is positive, so rank(TS) <= rank(S)

[nocturne jewel]

yeah but A and T are the same, so shouldn't it be dim(Im(S)) = dim(ker(T)) + dim(Im(T))?

[brisk fractal]

where did you get that relation from?

[nocturne jewel]

dimension formula?

[brisk fractal]

okay so here's the problem

you're thinking of T and S still as their original linear maps

but when we consider their restrictions, the formulas change

$T \mid_{\Im(S)} \neq T$

[stoic python]

bacono

[brisk fractal]

and similarly, while $\dim V = \mathrm{null} T + \rank T$

[stoic python]

bacono

[brisk fractal]

$\rank (S) = \rank (TS) + \mathrm{null}(T \mid_{\Im(S)}$)

[stoic python]

bacono

[brisk fractal]

$T \mid_{\Im(S)} : \Im(S) \to \Im(TS)$

[stoic python]

bacono

[brisk fractal]

$T : V \to W$

[stoic python]

bacono

[brisk fractal]

does that make sense?

[nocturne jewel]

Im just gonna google for an answer, cause idk

[acoustic path]

I wanna use 2x2 matrices to show they dont commute but axler hasnt gone through matrices yet. is there another way to prove this?

[wintry steppe]

matrices represent linear maps with respect to some basis, so you could choose a basis of V (note finite-dimensionality is assumed), pick linear maps that look like the matrices you have in mind, and then show that those don't commute

[acoustic path]

yea but he hasnt gone through matrices tho

[wintry steppe]

right, but with this solution you're not explicitly working with matrices

[acoustic path]

so just pick 2 linearly independent vectors for the basis of V

[wintry steppe]

well V could have higher dimension

so be careful

[acoustic path]

true. but hes just asking that there exists a case

so even one example could be a proof

or do i gotta work with dim V=n

to prove for all cases

ah ok

ok i got it. ty

[fallow jolt]

Can someone explain the last line?

I think they skipped some intermediate steps, so I'm a bit confused how they got it

I_n is an nxn identity matrix

[wintry steppe]

what's r(matrix)?

[fallow jolt]

rank

@wintry steppe

Oh is because rank(I_n) = n, then rank(AB) = rank(AB) obviously

so its just added?

Ok, so then why is this true?

<@&286206848099549185>

[wintry steppe]

Hmm

[dusky epoch]

ac + bd = 0 and c^2+d^2=1, if you wanna go that route

@stoic jungle

or you could just note by observation that $-b\ket{0} + a \ket{1}$ works

[stoic python]

Ann

[dusky epoch]

that won't work unless |a| = |b|

the inner product of your state with |v> is a^2 - b^2 but you want it to be 0

you want your state to have your norm do you not

uh

unit* norm

autocorrect is drunk

[fallow jolt]

What elementary row/col operation could be done to do this?

I_n is an nxn identity matrix, A is an mxn, and B is an nxp

Obviously you can switch the two columns, but I'm not sure how to get rid of the negative

[dusky epoch]

multiply column by -1

[fallow jolt]

Ohp

I forgot you can multiply a column by a scalar

oops

thank you

LOL

As a followup, why does this work?

Where r is rank

[fallow jolt]

and these still apply

[dusky epoch]

so that big block matrix is m+n by n+p

hmm

trying to think of a way to show this without too much index fuckery...

col(A) has a basis consisting of r(A) columns, and ditto for B

the corresponding cols in the big block matrix ought to be LI

[fallow jolt]

What’s Ll?

[dusky epoch]

linearly independent.

[fallow jolt]

Hm I don’t think you can assume that

If it helps, the original big block matrix was [[I_n 0], [0 AB]]

So it’d only be linearly independent if AB is

Which we don’t know

@dusky epoch

[dusky epoch]

uh

mind showing the original problem

[fallow jolt]

I did part a

But I'm stuck on proving the first in equality of part b

I know that rank(AB) + n = rank([[I_n 0], [0 AB]]) = rank([[B I_n], [0 A]])

So I assume you now need to show rank([[I_n 0], [0 AB]]) >= rank(A) + rank(B)

but idk how

@dusky epoch

[dusky epoch]

huh

wait

is it even possible to transform [I, 0; 0, AB] into [B, I; 0, A]??

[fallow jolt]

Yes it is

[dusky epoch]

i don't see it

[fallow jolt]

I proved it already

[dusky epoch]

also i'm gonna be busy for the next hour or so

[fallow jolt]

Rip

Can <@&286206848099549185> help me out?

I guess I'm not sure why this is true

Where r is rank

[pallid rampart]

You can apply row reduction to A so that it has r(A) linearly independent rows right?

That's the definition of r(A)

Similarly you can apply row reduction on r(B)

Now given these two sets of row reductions, you can apply them on the matrix on the left hand side

This is not a complete solution but a hint to how you can prove this

@fallow jolt

[fallow jolt]

How are you able to reorder AABB into BAAB?

[native rampart]

Property of tr

tr(AB)=tr(BA)

tr((AAB)B)=tr(B(AAB))

[fallow jolt]

thanks!

[pallid pumice]

#linear-algebra message

Hey everyone, This is a question I asked yesterday and I clarified with the professor that its not a typo and that its supposed to be like that. So since there are only infinitely many solutions. How would I solve for h and k.

I have hx2 + 6x3 = k - 4 with x3 = t but I am confused on where to go after that

[tame mural]

Try to work with the matrix equation

[1 0; 0 0] [x, y] = [2, 3]

[tame mural]

Are there some [x, y] which can help me reach [2, 3] with the matrix I provided?

[pallid pumice]

x would be 2 but wouldnt there be no solution for y that results in 3

[tame mural]

Yeah

But you can also imagine there are some problems you can perfectly solve

For example, if you were trying to reach [2, 0]

Then you'd have no problems

[pallid pumice]

yeah i understand that

but im just confused that since theres a free variable couldnt h and k be anything?

[tame mural]

[1 0; 0 h] [x, y] = [3, k]

This is how I interpret your problem

If you want unique solution

Then you know the RREF of your matrix should be identity

So you just have to have a choice of h which is good

[pallid pumice]

there is no unique solution

its only infinitely many solutions

im just confused why. shouldnt there be a value for h or k where it doesnt work

[tame mural]

Have you ever found the basis for a nullspace?

