linear-algebra

[bold sun]

it was 4 and 2.5 that worked?

[gleaming knot]

I mean the vector t is multiplied by

the "slope" if you will

[bold sun]

oh ok

so now i chnage the (2,1)

to see what fits (6,4)

12 and 8.5

[gleaming knot]

Now I think you have enough data points to figure out what's going on

[bold sun]

hmmm

times 2 and times 2 add 0.5

?? still dont see any pattern

[gleaming knot]

Think about the math of the situation

What is the app even drawing? Maybe that will help you see the right pattern

[bold sun]

line graphs?

[gleaming knot]

What is the app drawing and how are the numbers you're changing involved

in a precise manner

[bold sun]

the first cordinates i change determines the position of line if paralell or if part of same line

[gleaming knot]

that's qualitative

You can give a quantitative answer

[bold sun]

thats lambda (2,1) is parallel

dunno about same line

[gleaming knot]

Don't answer if you aren't sure though

Keep experimenting in that case

[bold sun]

its never gnna workkk......

[gleaming knot]

Basically experiment until you have a solid handle on what's going on

[bold sun]

i got it i think you pick a value for t and add it up

then the answer is part of that line

[gleaming knot]

yeaaaa

Exactly

[bold sun]

okkk that makes sense now.....thank you!!!

[gleaming knot]

Np!

[teal grotto]

because any eigen vector of T is perpendicular to (T-lambda I)*(x) for all x. it means that there are no components of v that ever lie in the image of (T-lambda I)*

[gleaming knot]

I think you just condensed the proof in the picture lol

[winter harbor]

Remember that since $V$ is finite dimensional and $\langle \cdot, \cdot \rangle$ is positive definite, then for any subspace $W \subset V$ we have an orthogonal decomposition:
$$
V = W \oplus W^{\perp}
$$
In particular, this means that $W \cap W^{\perp} = {0}$. Meaning that if we take an element $w \in W$ and $w \neq 0$, then $w \not\in W^{\perp}$.
\
\
Now let's go back to the proof. Notice that they have shown $\forall x \in V$, we have $\langle v, (T^{\ast} - \overline{\lambda})(x) \rangle$. This means that the image of the operator $T^{\ast} - \overline{\lambda} I$ is orthogonal to the subspace $\langle v \rangle$ generated by $v$. Thus, what they have shown is that $\text{Im}(T^{\ast} - \overline{\lambda} I) \subset \langle v \rangle^{\perp}$. I.e, the image of the operator $T^{\ast} - \overline{\lambda} I$ is a subset of the orthogonal complement of the subspace generated by $v$.
\
\
Remember how I have remarked how the orthogonal complement of a subspace and the subspace itself have trivial intersection? Well, $v \neq 0$ by definition, since it is an eigenvector. So it means that $v \not\in \langle v \rangle^{\perp}$ and thus, a fortiori, we have $v \not\in \text{Im}(T^{\ast} - \overline{\lambda} I)$ and it is not surjective.
\
\
As a final remark, since $T^{\ast} - \overline{\lambda} I : V \rightarrow V$ is a linear operator, if it is not surjective then it can't be injective and thus $\overline{\lambda}$ is an eigenvalue for $T^{\ast}$.

[gleaming knot]

How I'd explain it is that if a linear operator A is onto, then its range is the entire space, and in that case you won't find a vector orthogonal to the range, because no nonzero vector is orthogonal to the whole space. Therefore, if there's a vector orthogonal to the range of A, A is not onto

[stoic python]

MisterSystem

[tulip basalt]

what does rank([A b]) mean?

[winter harbor]

I suppose b is a column vector

so [A b] must denoted the augmented matrix obtained by adjoining A and b.

this is again a matrix

and we can talk about its rank

[tulip basalt]

Oh okay now that makes sense to me lol. thanks!

[winter harbor]

which is just the number of linearly independent rows/columns

[tulip basalt]

been ages since i took linear algebra

[winter harbor]

np

[bold sun]

hey sorry if this is a really silly question- im just seeing if i understand this and what its tryna say

but basically is this trying to say that a11+a12+...+a1n-1+a1n= b1

and a21+212+...+a2n-1+a2n= b2

or is it.....a11x1+a12x2+...+a1n-1xn-1+a1nxn= b1 with xs in it

[winter harbor]

No

[bold sun]

then?

[winter harbor]

An augmented matrix is literally just taking a n x m matrix A and taking an n dimensional column vector b (i.e an nx1 matrix) and forming a new (n x (m+1)) matrix

for which the first m columns are those of matrix A

and for which the (m+1)-th column is the vector b

[bold sun]

isnt it originally mxn matrix

[winter harbor]

we usually denote this by [A b]

[winter harbor]

Whatever, I am just using a different convention

but sure

Since in your example it is m x n

the augmented matrix would be m x (n+1)

[bold sun]

oh so what does the line show us?

like the seperators of a and b that line...

[winter harbor]

It just separates the columns of the original matrix A

and the column vector which was augmented

[bold sun]

oh so its 2 diff matrixes

multipled or what?

[winter harbor]

No, we have taken a matrix A

and a column vector b

and created a new matrix out of it

called the augmented matrix of A with the column vector b

[bold sun]

oooh okk so its the solution of the 2 matrixes like its a combined version?

[winter harbor]

somewhat

This is particularly useful

if you want to solve systems of equations

via gaussian elimination

for instance

If you have a system of equations Ax = b

[bold sun]

oooh okk havent learnt that yet but ive heard of it....i think we gnna cover it soon tho

[winter harbor]

you can augment A and b to form the matrix [A b]

and you perform elementary row/column operations

in order to solve the system of equations

and you need to augment the matrix A with b

[bold sun]

? wdym

[winter harbor]

You will prolly see these at some point

[bold sun]

yhh havent yet but i think its gnna be soon tho

im tryna catch up befre mondayyy

but thank you so much for the explanation and clarification on this!!

🙂

[winter harbor]

Np

[hollow finch]

this is a correct interpretation. MisterSystem is correct in that it is, in general, just a merging of two matrices. but it's primary (and more commonly used) purpose (especially in the beginning of a linear algebra course) is to compactly write a system of equations all in one matrix. in this system of equations context, the vertical line basically represents the equal signs.

so you do get the $m$ equations

$a_{i1}x_1+a_{i2}x_2+\ldots+a_{i(n-1)}x_{n-1}+a_{in}x_n=b_i$

for $i=1,2,\ldots,m-1,m$

[stoic python]

nix (reply please)

[hollow finch]

commonly, especially as you are learning linear algebra, you will switch between these contexts. going from a system of equations to an augmented matrix, and then sometimes going back to a system of equations once you have simplified the augmented matrix. this makes certain techniques more intuitive like parameterizing solutions.

and, no, it's not a silly question btw.

[bold sun]

ooooh okay then thank you that makes sense i get that thank you for the explanation

[hollow finch]

[nocturne jewel]

what have you tried?