It's a very similar problem

[pallid pumice]

nope

week 2 of class lol

very early on

[tame mural]

IMO, if you watch a video on calculating the basis for a nullspace, you will probably have your doubts fixed for dealing with dependent and independent variables

[pallid pumice]

cool. ill look into some videos. thanks

[wintry steppe]

the book says 23 is not in reduced row echelon form but isnt it?

[reef sleet]

it isnt

that's just row echelon

reduced row echelon is when every nonzero entry has zeroes in all rows above and below it (I don't think that's formally correct but I think it's simple enough to understand)

[wintry steppe]

Oh ty

[reef sleet]

oh wait

yeah that's still not rref

[little citrus]

Can anyone give me pointers for part 2

Like I am confused why they have given numbers to find eigenvalues

can soneone please explain

anyone?

[candid cosmos]

im assuming some of the eigenvalues have roots of 2 or 8 in them which you need an actual value for

@little citrus

[west shoal]

I have a question

It says to see if a b is a linear combination of a1, a2, and a3

Though I'm confused how b is a linear combination

I ended up with x^3 as the free variable, and x1 = 2 - 5^3, x^2 = 3 - 4x^3

[candid cosmos]

i dont think you should be getting any powers of x in your solutions, do you mean subscripts by "^"?

[west shoal]

the x's are the c's

I just changed them to that

[candid cosmos]

x1,x2,x3

[west shoal]

they are the constants

[candid cosmos]

is what you mean by x^1,x^2,x^3?

[west shoal]

yes

I meant x_1, x_2, and x_3

[stoic python]

Profound

[candid cosmos]

so you can write down an augumented matrix for this problem, did you do that?

[west shoal]

yes and that's what i ended up with

i just converted the matrix to a system of equations

after i made it reduced echelon form

[candid cosmos]

if that system has a solution then it is a linear combonation of them, since it is possible to make the vector b by adding up scaled versions of vectors a1,a2,a3

you said you had a free variable that gave a solution right?

[west shoal]

yeah the free variable was x_3

[candid cosmos]

so take any solution you want after picking the free variable(probably choose x_3 = 1) and double check in the original equation whether that vector sum is equal to b, if it is then you can say b is a linear combonation of the 3 a vectors

[west shoal]

wait a free variable is equal to 1?

[candid cosmos]

you can choose it to be anything

thats why its called free

[west shoal]

really i never knew that

[gray dust]

usually we solve a system to find how b is a linear combo of the a_i's, but it can be done by inspection here

[candid cosmos]

so if the augemented matrix has a solution or infinite solutions(free variable) you can say that the vector is a linear combination

[west shoal]

yeah i looked at it and just did the math across the vectors

[gray dust]

note b=a3-a2-3a1

[west shoal]

i see

[gray dust]

so indeed b is a linear combo of the a_i's

[candid cosmos]

because there is a way to add those three vectors up to get b

[west shoal]

thanks

[candid cosmos]

no problem

[rose umbra]

is this true?

[limber sierra]

this doesnt follow

i'm assuming your operation is associative but not commutative

you cant just juxtapose the B_2 and the I arbitrarily

you are able to go from (B_1A)B_2 to IB_2

and hence B_2

[rose umbra]

Why not? A*B_2 = I is given

[limber sierra]

right

so $B_1AB_2 = B_1(AB_2) = B_1 I$

[stoic python]

Namington

[limber sierra]

how do you get from that to $B_2 I$?

[stoic python]

Namington

[limber sierra]

wait

hold on

maybe im misreading your handwriting

is that supposed to be a B_2?

then the proof is invalid from the first step

oh wait

never mind

sorry i was misreading your handwriting entirely

thought the 2s were 1s

my bad!

in that case the argument seems fine

apologies

[rose umbra]

ah lol alright sry for the handwrite was writing with mouse

thanks anyway

[tame mural]

I heard that a linear form / functional V × F → F either maps to 0 or is surjective. Is that true? Or only in finite-dimensional case?

[wintry steppe]

hint: any field is a one-dimensional vector space over itself. what can you say about the rank of a linear map going into a field, then?

[tame mural]

it must be dimension 1

[wintry steppe]

that's not the only case

what are the possible dimensions of a subspace of a 1-dimensional vector space?

[tame mural]

ah { 0 } is 0 dimensional?

[wintry steppe]

yeah

[tame mural]

-_-a

but u do have a basis

[wintry steppe]

the basis of {0} is defined to have zero elements (it's the empty set)

we basically do that so we can say dim{0} = 0 lol

[tame mural]

I see @_@

lol thx

[wintry steppe]

and from that you have either 0 (the rank is zero) or surjective (the rank is 1)

the only subspaces of a field are the trivial one and the whole thing, after all

[wintry steppe]

Span(thing) = smallest subspace containing (thing)

[shy mango]

not sure exactly where this goes, but we're talking about univ properties of VS in linear algebra

supposedly the curvy arrow means inclusion map

can someone explain this in english (instead of weird math diagrams)

[shy mango]

?

LOL what

uhhh I guess my question is, what's the point of the inclusion map?

aren't T and f very similar maps?

[limber sierra]

what's B here?

a basis?

a subspace of V?

[shy mango]

ah sorry about that

so

B is a basis of V

T is a map,

V is subspace

[limber sierra]

okay so the idea is that rather than regarding f as a map from one space's basis to another space, we can construct a map T: V -> W that "acts like" f on B

the notion of an inclusion map captures this

also the map T is unique for a given f

thats important

bleh

got it backwards

fixed

[shy mango]

ah, what exactly is the point of that? like we can make some map T:V->W that acts on the basis vectors, which is a finite set

[limber sierra]

f can often be hard to explicitly describe if we only know what it does to a basis

this is saying we can describe f in terms of a linear transformation

which we call T

and there are many ways to describe linear transformations

e.g. matrices

[shy mango]

gotcha, why do we use the basis as the set of vectors?

[limber sierra]

also this "description" of f is unique

that is

there's only one linear map T that "captures" f's behaviour on a given basis

in other words: to figure out what a map does V -> W

it suffices to look at what it does to the basis of V

since each function from a basis of V to W has a unique corresponding linear transformation T

so if you know what T does to a basis of V

since it's unique

you know what T does to all of V

[limber sierra]

not sure exactly what you mean by this

this is kinda like, the whole point of a basis?

like

the ability to describe linear transformations just by looking at a basis

[shy mango]

okay right

[limber sierra]

in many ways you can view a basis as the "least possible amount of information" you'll ever need about a vector space

(ignoring additional structure like inner products or w/e)

and this should make sense: we know a basis can be used to recover the space itself

by taking lin combinations

[shy mango]

right

[limber sierra]

and it's "least" since it's the smallest such set

this is just a way of saaying "this basis-oriented perspective still works for maps from subspaces to the whole space"

[shy mango]

I think you said f is the map from one space's basis to another, can you clarify that a bit more

[limber sierra]

in other words, we can also study subspaces from their basis

like okay

lets say W = R^3

and we're considering the subspace of vectors of the form $\begin{pmatrix}a\b\0\end{pmatrix}$ for $a, b \in \bR$

idk where texit is 😦

vectors of the form (a b 0) for a, b real numbers

now lets say you have a function that you know maps (2 2 0) to (1 1 1), and (2 1 0) to (0 0 1)

as it turns out, {(2 2 0), (2 1 0)} is a basis of the subspace of vectors (a b 0)

so from that small piece of information

you know there can only be one linear function

that maps (2 2 0) to (1 1 1), and (2 1 0) to (0 0 1)

from this subspace to the whole space R^3

and you can actually work out what this linear function must be by solving a system of equations

now its rare that youll have situations "naturally" where you're just randomly given 2 vectors

but there ARE situations where you cant quite figure out the ENTIRE function

but you can plug in some test values

and kidna figure out whats going on

so if you plug in a "basis" to test

you can recover the linear function from that!

(this is the idea behind linear interpolation in statistics)

(and, well, most of machine learning tbh)

[shy mango]

intriguing

uh

[shy mango]

what exactly does this mean

[limber sierra]

like

if S = the set of vectors of the form (a b 0)

for a, b real numbers

then that's a map S -> R^3

and we can study that map entirely by looking at what it does to the basis

in this case, to (2 2 0) and (2 1 0)

but any basis would work

again this is just a way of formalizing that "studying subspaces is really the same thing as just studying their bases"

this might seem kinda "well, duh" for finite dimensional spaces

but its much less obvious for infinite dimensional ones

[shy mango]

so f is the (2,1,0)-> (1,1,0) and the other relation that you wrote
and T would be the actual linear equations you can use to get these things?

[limber sierra]

uh not quite

f(2, 1, 0) = (0, 0, 1) and f(2, 2, 0) = (1, 1, 1)

and i claim that just knowing this is enough

to construct a unique function T: S -> R^3

[whyd you delete the message, it was good 😦 ]

[shy mango]

^^

i was trying to read it

[lost dragon]

@shy mango

Here is what it's saying in english.
Our intuition stems from the categorical notion of naturality.

Suppose you have some vector spaces V, W.
Now in the categorical intuition, you know nothing about what the vector spaces are--
know nothing of what elements they contain.
Merely you know that they are vector spaces (structurally).
For example, in this way you know V and W must have identities,
and since their must exist a unique such element, you can call the identity a natural vector.
||(naturality arises as witnesses to existentially unique statements about abritrary objects of the category.
This is an intensional view of category theory.)||

Now supposing you are given B, the set of basis vectors of V.
Obviously B is just seen as a set. Not as a vector space.
The inclusion map inc from B to V, just assigns each element in B to iteslf.
This inclusion map is itself natural. After all you need no knowledge of any of the elements of B to specify inc.

Supposing you are also given a map f from B to W.
This is purely a set theoretic map. It is not a linear map.

Now don't you agree there must exist a unique linear map T from V to W, st. the diagram commutes?
After all a linear map is defined by the transformations of the basis elements.
To know how a linear map evaluates the basis, is to know how it evaluates any vector.

Therefore T could be said to be natural .

[shy mango]

tyty lemme decipher it

[shy mango]

S contains (2,1,0) and (2,2,0) right

[limber sierra]

right

[shy mango]

wait

[limber sierra]

S in my example is all vectors of the form (a, b, 0)

B = {(2, 1, 0), (2, 2, 0)} is a basis for S

(if you dont believe me, verify it yourself)

[shy mango]

i believe you lol

ok

[limber sierra]

so we can write any vector in S as a linear combination of (2, 1, 0) and (2, 2, 0)

[shy mango]

you claim that it's enough to construct that function such that T(2,1,0)=(0,0,1) and the other one right

[limber sierra]

i'm saying if we have a function f such that f(2, 1, 0) = (0, 0, 1) and f(2, 2, 0) = (1, 1, 1)

then there is a unique linear function T from S to R^3

that "agrees with" f

in other words, if we have a linear map T: S -> R^3

and we know T(2, 1, 0) = (0, 0, 1) and T(2, 2, 0) = (1, 1, 1)

we know everything about the map

we can determine what it does to any element of S

[shy mango]

oh right, and the important criteria is that we know f(v) for every v in the basis right?

[limber sierra]

right

[shy mango]

oh okay that clarifies what the point of that is hm
finally, the inclusion map is just to make the basis vector be treated as an element of V instead of B right?

[limber sierra]

basically yeah

since T needs to be a map S -> R^3

whereas f is a map B -> R^3

when we say a "diagram commutes", we mean if you take any element from the "start"

and follow a path of arrows

your element will end up "the same" at the "end" of the arrows

so if v is an element of your basis B

then f(v) and T(inc(v)) will be the same

but ofc T(inv(v)) is just T(v)

except now v is considered an element of V instead of B.

[shy mango]

right

okay that's pretty interesting, thanks!

had no idea what was going on from the notes lol

[shy mango]

right this makes sense too

thanks for typing all that out

[lost dragon]

@shy mango no problem. sorry for interuptting.

[restive wraith]

I have no clue how to go about this problem, does anyone have any ideas?

[wintry steppe]

well what's "the linear combination idea of ex 414 and 415?"

to write v as a linear combination of the eigenvectors of A?

you should probably try finding those

[steady fiber]

eigenbasis time

[wintry steppe]

that's always how you solve the "find this absurdly high power of matrix" problems

[steady fiber]

there's also the write A as PDP^{-1} where D is a diagonal matrix way of working with "absurdly high power of matrix" questions

which is just eigen stuff again

[wintry steppe]

same thing

[steady fiber]

but a slightly different process

[faint lintel]

Suppose we have a vector space V and a subspace of V called W. for any vector v in V, is v + W = {v + w | w in W} a subspace of V?

my answer is no because suppose v is not in W

then -v isn't in W

[restive wraith]

ive looked at those i fail to see how they are relevant

[faint lintel]

so then v + W doesn't have a zero element

but something feels off

[restive wraith]

this is whats on 414

[frosty vapor]

what if v is in W already

[steady fiber]

just let the subspace W be the subspace with only 0 in it, then it's obviously untrue (assuming v != 0)

[faint lintel]

oh

[faint lintel]

then v + W is W which is a subspace already

[frosty vapor]

alright

yeah that makes sense

[faint lintel]

well v != 0 if v isn't in W

cause 0 is in W

so if v isn't in W then v can't be 0

[steady fiber]

but ya, it's not true in general

[faint lintel]

ok dope ty

[wintry steppe]

that doesn't look useful. go with what i said, express v in terms of eigenvectors of A and compute

[steady fiber]

imagine if someone wrote like a script that generated latex code for doing the 2021 power by actual multiplications

and they hand in a many thousand page long document

actually, it isn't that bad to do

would "only" take like 11 matrix multiplications to do manually

[quartz compass]

well if you're gonna write a script to do that, might as well make it do one at a time for the memes lol

[pure tangle]

will rref and ref result in the same solutions?