@charred root

So... attempt it first

asking you to show that if the matrix of a transformation is invertible, then T is surjective and injective (onto and 1-1 respectively)

ie invertibility of a matrix implies bijectivity of the transformation

if you are 1-1, then you have that T[u]=T[v] implies u=v, ie every output is "hit" at most once
If you are onto, then the output space is identically the codomain, so every output is "hit" at least once

onto: for every b in the codomain, there exists a v in the domain such that T[v]=b

[storm canyon]

Some googling tells me you need minimum of 5 vectors to span r5?
is that still true for a subspace of r5? or could you span subspace of r5 with 3 vectors

[half ice]

Not still true

That is, if you take a subspace of some 5D vector space, it may not be 5D

[storm canyon]

mmm

alright thanks

[half ice]

Simple example, take a line in R5. Only needs one vector to span.

[wintry steppe]

what's an affine space? I see that sometimes in linear algebra stuff

like "affine space" and "affine transformation"

[winter harbor]

By an affine subspace it intuitively means that we have a subspace of a vector space that is "shifted/translated" by certain amount.

For instance

When we solve a system of linear equations Ax=b

The solutions are of the form x_0 + kernel (A)

Where x_0 is some particular solution

Another example is a plane in R^3 that doesn't pass through the origin

It is just a codimension 1 subspace of R^3 that was shifted by a certain amount

Ofc, every subspace of a vector space is an affine subspace, but the converse is not true.

Planes that do not pass through the origin are not subspaces of R^3.

Affine transformations are intuitively as simple

They are just linear transformation + constant amount / translation.

Say a map A : R^n -> R^m that is of the form Ax = Tx + b, where T : R^n -> R^m is linear.

Intuitively this is what it means when you hear the word affine.

[ionic laurel]

hi i have a question - what does it mean to say that if a matrix C is in the span of matrix A and B? Do I just make a linear combination x1 * A + x2 * B + x3 * C = 0 and figure out what the x1-x3 scalars are?

i am having trouble envisioning a matrix within a span of two matrices

[winter harbor]

It means that C = x_1 A + x_2 B for some scalars x_1, x_2

Are you familiar with fact that matrices form a vector space?

[ionic laurel]

Matrices form vector spaces? I'm not sure if we learned that yet

Or I might be more aware

if examples are provided

My class mostly talked about dimensions of real numbers as vector spaces

[winter harbor]

Are you familiar with the formal definition of a vector space?

[ionic laurel]

I know the conditions

[winter harbor]

Like, a set with an operation of sum and an operation of scalar multiplication and etc...

[ionic laurel]

yeah

[winter harbor]

So

[ionic laurel]

0 vector

closed under addition and scalar mult

[winter harbor]

Matrices with the usual sum and multiplication by scalars form a vector space.

[ionic laurel]

okay

[winter harbor]

If you didn't know this

You might want to check this by yourself

But the thing is that we can view matrices as vectors

[ionic laurel]

right

[winter harbor]

And we have a definition that is "general" and applies to any vector space when it comes to "span".

In any vector space

[ionic laurel]

because i was thinking it is similarl written as a linear combination

[winter harbor]

If we say a vector v is the span of the vectors u and w.

It means that I can write v as a linear combination of u and w.

[ionic laurel]

right

so it should apply for matrices as well

[winter harbor]

Matrices form a vector space too, so the same definition applies.

[ionic laurel]

okay so one more weird question

[winter harbor]

Exactly!

[ionic laurel]

would the better approach for the first and second matrix to equal the third

or first second third equal to 0

or does it not matter

like the first option if you find scalars then it is within the span and for the second option if the only solution is the trivial solution then it is not in the span

[winter harbor]

This, and the reason is because we take this to be the definition of spanning.

[ionic laurel]

Okay

right

[winter harbor]

Not really. I think you are getting a bit confused with the definition of linear independence and span.

[ionic laurel]

oh you are right

that is linear independence

well linear independence is closely related to span though

like the third matrix is linearly independent if it is not in the span and dependent if it is

but you are right nonetheless

thank you MisterSystem for clarifying my confusions

[winter harbor]

Np

[ionic laurel]

@winter harbor Last question if you don't mind

How would you go about visualizing a span of matrices?

would you just break them up into column vectors and then that space acts as a vector space?

[stable urchin]

i recommend you watch the essence of linear algebra

the youtube miniseries by 3blue1brown

[ionic laurel]

hm?

i guess i can watch it

[winter harbor]

I particularly don't know lmao.

[ionic laurel]

we just started vector spaces and this chapter is hard to understand so i have more questions than usual

[winter harbor]

2 things come to my mind tho

[ionic laurel]

i understand how to visualize vectors but not matrices lol

[stable urchin]

also it probably doesn't help to visualize a span of matrices

[ionic laurel]

i see

[winter harbor]

The space of n×m matrices is isomorphic to R^(nm)

[stable urchin]

the only way I can imagine is just every possible linear combination of matrices

[winter harbor]

So you can just think of the span of two n×m matrices

[ionic laurel]

alright

i was just wondering if it was important

it is easy to visualize a span of vectors

but i was just wondering how would do that with matrices

[winter harbor]

As the span of their corresponding vectors in R^(nm)

[ionic laurel]

👍

yup

[winter harbor]

That is probably not that useful

Because you see

For 2×2 matrices we would be trying to visualize vectors in R^4

Which is just not possible

At least in the strict sense of the word "visualize"

In any case

[ionic laurel]

i see

[winter harbor]

If you have a good feeling of linear combination in R^n, it shouldn't be so hard to just make an analogy.

Ofc, matrices are related to linear transformations

And you can try to visualize a linear combination of matrices

By how they act on vectors

This is another way to go

And if you have two n×m matrices and a vector b in R^n

We have (αA+βB)(x) = αA(x) + β B(x)

That's how a linear combination of matrices acts on vectors

And if you know how to visualize matrices as linear transformations

Vizualizing linear combinations of those

Shouldn't be that hard

[ionic laurel]

yep

Thanks again!!

Very helpful information

[stable urchin]

okay I just had a quick q about nullspaces

so like take the following subspace

$$W=\left{\begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{pmatrix}\colon x_1+x_2-2x_3-4x_4=0\right}$$

[stoic python]

timmmm

[stable urchin]

so like this is asking me to expess it as a nullspace

[teal grotto]

look at the coefficients on the x_i’s

[stable urchin]

nono like I know how to get there

the thing I'm not sure about

is like

how to express the nullspace

like can it be a span

or is it simply just the subspace of solution vectors to the augmented matrix given by the subspace W

this is what I expressed nul(A) as so far

$$\text{nul}(A)=\left{\begin{pmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{pmatrix}\colon \begin{pmatrix} 0 \ 0 \ 0 \ t \end{pmatrix},t\in\mathbb{R}\right}$$

[stoic python]

timmmm

[stable urchin]

there we go

[teal grotto]

um

[teal grotto]

how does the vector of x_i’s relate to (0,0,0,t)

[teal grotto]

if u knew how to get there from the coefficients then you wouldn’t be having the confusion lol

[stable urchin]

is a subspace by definition also a span

[teal grotto]

i have no idea what this set represents

like

it’s not well defined

name one vector in that set

[stable urchin]

<0,0,0,2>

[teal grotto]

okay well that’s not the set you’re trying to represent

it should be ${(0,0,0,t)\in\mathbb{R}^4:t\in\mathbb{R}}=\text{span}(0,0,0,1)$

[stoic python]

c squared

[teal grotto]

W is equal to the kernel of the linear map given by the matrix [1 1 -2 -4]

[stable urchin]

that actually makes a lot more sense

mb

[stable urchin]

right so I think I just got my definitions wrong

in the case of the initial subspace, would the corresponding nullspace be the values of x_i such that x1+x2-2x3-4x4=0

like the matrix $$\begin{pmatrix} 1 &1 &{-2} &{-4} \end{pmatrix}$$

[stoic python]

timmmm

[rich garden]

are change of basis matrices always invertible?

so the columns need to be linearly independent?

is the zero matrix a valid change of basis matrix then?