[limber sierra]

if done properly, yes

row operations dont affect solution sets

[pure tangle]

thanks so much just wanted to double check

[pseudo thicket]

Is this how you do it>

Given vector a:

array([  9.97566234, 130.67172705,  16.45635726])

x1 = np.array([0.846,1.324,1.150,3.037,3.984])

x2 = np.array([1,2,3,4,5])



yplane = a[0] + x1.dot(a[1]) + x2.dot(a[2]
)

The result of yplane:

[136.98030068 215.89774347 209.61722022 472.65112643 612.8536092 ]

[hybrid elk]

there's an explanation at the bottom of what's being done, but when i follow that I don't end up with the answer. Is there something else I need to do to get to the correct answer?

[dusky epoch]

what answer do you get when you follow the explanation?

surely not the zero matrix as written there

[hybrid elk]

oh i havent attempted this one, i did the example right before and messed up so i just clicked to see the answer on this and see what i should be doing

[dusky epoch]

...

[hybrid elk]

but then i just followed the explanation but i didnt get the right answer

[dusky epoch]

can i see the problem you attempted, its explanation, and what you get when you follow that

[hybrid elk]

ok im gonna need a minute to try another one because i cant pull up a previous problem

ill send it in a bit when i try

[dusky epoch]

ok

[hybrid elk]

oh my gosh im so stupid nvm

i didnt notice the last line of the explanation that you apply it to the identity matrix

thats why i was so confused

i got it thanks for reaching out tho

[main kite]

I’m in my second week of Lin algebra (proof heavy) & at struggling with the math syntax. Anyone recommend any good resources or tips on how to show proofs?

[dusky epoch]

how to prove it, by velleman.

[main kite]

Thank you!

[west shoal]

can somebody explain what the question is asking?

[marble lance]

Do you understand what the span is? @west shoal

[west shoal]

yeah, the span is the set of linear combinations of v_1 and v_2

[marble lance]

Okay, so you need to list five elements of the span, i.e. five linear combinations @west shoal

A linear combination looks like a1v1 + a2v2. By the weights, they mean the scalars a1 and a2

So they want you to show it as a linear combination and then simplify it to show its coordinates

[west shoal]

i see

[west shoal]

so i can use anything for the weights

like 5v_1 + 2v_2

it doesn't matter?

and then i just do that again 5 times with different scalar values?

[marble lance]

@west shoal yep

[west shoal]

Alright thanks

[west shoal]

I don't think a is asking if b is a linear combination of the span (a_1, a_2, a_3) because b is a linear combination

i'm not sure what to do

[wintry steppe]

lmao that's a troll question

you're right, it's not asking if b is in the span

it's asking if it's in the set of columns

[steady fiber]

"how many vectors are in {a_1, a_2, a_3}"

thonk

[west shoal]

oh obv not lmao

i feel dumb 🤣

thanks

[dusky epoch]

nice question

[pallid rampart]

This is a very difficult question

You have to prove 1 is not 0 and 0 is not 3 etc

Which requires advanced techniques from peano axioms which only a handful of people on this world knows

[reef sleet]

is my professor's key wrong or am I calculating the determinant wrong??

I keep getting -14

he got 14??

[limber sierra]

-14 is correct.

presumably it's a typo

[reef sleet]

pain 😔

so every time I see LA I see determinant, I don't even really know how to calculate it (just visually when finding cross product) or what it is

what does the determinant say about a system?

[limber sierra]

okay this is kinda a rabbit hole

the very informal explanation i like to give is that the determinant is a way of describing "multiplicative information" about a matrix

what do i mean by this? well, we know det(AB) = det(A)det(B)

so the determinant behaves "nicely" with matrix multiplication

but from this behaviour we can extrapolate some things

for example (here i'm gonna assume we're working in a real vector space)

the determinant will always be a real number, right?

which is obviosuly "easier" to work with than matrices in many ways

one of these is that we know that almost all real numbers are invertible

that is to say, it's possible to "divide" by almost ANY real number

the ONLY exception is 0

and so from the formula det(AB) = det(A)det(B) we can extrapolate that

if a matrix's determinant is 0, then it is NOT possible to "divide by" it

that is to say, it doesnt have an inverse matrix

whereas if it is 0, then it IS possible to "divide by" it

so it WILL have an inverse matrix

hence a matrix has determinant 0 iff it is noninvertible

this is just one immediate connection, there's a whole bunch of similar connections

but they all stem from this idea of the determinant "encoding multiplicative information"

now, as for how the determinant does this? the honest answer is that it's a bit of a pain

the whole reason we use the determinant instead of looking at matrices themselves is because matrix multiplication is a bit gnarly

its well-behaved but its hard to "look at" a matrix and figure out what it'll do

when you multiply by it

without breaking out the pen and paper and doing it yourself

it stands to reason, then, that the determinant ought to also be kinda "weird" to compute

since it's hard to tell how a matrix will behave multiplicatively, and the determinant tells us how a matrix behaves multiplicatively

we are lucky in that there's an explicit formula for the determinant

but the formula might seem pulled "out of the aether"

you can prove that it has all the desired properties, if you want

in fact, the determinant is the ONLY multilinear operator with the desired properties

ie det(identity matrix) = 1 and det(AB) = det(A)det(B)

this can also be proven

but im not sure the proof gives great intuition for the computational side

the honest answer is that i wouldnt think too hard about "why" the determinant is the way it is

in terms of how you compute it

just know that its something you can compute to capture the multiplicative behaviour of a given matrix

as for what it says about a SYSTEM, well

applying matrix multiplication is just substituting in systems to each other

[tame mural]

it feels like a very elegant metaphor is just within reach for the determinant's properties

[limber sierra]

so... yeah

theres your connection

it says how systems "interact"

[native rampart]

Are you sure about that?