[marble lance]

The zero matrix is not a change of basis matrix, because it sends everything to the zero vector

And yes CoB matrices are invertible

[rich garden]

i saw this on wikipedia and was confused

[lavish jewel]

with a change of basis, one wishes to represent vectors in a vector space V with a different basis than the original one. since they are both bases for V, they need a full set of linearly independent vectors ,which when set as columns of a matrix, mean the matrix is invertible

[rich garden]

it seems like every square matrix is a change of basis matrix

[lavish jewel]

in fact the change of basis is rather the inverse

any square invertible matrix can be interpreted as a change of basis, yes

you don't NEED to interpret it that way, but it can be done

[rich garden]

so invertible <=> change of basis

but if the columns don't form a basis it's not a change of basis matrix i'm guessing

[lavish jewel]

the columns will always form a basis, the question is "of what?"

it has to be a basis of the whole vector space V

[rich garden]

what if they're linearly dependent?

[lavish jewel]

then they can be interpreted as a spanning set of a subspace of V

and can be made into a basis by something like checking the pivots of the matrix

so the invertibility part is the most important

[rich garden]

ahh okay

so the span of the columns cannot decrease

it still has to be a basis of the original space V for it to be a change of basis matrix

which implies invertibility

thank you so much!

[sour cloud]

quick question if I have AB = 0 and i have the A matrix and i need to find matrix B how do i do that? And B is not equal to 0.

[lavish jewel]

the columns of B must be in the null space of A

[sour cloud]

wdym?

[lavish jewel]

unfortunately i have already explained the topic more than once to you before, and recommended that you go review the definitions

so i can't give you any more help

[sour cloud]

so i have matrix A : ```
[
-4 -1 -2
0 -4 -8
2 2 2
]

for null space i got [1, -2, 1]

what do i do with the null space ?

[dusky epoch]

first off, the nullspace isn't a single vector.

and second, what are you asked to do?

[sour cloud]

I have to find matrix B. I have this equation AB = 0. Im given matrix A

[dusky epoch]

so you want to find all matrices B such that AB = 0?

what size is B?

[sour cloud]

3x3

[dusky epoch]

and you got that the nullspace is span( [1; -2; 1] ), yes?

[sour cloud]

yes

[dusky epoch]

well then

clearly AB = 0 if and only if every column of B is in null(A).

[sour cloud]

yea but i dont understand what i need to do with the null space

[dusky epoch]

do you understand what i just said?

[lavish jewel]

they don't. maybe you can give them an explanation of the null space that they can grasp, since mine didn't do the trick

and of linearity, for that matter

[dusky epoch]

you've tried to help them before?

[sour cloud]

i got it thank you. sorry i was confused before

[lavish jewel]

yes

#linear-algebra message here

[sour cloud]

yea so i just add the null space to each column of A to get matrix B

[lavish jewel]

[sour cloud]

nvm lol

[lavish jewel]

stop guessing stuff, there is a really easy way to do it if you understand what you're doing

please go read the definitions and the explanations ann and i have given you

[sour cloud]

i read the definitions tho

[lavish jewel]

then the next step is to understand them

[sour cloud]

ok

[bold sun]

hey a random question basically is what is a linear system ??? is it the same as linear equation - in the lect notes we have the def of a linear equation and a system of linear equastions but not linear system? so l wanned to know if its the same as linear equation?

[dusky epoch]

"linear system" is another word for "system of linear equations"

[bold sun]

oooh okk then

Thank you!!

[wintry steppe]

how do I proceed

[lavish jewel]

try the definition of "consistent"

[strange delta]

on the 3rd equality what did they do?

[winter harbor]

They just used the definition of conjugate transpose/ adjoint of an operator.

$N^{\ast} \in \mathcal{M}_{n}(\mathbb{C})$ is the unique matrix for which $\forall x,y \in \mathbb{C}^{n}$ we have:
$$
\langle Nx, y \rangle = \langle x, N^{\ast} y \rangle
$$

[stoic python]

MisterSystem

[strange delta]

yeah i get that but then i try to do that for the proof and it doesn't work

[winter harbor]

We also have (N*)^* = N, so...

[winter harbor]

Wdym?

[strange delta]

lemme write it down one second

i tried using the adjoint property by just subbing it in and i just get that

[winter harbor]

Notice that the adjoint of N* is N itself

This means that
$$
\langle N^{\ast} x, y \rangle = \langle x, Ny \rangle
$$

[stoic python]

MisterSystem

[winter harbor]

This is why we can "put N*" on the left hand side of the inner product.

Idk if that's a good way to phrase that.

[winter harbor]

In any case, they are using this.

The fact that
$$
(N^{\ast})^{\ast} = N
$$

[stoic python]

MisterSystem

[strange delta]

yeah i get that

[winter harbor]

So do you get what they did now?

[strange delta]

so for this part did they just switch the positions of x and n*nx

[winter harbor]

Yeah, they just used this.

$\langle Nx, Nx \rangle = \langle N^{\ast}(Nx), x \rangle$

[stoic python]

MisterSystem

[winter harbor]

As a consequence of this.

[strange delta]

hmm

i think i got it

does this seem right?

this is the whole equality

[winter harbor]

Yeah

For the 4th equality they used the fact that N* and N commute

So we have

[winter harbor]

And for the 5th they just used this.

[strange delta]

i see, thanks

[ionic belfry]

hello

can anybody help me with linear algebra?

i am kinda confuse about some theory concepts

[winter harbor]

You can just ask

[nocturne jewel]

👻

[torpid swift]

Hello 🙋🏻‍♂️, I am a freshman in college and need some assistance. How do I prove this? A ∪ B = B ⇒ A ⊂ B

[lavish jewel]

this isn't linalg, but try drawing some venn diagrams

[torpid swift]

Is it analysis?

[lavish jewel]

no

[golden reef]

is <-2, 2, -3> parallel but in opposite direction to <2, -2, 3>

[lavish jewel]

sounds about right

[golden reef]

if i have a plane equation of 10x + 10y + 10z + 20 =0

can i divide the whole thing by 10?

[lavish jewel]

sure thing

[granite kraken]

Can someone pls help with this

[torpid swift]

How do I prove this without a Venn diagram? I am supposed to "Prove the true statements and disprove the false ones".

[lavish jewel]

this is not allowed

[nocturne jewel]

sets are apart of discrete math

[granite kraken]

Ok

[wintry steppe]

Wouldn’t the latter imply that w is in $span(v_{j+1}, … v_{n})$?

[lavish jewel]

as mosh says, maybe someone in #discrete-math can help, since set theory is often covered there for the first time

[stoic python]

justini

[granite kraken]

Does anyone know how to do proofs for invertible linear transformation?

[marble lance]

@charred root what do you mean?

What do you need help with?