That det is the only multilinear map which takes rows of a matrix as inputs such that det(AB)=det(A)det(B)

[quartz compass]

there are more properties than that

[native rampart]

nvm,So Nami meant unique alternating multilinear map?

[limber sierra]

i mean what i said

oh

never mind

i didnt say alternating

i dont mean what i said

i mean what you said

but you can phraes it alternatively

[native rampart]

[limber sierra]

since thats actually a stronger condition than necessary

"unique nontrivial multilinear map with det(I) = 1 and det(AB) = det(A)det(B)" suffices

where here "nontrivial" means "image is all of R"

(otherwise you could just let it be x |-> 1, or x |-> +-1)

it may be a good exercise to try proving this

but its hard

[willow patio]

hello i'm a wannabe economist hoping to understand linear algebra

is this a good place to ask questions? a sample is:

why might we be interested in finding a new basis given a previously determined spanning set of linearly independent vectors? why might we be interested in using the gram-schmidt process? (i think these are the same question, just asked in different way)

[steady fiber]

because problems can often be much cleaner to deal with in different bases

there's sometimes special bases like an eigenbasis which are incredibly useful for a wide range of problems

and yes, this is a place to ask questions on linear algebra

[willow patio]

i guess a very simple example would be transferring from a basis of orthogonal vectors that are linearly independent in R^3 to your standard i, j, k? but i imagine i'd be going the other way mostly

thank you, probably will be mostly clear as i progress

but that's a good answer for me for now

[steady fiber]

just as a somewhat tangential problem, we could theoretically do all high school physics by only thinking in the 3D space we live in

but most high school physics problems simply can be framed as a 1D problem

along one axis

what if you come across a problem where you don't have such a clean one axis to work along

just rotate yourself around and work with a new basis

so that your problem is trivial in 2 of the dimensions, and so it's reduced to a 1D problem

[willow patio]

ok got it

is that why this immediately follows the section on colspace/rowspace/nullspace? because my understanding of the uses of that are that, as economists we use matrices to both store data and store systems of equations, depending on the problem and model,

but maybe i'm getting ahead of myself

but the use that seems most apparent from my limited understanding is explaining how many free and how many constrained variables you have in a system of equations

i should probably read ahead and get back to you lol, else i'll probably have you explaining all of linear algebra to me

[tame mural]

I would say Linear Algebra is the lingua franca of technical multi-disciplinary communication

It has many applications, of course, but if you're trying to get into grad school for econ, you'll be expected to take up to at least Analysis to be competitive

Many algorithms or procedures you learn will be pedagogical

Or seemingly inwardly focused

[reef sleet]

@limber sierra
What does it mean for a matrix to be invertible? For a number to be invertible, does that mean you can divide (why is it in quotes) by any other real number? Or, for negative numbers, add a real number to it? What does it mean for a matrix to be invertible? How do you divide by a matrix?? Or multiply by one?!
Is multilinearity similar to bilinearity?

if that was too much just ignore it LOL I'm sure I'll learn it all eventually

[limber sierra]

oh you arent familiar with "invertible"?

[reef sleet]

No sir

[limber sierra]

weird that you'd be introduced to determinants before invertibility

my bad then

[tame mural]

A matrix is invertible if you can undo the transformation.

[willow patio]

yeah meow, i wish i had taken it in college, i majored in econ/international relations at a compeetitive school and did very well in econometrics but never learned linear algebra/real analysis in a classroom setting

[limber sierra]

a matrix $A$ is invertible iff there exists a matrix $B$ with $AB = BA = I$

[stoic python]

Namington

[limber sierra]

where I denotes the identity matrix

[reef sleet]

I only know about it/how to calculate it (and only in a 3x3 matrix) because of cross product shenanigans

[limber sierra]

(ie the matrix with 1s on the diagonal, 0s everywhere else)

we call B the "inverse" of A

[willow patio]

you should check out my textbook, it has an entire section of conditions equivalent to saying a matrix is invertible

thank you meow and poros

[reef sleet]

So a matrix is invertible if A * B = B * A? So matrix multiplication is not always commutative (that's what it means if we can switch the terms around and get the same result right?)?

[limber sierra]

AB = BA = I

thats the important bit

[reef sleet]

Oops yeah was about to edit

[limber sierra]

the fact that we have a matrix B that makes AB equal to the identity

matrix multiplication is not always commutative, but if AB = I, then BA = I as well

assuming A is square

[reef sleet]

Interesting

Square as in size nxn?

[limber sierra]

yes.

"invertible" doesnt make sense for nonsquare matrices

[reef sleet]

Because you can't have an identity matrix then

Yes?

[limber sierra]

well

[willow patio]

and it is different than the transpose (which sounds obvious, but if you are new, it is a very important distinction)

[limber sierra]

if A is not square, then AB and BA will necessarily produce different matrices

[reef sleet]

I am new 😛

Wow how does matrix multiplication work?

This is the coolest thing ever LOL

[tame mural]

3B1B may be a good place to get intuition

[reef sleet]

Yesss I will begin watching his LA series once I get deeper into it

[tame mural]

He is good almost right away as you begin your textbook

[willow patio]

i am reading Differential Equations and Linear Algebra, and the first chapters are just on matrices or vectors, and it is a relatively cheap/free book depending on where you get it

[reef sleet]

Oh, cool, okay

[willow patio]

i should check out that

[reef sleet]

I haven't started DE yet

[willow patio]

yeah, if you do ch2 (matrices) ch3 (determinants) ch4 (vector spaces) that is all about matrices and their invertability and their properties

thats just the book my school used, where i got a syllabus from to teach myself, but there are others too

by goode and annin

[tame mural]

Any textbook will address these topics because they are very fundamental

[willow patio]

yeah

hence why i, who cares about economics, am very interested in understanding them

now i just need to shell out an insane amount of money to get credit so i can actually be competitive in phd applications...

[tame mural]

Also I think an injective linear function is left-invertible

And a surjective linear function is right-invertible

[willow patio]

that, uh, sounds terrifying

[wintry steppe]

it's literally just the definition of injective and surjective functions

[limber sierra]

no it isnt

[wintry steppe]

[limber sierra]

tterra epic does not imply surjective

look at the category of rings

[wintry steppe]

[limber sierra]

er

oops

surjective does not imply epic*

[willow patio]

stop you're scaring me wtf

[wintry steppe]

[tame mural]

And I believe in a finite-dimensional case for V → V, injective = surjective = bijective

[limber sierra]

wait no

i was right the first time

wtf am i drunk

sorry

[wintry steppe]

when the domain and codomain have the same finite dimension

[limber sierra]

epic does not imply surjective and the category of rings is a counterexample

im just a dumbass

my bad

[tame mural]

oh thx

[wintry steppe]

it's okay nami

[limber sierra]

every surjection is an epimorphism however

as long as your category has a separating terminal object

[wintry steppe]

[limber sierra]

but tterra my point is that this isnt the definition

😠

[wintry steppe]

[limber sierra]

since it falls apart if you try and do it outside an LA context

[wintry steppe]

it's fine if we're doing LA though

this is a vacuum

no mathematics other than LA exists in the #linear-algebra channel

[reef sleet]

I've had to use calc for a LA problem once ;^)

[limber sierra]

but i mean

tterra

if youre gonna tell people that left and right invertible are equivalent to inj/surj

theyll start thinking that injective and surjective are the same thing for maps R^n -> R^n

which they are if your function is linear

but many LA courses ask questions like

[wintry steppe]

i understand your point and you are correct

but

[tame mural]

ah yes, but we specified that context

[wintry steppe]

im going to pointlessly defend myself anyways

[limber sierra]

"Determine if the following transformations are (i) linear (ii) injective (iii) surjective"

or whatever

[willow patio]

i have a question on notation

in the context of least squares

[lavish jewel]

what are these X, beta, and y?

[willow patio]

X is the nxk matrix of independent observations, beta is the kx1 vector of the fitted coefficients, and y is the nx1 vector of observed outcomes

that's how they do it in econometrics

[lavish jewel]

seems about right

well

beta would be x_0

yeah

[willow patio]

yeah, thats what im thinking

i just wanted to check because i was confused

[lavish jewel]

idk why i read x before, but yes, that'S right

[willow patio]

,tex what would $ \vec{x} $ be? is that just a placeholder vector for undetermined values of$ \vec{x_{0}} $?

[stoic python]

Paname

[lavish jewel]

how much linear algebra do you know?

[willow patio]

basic matrices, vectors, inner product spaces, change of bases, determinants, and whatever econometrics means

[lavish jewel]

aight

well, in general your matrix A will not be invertible

this means it has a null space

[willow patio]

yeah because it's n x k, so you have to multiply it by transpose to make it a square?

[willow patio]

that's helpful i didnt think of it like this

[lavish jewel]

yep, but the result of doing this is a projection onto the row space

there is a part of the vector x that is forever lost to the null space

x = x_0 + x_null

the null part, you can't get back

the x_0 part is what least squares gives you

[willow patio]

got it, that makes everything make much more sense

thank you

[lavish jewel]

aight, no prob

[grand imp]

oh wait I should share the full context

So I'm trying to prove part c. here

the answer to b. is that the change of basis matrix $\tilde{C}$ is given by replacing all the $c_{i, j}$s with blocks $\begin{bmatrix}\Re(c_{i, j}) & -\Im(c_{i, j})\ \Im(c_{i, j}) & \Re(c_{i, j})\end{bmatrix}$

[stoic python]

~S^1

[grand imp]

however, it's unclear to me how to show generally that $\det \tilde{C} = |\det C|^2$, even though this was definitely true for $n=1$ and $n=2$ (n=2 was already quite tedious and didn't feel enlightening)

[stoic python]

~S^1

[grand imp]

it seems like maybe I should try induction? but idk how I would set that up

[jaunty sage]

can anybody tell me what exactly i am to do use in the b part? gaussian or gauss jordan elimination( i have already calculated the inverse using gauss jordan and rank by making it in ref)

[nocturne jewel]

Notice the linear system is equivalent to Ax = [-2,1,6]^T

@jaunty sage

[jaunty sage]

yes

i get that

so am I to use Gaussian elimination or gauss Jordan for this questions @nocturne jewel

those are the only 2 methods I am allowed to use btw

[nocturne jewel]

What's $A^{-1}A$?

[stoic python]

moshill1

[jaunty sage]

identity matrix

I

[nocturne jewel]

yes

$Ax = b \to A^{-1}Ax = A^{-1}b$

[stoic python]

moshill1

[jaunty sage]

ah I see

so the answer would be augmented matrix of a inverse with b(-2,1,6)?

@nocturne jewel

[nocturne jewel]

yes, $x=A^{-1}b$

[stoic python]

moshill1

[jaunty sage]

ah so I need to solve using that

and then I will get the answers for the variables yes?

[nocturne jewel]

yes

[jaunty sage]

thanks a lot mate

one more final thing

is ir better to use Gaussian or gauss Jordan here?

@nocturne jewel

[nocturne jewel]

neither, you're suppose to compute A^-1 b

"using part a..."

[jaunty sage]

Ahhhh

i will try it

now

[nocturne jewel]

(technically you can use gauss jordan, but it's quicker to use inverse matrix)

[jaunty sage]

oh okay I already have it so after I augment it

or do I do matrix multiplication of a inverse and b?

@nocturne jewel that is what you mean yes?

[nocturne jewel]

matrix multiply inverse by b

[jaunty sage]

OH OKAY well that clarifies it

completely

[nocturne jewel]

yeah you get a column vector = the x column vector

so the entries of A inverse * b = x entries

[jaunty sage]

Bro you are a legend thanks a lot

[hoary osprey]

Erin drip

[reef sleet]

So the solution obviously isn't 1/2 * distance between AB and whatnot

I have ot use vectors to solve this somehow but I'm not sure how?

Dot product could give me the angle between vectors but I don't think that tells me anything about the area

I don't know what cross product owuld do here

(like its purpose)

[wintry steppe]

so does the n in R^n denote the number of entries in a vector?

yes

thanks

i don't understand point b

does point b imply that there is a vector v_k that is a linear combination of the remaining vectors in S?

[limber sierra]

not necessarily

"subset" doesnt mean "proper subset"

[wintry steppe]

how can a subset of S still be a basis for H

[limber sierra]

its possible the "subset" is S itself

[wintry steppe]

oh

i see

[limber sierra]

think of "subset" like a ≤

rather than like a <

[wintry steppe]

ok

if they meant proper subset, would statement b be false?

oh wait

i get it now

[limber sierra]

it would be false, yes

in that case

the reason we need a subset is that S might not be linearly independent

thats why we need to consider a subset of S

that way, if its NOT linearly independent, we can "delete vectors" from it until it is

and the resulting set will be a subset

[nocturne jewel]

do you know the trig formula for area of a triangle?