If x is the vector (x1, x2, x3) and u is the vector (b,c,d) then u dot x = bx1 + cx2 + dx3

x is the variable

Like in f(x) = x^2, x doesn't have a specific value

It's the input to the function

Where did you get these numbers from?

They are not in the question

[nocturne jewel]

there's no numbers

[marble lance]

How do you know those are the values of b c and d?

They are not given

I don't think you are supposed to

Okay

[nocturne jewel]

Ok well you dont need numbers anyway

[marble lance]

Then that's fine

Tu(x, y, z) = ax + by + cz where u = (a,b,c) and (x,y,z) is the vector x that is inputted into the function

So (x,y,z) is a variable

It can be any vector

And different inputs give different outputs

Now you have to show that if x and y are vectors and m and n are scalars, then Tu (mx + ny) = m Tu (x) + n Tu(y)

[nocturne jewel]

no, it's a generic vector

[marble lance]

No

Leave it as arbitrary

So let x = (x1, x2, x3) and let y = (y1, y2, y3) but don't give them specific values because what you want to show has to be true for all vectors

And let m and n be any real numbers

Then work out what Tu (mx + ny) is equal to

And what m Tu (x) + n Tu( y) is equal to

And check that they are equal

When I say check what it is equal to, you won't get a number. It will depend on x1,x2,x3, y1, y2, y3 and m and n

[nocturne jewel]

Yes, those are the requirements for T to be linear.. Lunasong just combined them into applying T to a linear combination of vectors and getting a linear combination of the output vectors

no

[marble lance]

(0.81, 0.34, 0.46) is NOT the input to the function

The input to the function should be arbitrary

The numbers are part of the function definition and will appear when you evaluate

[silent dune]

is it wrong to write the original 3 rows as the basis for the row space, in a question like this?

instead of the ref rows

[lavish jewel]

they should both work

[marble lance]

It won't work if you swapped rows, which you did

[lavish jewel]

oops i didn't notice

the corresponding rows need to match

[sick sandal]

just like you would prove a invertible function :3

[silent dune]

ohh

so since the rows were swapped would I have to include the swapped row instead or I can't use the original at all

[marble lance]

You can use the original if you keep track of what you swapped

So here it would be the first, third and fourth row since the fourth row corresponds to the second row in RRE form

[lavish jewel]

the swapped rows would do, as luna says

[silent dune]

kk thanks

[granite kraken]

So can anyone help with this

Here's the hint to do it too

But I'm stuck even with the hint

[winter harbor]

Proving the first implication is trivial

So we need to prove that if an endomorphism of a finite dimensional vector space is is bijective iff it is surjective

[nocturne jewel]

I'd suggest using simpler language for people learning tbh

[winter harbor]

I actually don't know how to prove this result using the hint tbh. At least not immediately.

This is a corollary of the rank-nullity theorem

And can also be proved using the following observation

[sick sandal]

ye thats what i was thinking as well

[gleaming knot]

Is this like a problem you do to prove that left invertible is equivalent to right invertible, and with the same inverse?

[winter harbor]

Yeah, it is exactly the same problem

[gleaming knot]

If we're using this hint then it's almost immediate

Suppose the hint; we're given TS = I, therefore STS = S. S is a bijection by the hint, and STS = S says that ST is the identity on the range of S, so ST is the identity

[sick sandal]

have you perhaps proved that rank(T^2)=rank(T) implies range and null space are disjoint?

[lament bluff]

guys, how do we find the hermitian conjugate of a continuous operator?

let's say, knowing these:

we needed to find the hermitian conjugate of operator a_l

[sick sandal]

since its been a while im assuming question is over so here is my own attempt at the other side of this equivalence , do point out any flaws in my argument

exuse the handwriting

i always have some idea in mind but its hard to formally write it

[lavish jewel]

this looks like a graded evaluation

[median ocean]

fff

[wintry steppe]

screw it, I’m just going to trust my instincts

[vital pagoda]

if i have two matrices A and B and ive been asked to determine whether they are similar or not (their determinants, traces and eigenvalues are the same), is it enough for me to just show that both are diagonalisable to determine if they are similar?

or would i have to do the whole shpiel of finding an S such that B = (S^-1)AS

[winter harbor]

No

Think about nilpotent matrices

They have the same eigenvalues (all of them are 0), they have the same trace (they are all zero too) and they have the same determinant (it vanishes too).

But they are not diagonalizable (except for the 0 matrix) so meh...

[gleaming knot]

Compute Jordan canonical form of both; gives you a deterministic algorithm!

[vital pagoda]

im sorry im only at the beginning of my matrix analysis course i dont know what a jordan canonical form is

[winter harbor]

Basically, any matrix over the complex numbers, or an algebraically closed field, is similar to a Jordan Canonical Form. And a Jordan Canonical Form is basically a matrix consisting of Jordan blocks, which give you information about the eigenvalues of your original matrix and their algebraic and geometric multiplicities.

It is quite a remarkable result

But if we can't use this result now, that's ok.

[vital pagoda]

hmmmmm

like whats wrong with this line of logic for these two matrices? Since both have distinct eigenvalues, it means they are diagonalisable, if they are diagonalisable it means that A = SBS^-1 right?

my first attempt was to just set a 2x2 matrix S with entries a,b,c,d then solve the system of equations A - SBS^-1 = 0

but the algebra got nasty really quickly which made me think i was doing something wrong

[winter harbor]

Ah

[gleaming knot]

Solve AS = SB instead!!!!!!

Very neat trick

[vital pagoda]

😮

[gleaming knot]

One of the questions on the linear algebra final I TAed for last year required knowing that trick!

[winter harbor]

For 2 x 2 matrices finding whether two matrices are similar is actually simpler.

Namely, 2x2 matrices with same minimal and characteristic polynomial are actually similar. This also works for 3x3 matrices too, but fails for n>3 in general.

So we need only to check that.

[vital pagoda]

😮

[winter harbor]

You actually use Jordan Canonical Form to prove this lmao

But it is a neat result

[winter harbor]

At least that's the proof that immediately comes to mind rn.

[vital pagoda]

thank you both of yous!

[nocturne jewel]

Asked about this a bit ago, more for clarifiying my understanding in a, however I honestly have no clue how to use matrix product rule.. so more just need it explained so I can give a better attempt at b (and subsequently c)

Intuitively I know it's because g is the linerization of f at t=0, so their derivatives must match, but I'd like to show that properly

[gleaming knot]

Isn't $(A+tB)f(A+tB)$ identically the identity matrix?

[stoic python]

Icy001

[nocturne jewel]

Yeah, I asked about part b

Xf(X)=I, and d/dt I = 0

[gleaming knot]

Is g even defined in the problem?

[nocturne jewel]

yes

2nd part of the preamble

[gleaming knot]

Is it defined because the concept of linear approximation is defined?

[nocturne jewel]

it's defined as the linear approximation of f

[gleaming knot]

So by definition, g and f have the same gradient at A

Then (b) follows by the matrix product rule

[nocturne jewel]

Ok, so what I asked

was how to use the matrix product rule.

[nocturne jewel]

.

I also already know why the answer is what it is, I want to understand the actual algebra behind it.

[gleaming knot]

I'm not sure where you're getting stuck using the matrix product rule

[stable urchin]

So I figured out the first two

but for the nullspace

how do I get a matrix when given a nullspace

[gleaming knot]

I guess it's because the product rule is in terms of gradients? Just take both sides of the equation and apply it to B which is the direction vector

[nocturne jewel]

what's W?

[gleaming knot]

W is the "direction" vector

which is a matrix in this case

but interpreted as a vector

[nocturne jewel]

so W=A+tB in this context?

[gleaming knot]

No, just B

[nocturne jewel]

why

oh wait

[gleaming knot]

d/dt (A+tB) at t=0 is equal to the gradient of X -> A+X at 0, multiplied by B

[stable urchin]

$$\text{if}\ \text{nul}(C)=\text{span}\left{\begin{pmatrix} -1 \ 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 2 \ 0 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 4 \ 0 \ 0 \ 1 \end{pmatrix}\right}, \ \text{what is} \ C\text{?}$$

[nocturne jewel]

so DL(W) becomes 1 right?

[stoic python]

timmmm

[gleaming knot]

I think that's right

[nocturne jewel]

since L(x)=x, so we evaluate 1 at x=B

[gleaming knot]

L(X) is A + X is it not? And M(X) is f(A+X)

[nocturne jewel]

idk

[gleaming knot]

Hmm let me think

[nocturne jewel]

it's h(x)=xg(x)

at x=A+tB

[gleaming knot]

Maybe it's possible to treat L as a function R -> Matrices

instead of a function Matrices -> Matrices

How were you treating L?

[nocturne jewel]

wdym

L(x)=x in this case I thought

[gleaming knot]

In the product rule, what type of thing is L?

[nocturne jewel]

a function...

[gleaming knot]

from what to what

I can see 2 possible interpretations

I think both work

We can have $L\colon \bR\to\bR^{n^2}$ or $L\colon \bR^{n^2}\to\bR^{n^2}$

[stoic python]

Icy001

[gleaming knot]

In the first case, $DL$ is an $n^2\times 1$ matrix, in the second case $DL$ is an $n^2\times n^2$ matrix

[stoic python]

Icy001

[nocturne jewel]

yeah

[gleaming knot]

In the first interpretation, where x is a real number, it makes sense to let L(x) = A + xB

In the second interpretation, where x is a matrix, it makes sense to let L(x) = A + x

In the first interpretation, W is just an element of $\bR^1$, so just a scalar, so it doesn't matter what W is

[stoic python]

Icy001

[gleaming knot]

In the second interpretation, we want to choose $W=B$ to capture directional derivative in the $B$ direction

[stoic python]

Icy001

[nocturne jewel]

so in the n^2 to n^2 interpretation, is A+tB L(tB) or L(B)?

the prior right?

[gleaming knot]

L(tB) yeah

[nocturne jewel]

Ok

Let me try it again after I eat

[sly shuttle]

Need some help setting up a model for this

[silent dune]

so did this a while ago and had a question, after I do guass elim on the matrix A, do I still need to do col(A) = row(A^T) or can I just say the column space of A is the columns in A with pivots? for (b)

[wintry steppe]

I'm trying to solve this system of equations in terms of $x_{5}$ and I ran into this; can someone help clarify what this means?

[stoic python]

justini

[wintry steppe]

ah I see, thanks

[dense wave]

$epsilon$

[stoic python]

MachTurtle

[dense wave]

$:epsilon:$

[stoic python]

MachTurtle

[dense wave]

$\epsilon$

[stoic python]

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

[dense wave]

$\epsilon$

[stoic python]

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

[dense wave]

Sorry for the spam

[stoic python]

MachTurtle

[dense wave]

$\epsilon_$

[stoic python]

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

[loud crow]

Even if its messed up I still need to solve it somehow too.

Thnaks.

[hollow finch]

@dense wave there's a #latex-testing channel...

[hollow finch]

Once you get rref, the nonzero rows of the rref matrix are a basis for row(A), and the pivot columns of the original matrix are a basis for col(A).
a basis is a minimal spanning set so same thing.

[minor vapor]

Can someone please help me with this!

[digital hazel]

I guess you can check whether the three are linearly independent?

[minor vapor]

I’m not sure at all! I don’t even know how to set up the matrix tbh
: (

[slate snow]

Hi, I need an help about a Jordan form and a minimal polynomial

I have to find Jordan forms from a minimal polynomial and…I can’t do it. It’s an old exercise. I study maths at UniPD in Italy.

[digital hazel]

https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3Doh19hECEz-0&psig=AOvVaw1FY4de37EbJ1_x2OsdXuOo&ust=1635233470113000&source=images&cd=vfe&ved=0CAwQjhxqFwoTCNCq38KF5fMCFQAAAAAdAAAAABAI

I think this vid can help you

[minor vapor]

Thanks I’ll check it out

[digital hazel]

But i will leave the gods here to verify whether we need to do that

Cuz my linear alg is rusty af

[minor vapor]

Wdym who are the gods lol

[lavish jewel]

put the vectors as rows or cols of a matrix and RREF

then check the nr of pivots

[minor vapor]

I’m not sure if I should put them in rows or columns?! Does it matter which one I do?

[lavish jewel]

no, it doesn't matter if all you want is to show linear independence

[minor vapor]

Cause I did this and I thought I put them in wrong

@lavish jewel Could you please help me with the next step? Sorry I’m not sure what you meant earlier by “RREF”

[lavish jewel]

watch the video, it explains it

[minor vapor]

Oh ok thank you for your help

[serene solstice]

Still can't make heads or tails of it.

OK folks, never mind. I was sniffing glue and had no idea what the hell I was saying.

What about (b), though?

[teal grotto]

@serene solstice

[stoic python]

c squared

[grizzled seal]

I just started linear algebra (yr10) it seems like a really complex subject does anyone have any suggestions to how to start?

[stoic python]

c squared

[teal grotto]

probably watching 3blue 1brown and kahn academy.

then probably pick up axlers lin alg book

[grizzled seal]

ok thanks

[teal grotto]

i’m partial to hoffman and kunze’s lin alg but wouldn’t recommend it as an intro book. still a really good book tho

[dusky epoch]

triangle inequality

[vital pagoda]

🤯 why didn’t that come into my head instantly

[vital pagoda]

@dusky epoch sorry, i've been doing this for a while now and im actually still stuck, how can i set up a triangle inequality with the trace function if ltr(a+b)l isn't necessarily smaller than ltr(a)l + ltr(b)l

[dusky epoch]

since $U$ is unitary, you have $|u_{jk}| \leq 1$ for all $j, k \in 1:n$

[stoic python]

Ann

[dusky epoch]

@vital pagoda

[vital pagoda]

so if every element of the matrix has modulus leq 1 isn't it obvious that the trace must be leq n?

[dusky epoch]

yup

[zinc timber]

these questions fucking trigger me,

[silent dune]

I don't need to get rref though do I? just ref?

for the basis I would, like for this I'd do rref

but the question before I didn't do ref

[fallow atlas]

Does anyone know how to determine the values of x1.....2500?

[lavish jewel]

each column is the vector of x_i's multiplied by a scalar y_j

e.g. $\begin{bmatrix} y_1 \boldsymbol{x} && y_2 \boldsymbol{x} && \dots && y_M \boldsymbol{x} \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

where $\boldsymbol{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_N \end{bmatrix}$

[stoic python]

Edd

[signal mulch]

Can i ask a question?