[limber sierra]

but if it is linear independent, then it's already a basis

so we're good

no modification needed

[reef sleet]

Uhhh no sir/ma'am, but someone gave me a hint and I think I figured out how to solve it?

apparently if you extend dimensions to 3 (with z = 0) you can use cross product to make a parallelogram then take half its area

[nocturne jewel]

$A = \frac{1}{2}||\vec{a} \cross \vec{b}||$

[stoic python]

moshill1

[nocturne jewel]

yeah, and that also leads to the trig area formula

since magnitude of a x b = |a||b|sin(theta)

[tame mural]

oh huh?

[reef sleet]

that 1/2 a cross b is the trig formula?

[tame mural]

that looks so similar to the outer product

[reef sleet]

what trigonometry is there?

OH

Wait a second

I see it now

wow interesting

[nocturne jewel]

$||\vec{a} \cross \vec{b}|| = ||\vec{a}|| \cdot ||\vec{b}|| \sin{\theta}$

[stoic python]

moshill1

[digital bough]

Hmm, neat

[jaunty sage]

can anybody explain as to what i do

im confused

[limber sierra]

are you unfamiliar with the definition of linear independence?

if so, review that.

[jaunty sage]

i am

but

my concern

is

[jaunty sage]

basically a=b=c=d=0 yes?

i dont know how to solve it in terms of matrix

i dont understand free variables

and how to know which variable to use as the free variable

[sleek spruce]

so

i don't understand why identity matrix is unit matrix

Let T be a transformation

T(I) only yields two coordinates

how is the identity matrix a unit matrix if identity matrix only has two coordinates while a unit matrix

has 4 coordinates

(0,0) (0,1) (1,1) (1,0)

[lavish jewel]

@sleek spruce the number of entries in the matrix has nothing to do with it

and anyway, the identity matrix also has 4 entries

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

[stoic python]

Edd

[hollow finch]

Can all Hermitian matrices be written as S+iQ where S and Q are real symmetric and skew symmetric matrices respectively?

[dusky epoch]

yes? just take the real and imaginary parts of your matrix

[tame mural]

w h a t i s t h e m a t r i x ?

[dusky epoch]

w h y a r e y o u w r i t i n g w i t h s p a c e s b e t w e e n e a c h l e t t e r l i k e t h i s ?

[hollow finch]

Just checking. It seems obvious but I've been wrong about things I thought were cut and dry before

[zealous junco]

i want to revise the main thms of lin alg

but hoffman kunze book is 2 dense for exam revision, any recommendation for like the main compiled thms

it's not lin alg exam, but an exam with some lin alg in it

cuz i forgot pretty much everything

[sleek spruce]

anybody know where to understand linear algebra

cause the book im reading is confusing me

all the weird notations, fancy languages, and phrasing is totally confusing me

[marble lance]

If the notation is standard and you just don't like it, you'll just have to get used to it

What book are you reading?

[sleek spruce]

Linear Algebra Fifth Edition
David C. Lay, Steven R. Lay, and Judi J. McDonald

it's so out of place that even doing simple things like dot product

confuses me

because the book jumps randomly

and introduces concepts then say we're going into them later

might get another book

since this is digital

[digital bough]

Sounds like the book is speedrunning

[zealous junco]

speaking of speedrun anyone have a good source to speedrun LA review

[west shoal]

what does it mean to say R^4?

Does that state that the vector has 4 dimensions?

[native rampart]

It means R x R x R x R

[dusky epoch]

R^4 is 4-dimensional euclidean space yes

[west shoal]

Thanks and one more thing

If I say b spans {a1, a2} that means that I'm stating that b is a linear combo of a1 and a2 correct?

[marble lance]

No

[west shoal]

Oh...

[marble lance]

You should b is in span{a1, a2}

[west shoal]

Oh I see

[marble lance]

You say a set spans a vector space if the span of the set equals the whole space

[zealous junco]

for LU decomposition, is the fastest way to do it to record all permutation and lower trangulars..

during gaussian

if dong by hand

[orchid harbor]

Does anyone knows a very good book to learn linear algebra from zero?

I mean with Calculus 1,2 background

[pseudo cobalt]

@orchid harbor perhaps #resources could help?

also

for $\hat{i}$ and $\hat{j}$, should I replace the point with a hat, our put the hat on top of the point? Because I've seen it done both ways.

[stoic python]

Ninja Tuna

[tame mural]

I'd pick any famous textbook that has gathered plenty of discussion. That way it's always easy to Google for many answers.

Linear Algebra Done Right is one possibility

[native rampart]

Insel is good,I think

[orchid harbor]

hmmm

Like i need conpletely from zero i only know what are vectors and matrix

like very simple things

gaussian el etc

[tame mural]

Many linear algebra textbooks are "foundational" ish

They are many students first/second course with foundational flavor

[orchid harbor]

Lemme actually show u what i need to study

[tame mural]

Gilbert Strang's Linear Algebra textbook + course is also famous

So there will be plenty of surrounding material if you ever get blocked

Any basic textbook will never miss this

Most first Linear Algebra texts are very foundational

They are mostly self-contained

[orchid harbor]

ok @tame mural

Thanks man

[floral oasis]

I'm trying to nail down my intuition - a defective matrix (linearly dependent eigenvectors/mapped to lower dim eigenspace) is equivalent to a matrix being singular/non-invertible right?

[wintry sphinx]

no

[floral oasis]

oh?

[wintry sphinx]

there are singular matrices that are diagonalizable

and the reverse doesn't hold either

[floral oasis]

time to do more reading

so then a linearly dependent column space is not equivalent to linearly dependent eigenspace/eigenvectors?

I thought if A is a linear transform to say an N dim vector space, the eigenspace is also within that vector space, so they're essentially the same space no?*

[wintry steppe]

does this set still span R^3?