[lavish jewel]

sure

[signal mulch]

perfect

theres a typo in the question. det(A-xI3) = -[(x-a)(x-b)(x-c)]

so techincally det(A-xI3) = (a-x)(b-x)(c-x)

any idea how to do this?

[hollow finch]

Ref is fine yeah

[hollow finch]

Well the the question has determinants in it. Can you relate the determinant to whether or not a transformation is one to one?

[lavish jewel]

show that the matrix needs to have full rank for it to have unique solutions to Ax = b

[hollow finch]

The answer is yes btw

[signal mulch]

How would i do that tho?

[hollow finch]

Think about determinant properties? Equivalent ways to say a transformation is one to one?

You have to get creative with proofs. Maybe not necessarily think outside the box, but try to make as many connections as you can between things in the problem. There's no one way to do every single proof.

[signal mulch]

we havent really learned about many determinant properties yet, just things like A(A^-1) = I

or detA^t = detA

detAB = detA(detB)

[hollow finch]

Well that's not really a determinant property but I mean more what can the determinant tell you?
Especially with respect to transformations being 1-1

(or with respect to things adjacent to transformations being 1-1)

[errant mist]

I'm unsure how to approach this problem as I just started learning about kernels. Any ideas?

[signal mulch]

essentially all of this would be true

[hollow finch]

I'm guessing it means $phi((x,y,z)^T)$ right?

[stoic python]

nix

[errant mist]

yes

[hollow finch]

If the determinant is nonzero yes. Unfortunately thats a very incomplete list... :/

Which isn't your fault

[signal mulch]

oh shoot... thats the only list we have been given

[marble lance]

It's pretty complete if you haven't covered the determinant yet

[hollow finch]

But three more things that could be added to the list are

1. det (A) is nonzero
2. The transformation of A is 1-1
3. The transformation of A is onto
   Technically most of the stuff in there is equivalent to those ofc but these are true as well

[marble lance]

It has 2 and 3

It only doesn't mention the determinant

[hollow finch]

Yep I'm blind.

Kinda rushing bc i have class rn lol

[marble lance]

Np np

Gl with class

[errant mist]

Could I get some help with my problem?

I think one can argue that the postive definiteness can be seen from the properties of the inner product

[signal mulch]

so if i do this:
since determinant is non zero; (a-x)(b-x)(c-x) is not equal to 0
which mean (a-x) or (B-x) or (c-x) are not equal to 0

would that be on the right track?

[silent dune]

Can a Matrix have neither left nor right inverse?

[winter harbor]

Yup, consider:
$$
\begin{bmatrix}
0 & 1 \
0 & 0
\end{bmatrix}
$$

[stoic python]

MisterSystem

[winter harbor]

Or any nilpotent matrix for that matter.

[silent dune]

ok word had a matrix where the rank wasn't equal to the columns or rows so I just said it had neither left or right inverse

[winter harbor]

Yeah, square matrix that do not have full rank are not invertible.

[silent dune]

it doesn't have to be a square matrix though right?

[winter harbor]

I mean, there's no concept of invertibility for non square matrices.

It doesn't make sense to talk about invertible matrices that are non square.

[silent dune]

I'm referring to questions like this

[winter harbor]

Ah

I thought we were talking about square matrices

[silent dune]

ah no

I just applied these theorems

[winter harbor]

In any case, yeah that's correct.

[silent dune]

okay nice thanks

[signal mulch]

Could you help me answer my question?

[winter harbor]

Actually, uh. This matrix has full column rank. Which corresponds to a linear map A : R^2 -> R^3 that is injective. So it has a left inverse.

The theorem is correct ofc

But you have applied it incorrectly

[silent dune]

oh yes

[silent dune]

for that specific question it does have a left inverse

but the one I just did was different

[winter harbor]

Ahhh

Ok

[silent dune]

I was just using that as an example

[winter harbor]

I am sorry, I thought you were referring to this one.

[silent dune]

ah no all good

[winter harbor]

Sure

[errant mist]

@winter harbor me as well if u have time)

[signal mulch]

So here is the question, im not sure how to start it

[winter harbor]

Ah

Are you familiar with eigenvalues?

[signal mulch]

nope not at all

[winter harbor]

Ok, I will give the definition and a brief overview of the first results.

Let $A \in \mathcal{M}_{n}(\maththbb{K})$ be an $n \times n$ matrix over a field $\mathbb{K}$ (take the reals or complex numbers if you are not familiar with general fields). Then, we say that $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if $\exists x \in \mathbb{K}^{n}$ a non zero vector for which:
$$
Ax = \lambda x
$$

[stoic python]

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

[winter harbor]

In this case, we say that $x$ is an eigenvector for the eigenvalue $\lambda$.

[stoic python]

MisterSystem

[winter harbor]

So, the definition is quite simple, right?

Notice that we can calculate eigenvalues fairly easily using determinants.

If $\lambda \in \mathbb{K}$ is an eigenvalue for $A$ and $x \in \mathbb{K}^{n}$ is an eigenvector for $\lambda$ we have that:
$$
Ax = \lambda x \iff Ax + \lambda x = 0 \iff (A- \lambda I_{n})x = 0
$$
Notice that in this case, the transformation $A - \lambda I_{n}$ has a non zero vector for which it vanishes, i.e it is not injective/bijective; thus has non zero determinant. And indeed, the eigenvalues of $A$ are all of the scalars $\lambda \in \mathbb{K}$ for which:
$$
\text{det}(A-\lambda I_{n}) = 0
$$

[stoic python]

MisterSystem

[winter harbor]

The polynomial $\text{det}(A - \lambda I_{n})$ has a name, it is called the characteristic polynomial of $A$, denoted:
$$
\chi_{A}(\lambda) = \text{det}(A - \lambda I_{n})
$$

[stoic python]

MisterSystem

[winter harbor]

And ofc, in this case the zeroes of the characteristic polynomial will be the eigenvalues of $A$.

[stoic python]

MisterSystem

[winter harbor]

Notice that what this problem is giving us

Is the characteristic polynomial of A

And the eigenvalues of $A$ too

[stoic python]

MisterSystem

[winter harbor]

Those would be the zeroes of the characteristic polynomial, which in this case are a,b and c.

So what the problem boils down to.

Is proving that when a matrix has a zero eigenvalue, then it is not invertible. And the converse too, i.e if a matrix is not invertible then it has a zero eigenvalue.

[winter harbor]

And this is easy to verify.

Because notice that if $A$ has a zero eigenvalue, then there exists $x \neq 0$ for which $Ax = 0 \cdot x = 0$ so $A$ has non trivial kernel, thus not invertible. Conversely, if $A$ is not invertible, then there exists $x \neq 0$ for which $Ax = 0 = 0 \cdot x$ and $0$ is an eigenvalue of $A$.

[stoic python]

MisterSystem

[winter harbor]

And we are done.

Since you haven't covered eigenvalues and eigenvectors yet.

[winter harbor]

Since you haven't covered eigenvalues and eigenvectors yet, it would be good to verify that the eigenvalues of $A$ are precisely the zeroes of the characteristic polynomial.

[stoic python]

MisterSystem

[winter harbor]

But that's the whole idea of the proof tho.

If you get stuck proving that the roots of the characteristic polynomial are precisely the eigenvalues of A, even tho I have hinted how to prove this, here's a stack exchange post:

https://math.stackexchange.com/questions/2567988/proof-of-roots-of-characteristic-polynomial-are-eigenvalues/2568001

[errant mist]

How can I create a Kernel function from this feature mapping?

[winter harbor]

This is Machine Learning stuff, right?

[errant mist]

yes, thats right

[winter harbor]

Idk anything about data science/machine learning lol.

Idk what the definition of a feature map or the kernel function of a feature map is

[errant mist]

haha, was unsure where to post it

[winter harbor]

I could take a look tho

[winter harbor]

Maybe #numerical-analysis ?

[fleet sage]

@winter harbor They just want an inner product that produces $\phi$

[stoic python]

StellaAthena

[errant mist]

right, so we give some two vectors x,y,z which produces the inner product?

[errant mist]

never even seen this one!

[winter harbor]

Ah, so you mean finding a positive define bilinear map
$$
\mathcal{K} : \mathbb{R}^{3} \times \mathbb{R}^{3} \rightarrow \mathbb{R}
$$
For which:
$$
\mathcal{K}(u,v) = \langle \phi(u), \phi(v) \rangle
$$

Is this the definition?

[stoic python]

MisterSystem

[fleet sage]

Yup. The idea is that a kernel allows you view the data as existing in some larger latent space, which you are only seeing a particular slice of. Often times $K$ is easier to compute than $\phi$, or at least easier to learn from data

[stoic python]

StellaAthena

[winter harbor]

Ah, alright. I am particularly ignorant about data science stuff. So I wouldn't be able to know if that was correct or not myself.

Thanks Stella!

[fleet sage]

Any day I can clarify a ML topic mathematically is a good day, lol. I've spent the last couple years fighting through pedagogical resources designed for people with undergrad CS degrees rather than graduate math degrees in my quest to figure things out.

[errant mist]

Right, so if you have some non-linear data then there may be way to represent this in some easier fashion i.e as inner products in a higher dimensional space, but if you just have linear data then it may not be as useful and perhaps take a longer time to compute?

[fleet sage]

If your data is linear, use a linear regression and call it a day

[winter harbor]

@errant mist If what I understand is correct, then you will need to compute this for a given choice of inner product $\langle \cdot, \cdot \rangle$ on $\mathbb{R}^{5}$. So for instance, you could take the Euclidean inner product on $\mathbb{R}^{5}$ and then compute for a given $(x,y,z) \in \mathbb{R}^{3}$ and $(x',y',z') \in \mathbb{R}^{3}$ the following:
\
\
$
\mathcal{K}( , (x,y,z), (x',y',z') ,) =
\
\
\langle \phi(x,y,z), \phi(x',y',z') \rangle
\
\

\
\
\langle (x,y,z,xy,xz), (x',y',z',x'y',x'z') \rangle

\
\
xx' + yy' + zz' + xyx'y' + xzx'z'
$

Damn

[stoic python]

MisterSystem

[wintry steppe]

interesting formatting

[winter harbor]

Horrible, right?

It's that I can't use $$ $$ ffs

[fleet sage]

Why can't you?

[winter harbor]

It appears that what I was inputing inside the double $$ was too big.

And it is not formatting correctly

I don't want to think about this too much tho lmao

This is good enough I guess

[errant mist]

never mind the formatting , tnx for the answer. )

[winter harbor]

Yeah, from what I understand you can also take different inner products and this gives you different kernels for your featuring map.

And sometimes taking different inner products may be easier to vizualize data I guess??

[errant mist]

Im also asked to give an example of a binary function that is not a positive definite Kernel. Any ideas how to approach this?

[fleet sage]

@winter harbor Typically we choose the inner produce in a way that induces a geometry that is useful for the problem. The naive embedding in R^k that our measurements produced likely obscures aspects of the data

[winter harbor]

Just find an example of a matrix $A$ that is not positive definite and define:
$$
\mathcal{K}(u,v) = \varphi(u)^{t} A \varphi(v)
$$

[stoic python]

MisterSystem

[winter harbor]

I think this might do the job.

[winter harbor]

I see

[left snow]

\begin{align*}
\mathcal{K} ( (x,y,z), (x', y', z') ) &= \langle \phi(x, y, z) , \phi(x', y', z') \rangle \\
&= \langle (x,y,z,xy,xz), (x',y',z',x'y',x'z') \rangle \\
&= xx' + yy' + zz' + xyx'y' + xzx'z'.
\end{align*}

[stoic python]

ricey

[fleet sage]

Woah, I didn't lnow you could do that

```latex
\begin{align*}
\mathcal{K} ( (x,y,z), (x', y', z') ) &= \langle \phi(x, y, z) , \phi(x', y', z') \rangle \
&= \langle (x,y,z,xy,xz), (x',y',z',x'y',x'z') \rangle \
&= xx' + yy' + zz' + xyx'y' + xzx'z'.
\end{align*}
```

amazing

[errant mist]

@winter harbor so the feature space may be comprised of some complex inner space in which we define a function?

[winter harbor]

Oh shit, thanks.

[errant mist]

but we need to involve some inner product $\phi(u) \phi(v)$?

[stoic python]

Fredrikpiano

[winter harbor]

I don't know exactly what you mean by that.

But I just gave an example of how to construct a map that is not positive definite.

[errant mist]

Im just trying to get some intuition behind what properties the functions must fulfill in the higher dimensional space.

I.e why we could insert the matrix you gave me in the example?

[winter harbor]

Because if we have a finite dimensional vector space over a field $\mathbb{K}$, which is our case, endowed with a bilinear map $\psi : V \times V \rightarrow \mathbb{K}$, then we can always construct a square matrix $A$ for which:
$$
\psi(u,v) = u^{t} A v
$$
In particular, if we have a real vector space endowed with an inner product $\langle \cdot, \cdot \rangle$, then $A$ must be a symmetric positive definite matrix. Conversely, for any given symmetric and positive definite real matrix we can define an inner product via $(u,v) \mapsto u^{t} A v$.
\
\
Therefore, there's a correspondence between the inner products on a finite dimensional real inner product space and symmetric positive definite real matrices. So, in particular, if we take any matrix that is not positive definite, then $(u,v) \mapsto u^{t} A v$ does not define an inner product on $V$.

[stoic python]

MisterSystem

[errant mist]

so this then entails that our kernel function will not be comprised of an inner product choosing this mapping?

[winter harbor]

Yeah, I guess that's a way to put it.

[errant mist]

thanks, became a lot clearer once I looked up bilinear map on wikipedia. Binary function just means in this context a function that takes two vectors as input right?

[winter harbor]

I guess so, tbf I don't see a lot of mathematicians using this terminology tho.

[errant mist]

So we have two vectors in our kernel function $K(x,y)$ which are non separable which may be linearly separable in higher dimensions which in some cases might result in an easier result to interpret?

[stoic python]

Fredrikpiano

[winter harbor]

But it usually means a function that takes two inputs, yeah.

[winter harbor]

Idk that much about the interpretations tbh

Can't help with that.

[errant mist]

thats ok. You helped a lot

would numerical linear algebra go in this thread BTW?