[zealous junco]

A:V->W, dimV = n, dimW = n, then A's eigenspace is a subspace of V

[wintry steppe]

if though there is a zero vector there

[zealous junco]

not necessarily in W

[floral oasis]

you mean R^4?

then yes

[zealous junco]

though if we let V = W = R^n then yes that's the case they are all subspaces of R^n, but eigenspace is not equal column space

[wintry steppe]

@floral oasis ?

why would it be R^4

[floral oasis]

isn't it obvious it spans R^3?

am I missing the point x_x

[wintry steppe]

you said R^4

[frozen compass]

Can someone help me find the x values for this matrice? (this is linear algebra)

[wintry steppe]

i think so

[floral oasis]

aaa

[zealous junco]

well you know LU(x) = b right?

do forward substitution for L(Ux) = b, then you find the vector equal Ux, so Ux = a, then do backward substitution

[frozen compass]

ok, I can try that. thank you

[wintry steppe]

@zealous junco help?

[zealous junco]

i'm not sure lol, does this say the column space span R^3

if it's refer to clumn space then yea

[floral oasis]

I'll do a bit more searching through math stackexchange

I also might not really understand the definitions which is tripping me up

[wintry steppe]

no problems

[west shoal]

Not sure how to answer the question since there are two values

I ended up with x_1 = b_1 - x_2
0 = 3b_1 + b_2

[uncut fjord]

what does your last line imply for the values of b_1 and b_2?

[west shoal]

im not sure what ur asking

[uncut fjord]

so you got the equation 0 = 3b_1 + b_2, what happens to this equation when lets say, the b_1 = 1 and b_2 = 4?

[west shoal]

it equals 7

[uncut fjord]

which is not 0 therefore the equation breaks, right?

[west shoal]

correct

[uncut fjord]

so your equation is basically telling you for what values of b_1 and b_2 does your Ax=b still work

[west shoal]

When b_1 and b_2 are 0?

[uncut fjord]

3b_1 + b_2 = 0 becomes a constraint to keep the equation "consistent"

[west shoal]

I'm confused because if b_1 is 1 and b_2 is -3 it still keeps the equation consistent

[uncut fjord]

well if b_1 and b_2 are both 0, then you have a homogeneous equation system

[west shoal]

homogeneous? i think that's the next section we go over

[uncut fjord]

no problem, the idea is if b_1 and b_2 are both 0, we don't really have an interesting problem because that always has a solution (all zeros), what we care about are when b_1 and b_2 are nonzero

and you're on the right track, we see that for some values of b_1 and b_2, the equation still makes sense, but for others it doesn't

[west shoal]

So when b_1 and b_2 are non zero constants most of the time the equation is inconsistent except for a few outliers

[uncut fjord]

yep, so that's your first part, to show that Ax=b doesn't have a solution for all b, only the values that satisfy your equation

for part 2, you need to interpret what that equation means

[west shoal]

oh

i read the first part wrong

[uncut fjord]

what does 3b_1 + b_2 = 0 look like?

[west shoal]

i feel like a dumbass

[uncut fjord]

sounds like you got it

so geometrically, what is 3b_1 + b_2 = 0

or maybe graphically

[west shoal]

hmm...

a diagonal line through the origin?

[uncut fjord]

exactly

it's a line that passes through the origin

those are all of the values of b that give a solution

[west shoal]

oh

[uncut fjord]

if you want to look into the problem further, look at the slope of that line and how that relates to the column vectors of A

[west shoal]

i thought it would have a solution if it intersected with another line i'd assume

not just the line b itself

maybe i'm over thinking this

brb i'll look at the relation on a graph

so i guess for the second part i could just say that the set for all b for which Ax=b does have a solution is when the vectors of A lie on the line of b?

@uncut fjord ?

[uncut fjord]

almost, it's the inverse

b must lie in the column space spanned by the vectors of A

in your case, the set is all b_1, b_2 in R (or C) s.t. 3b_1+b_2 = 0

[west shoal]

The set for all bs... are when b_1 and b_2 lie in the column place spanned by vectors of matrix A?

[uncut fjord]

yeah so basically the idea is that the solution x_1 and x_2 in Ax=b are coefficients for a linear combination of the column vectors of A to create b. That's what it means for b to be in the span of the column vectors. If you haven't come across this terminology, then don't worry about it. You can look into it further if you're really interested.

it's really just another connection/way of looking at the problem, although it's a bit more of a side note to the problem you posted

[west shoal]

It’s fine I already know about vector equations l

Basically just saying that the vectors of matrix A have a linear combination that produces the vector (b1,b2)

[uncut fjord]

yeah

[wintry steppe]

[ember matrix]

What was that theorem of 2 commutative maps having same eigenbasis?
Was it for Hermitian? Is this theorem "if and only if"

ping me oWO

[gray dust]

@ember matrix 2 self-adjoint (R/C) operators have a common eigenbasis iff they commute

[pulsar lily]

if two vectors are linearly independent, they are able to span all of R2, right?

[wintry steppe]

yes.

[pulsar lily]

but then how would you reach negative numbers with a linear combination? scalars can only be positive, right?

*if the two vectors were positive

[wintry steppe]

the scalars may be negative

it doesn't make sense to call a vector positive or negative

[pulsar lily]

oh okay

would it make sense to say both values in the vector are positive

rather than saying the vector is positive

[wintry steppe]

sure, it would make sense

[pulsar lily]

thanks

[ember matrix]

thanku

[zealous junco]

when people say P = {polynomial over R} is vector space

p in P

that arbitrary p can be represented by

p = sum from 0 to n of a_i * (x)^i right

like each p is finite

[native rampart]

Yes

[zealous junco]

cool

thx

[zealous junco]

wait so picking any basis for fin dimensional space

say {a1, ... an}

[native rampart]

Defines the vector space

[zealous junco]

then the eigenbasis is usually not a subset of {a1, ... an}

[native rampart]

May or may not be

[zealous junco]

yep ok

so for instance the d/dx

is a special case where then eigenbasis is {1}

[native rampart]

Yes

[zealous junco]

for Pn->Pn

ok thanks

[reef sleet]

What does this second bit mean?

Does it mean that x1a11 + x2a12 = a12?

That, for the same x1 & x2, x1a21 + x2a22 = a22?

And so on?

[slender patio]

That would be x1a11 + x2a12 = b1
, x1a21 + x2a22 = b2 instead
But yeah this is the idea

[reef sleet]

oh yeah oops haha

Thank you!

[slender patio]

Np ^^

[supple berry]

shouldn't it be probability of moving from state i to state j

[fallow jolt]

Any idea how I can go about proving this?

[stoic python]

Momo

[fallow jolt]

For M^1/2 I can use spectral decomposition, but what about M^-1/2?

[trail dirge]

Hi can a matrix with all the same elements be diagonlized?