[split jewel]

A base for this vectorial space?

basis"

[left snow]

[ \beta = \sbn{\cvec{0 & 1 \ 0 & 0}, \cvec{0 & 0 \ 1 & 0} } ]
Does this work?

[stoic python]

ricey

[split jewel]

These are the options:

[winter harbor]

I think so, even tho I suppose #numerical-analysis would be more helpful for this sort of thing.

[errant mist]

@winter harbor got it, thanks again!

[rotund verge]

Hello

The eigen vectors are imaginary

What will be its effect on graph

Can someone help me graoh it🤧

I dont have any idea

[winter harbor]

Are you familiar with the fact that square matrices M correspond to linear maps M : R^n -> R^n

The interpretation of imaginary eigenvalues is usually that this linear map acts on the space R^n as a sort of "rotation"

And this is exactly what is happening here

[rotund verge]

yeah i notice that it rotates 90 degree

but what will be its implicationon graph?

[winter harbor]

Notice that $0 = \text{cos}(\pi/2)$ and $1 = \text{sin}(\pi/2)$. So equivalently, we have that:
$$
M

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

\begin{bmatrix}
\text{cos}(\pi/2) & \text{sin}(\pi/2) \

• \text{sin}(\pi/2) & \text{cos}(\pi/2)
  \end{bmatrix}
  $$

[rotund verge]

oh yes

thtt

[winter harbor]

So M corresponds to the matrix that rotates the plane by 90° degrees

[rotund verge]

ahhhhhhhhhhhhh

[lavish jewel]

careful with the 1 on the lower right

[winter harbor]

Oh

Sorry

[stoic python]

MisterSystem

[winter harbor]

Corrected

[rotund verge]

omggghuhuhu

[winter harbor]

But the idea is that finding the eigenvalues gives us the vectors which are only stretched by the transformation.

None of them is being stretched because they are all being rotated by 90°

That's the idea, basically.

Notice how multiplying a complex number by "i" is also thought of as rotating that complex number by 90°

So all these notions are related

[rotund verge]

ahhhhh i see, now its clear to mee, so its allrotating andnotbeing stretched

THANK U SOMUCH , OUR TEACHER DIDNT TEACH US THIS😩

[winter harbor]

Np

[split jewel]

Okey, forget about my last question, i need help with this

What is the dimension of this vector subspace?

Options: 1,2,3 or 4

[marble lance]

Try to write what a matrix in this subspace looks like

How many arbitrary parameters does it depend on?

Then you can expand with those parameters to get a basis

[split jewel]

it depends of 2 parameters: a11 and a22

i call them x and y

then do: 44x + y = 0

x(44,0)+y(1,0)=0

So i think the answer is 2, am i wrong?

[marble lance]

Yes

A matrix looks like

$$A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}$$

[stoic python]

Lunasong the Supergay

[marble lance]

But based on the conditions they give

a11 and a22 are related

So we should replace one of them

[split jewel]

but, for example:

$$A = \begin{bmatrix} 1 & 0 \ 0 & -44 \end{bmatrix}$$

[marble lance]

Two \ to go to next line

[split jewel]

$$A = \begin{bmatrix} 1 & 0 \ 0 & -44 \end{bmatrix}$$

[stoic python]

Tacos

[split jewel]

This matrix for example, you can multiplicate by any number and still part of the subspace

[marble lance]

Yes

[split jewel]

So, the thing is, for any a11, a11 * 44 - a22 have to be = 0 and that is enough for being part of the supspace

subspace"

So you can not determinate if a matrix is part of this or not just with 1 parameter

you need a11, and a22, right?

[marble lance]

No

You only need one of those

Because the one determines the other one

By that equation

You also need a12 and a21

[split jewel]

So at the end, maybe the dimension of this subspace is 1, because with 1 of this parameters we can determinate if it is part or not of the subspace?

[marble lance]

No

Matrices in the subspace are those of the form

\begin{align*}
A
& = \begin{bmatrix} a{11} & a{12} \ a{21} & -44a_{11}\end{bmatrix} \
& = a_{11} \begin{bmatrix} 1 & 0 \ 0 & -44 \end{bmatrix} + a_{12} \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} + a_{21} \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}
\end{align*}

[stoic python]

Lunasong the Supergay

[split jewel]

Oooo!! i see, because a22 needs to be = -a11 so it can be part of the subspace right?

i mean

=-44a11

[marble lance]

This means that our subspace is the span of those three matrices at the bottom

Yes

And since those 3 matrices are linearly independent, they form a basis

[split jewel]

So by definition, we have 3 vectors, so the dimension is 3 right?

[marble lance]

Yes

[split jewel]

Thank you! Now i understand

[marble lance]

[wintry steppe]

Anyone know the answer to this? I thought it was true by gram-schmidt but apparently there exists a counterexample somehow

the question doesnt even make sense

what is a "basis for these vectors"

basis containing those vectors?

[stable kindle]

use euler's identity

[rotund verge]

Does someone have an idea what properties should i use this is so complex

I tried findinf the determinant of the left hand side of the equation

And i got this

Now my problem is how would i equate that into the right hand side fo the equation

[winter harbor]

damn lmao

You don't ned this

[rotund verge]

So complicated😭

Really

That is why im wondering what property to be use and there are so many

[winter harbor]

Are you familiar with the fact that the determinant is alternating multilinear with respect to rows/columns?

[rotund verge]

Oobb

I didnt

[winter harbor]

you probably haven't heard about this property with these terms

[rotund verge]

I never heard

[winter harbor]

https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/properties-of-determinants/MIT18_06SCF11_Ses2.5sum.pdf

But basically the first 3 properties that appear here

I am very lazy rn so I don't want to write down the properties on LaTeX lol

But basically

[winter harbor]

You will have to use these in order to simplify this determinant

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

This is a nice video that presents these properties of the determinant

an indeed, I think Peyam even proves that the determinant is the unique multilinear alternating form on the rows of a square matrix such that on the identity it equals 1.

This is what defines the determinant btw

But if you want to watch th video just to get familiar with these properties, it is good enough.

[rotund verge]

Ooohhhh okiee okieee thankk u so muchhhh ksksks

[rotund verge]

Is set of all orser 5 scalar matrices a vector space?

Order

[limber sierra]

sorry, what does order of a matrix mean?

[rotund verge]

Order means the size or dimension

[spare widget]

You mean 5x5 matrices?

[limber sierra]

thats what i thought, but then "order 5" makes no sense

do you mean 5 x 5?

[spare widget]

Mmxn(F) over a field F is a vector space

[limber sierra]

in any case, yes; matrices of a fixed size over a field form a vector space

to do this, note $M_{a\times b}(F) \cong F^{ab}$ by an easy map

[stoic python]

Namington

[rotund verge]

Oohh

How about

[limber sierra]

so $M_{5\times 5}(\mathbb{R})$, for example, is isomorphic to $\mathbb{R}^{25}$

[stoic python]

Namington

[limber sierra]

all we did is write the vectors as squares instead of columns

[rotund verge]

Does this also meams that a set of order 3 transition matrices is a vector space too?

[limber sierra]

is the set of transition matrices closed under addition and scalar multiplication?

[spare widget]

If you can prove that a subset is closed under addition and sc mult and contains 0 then it is a vector space

[limber sierra]

contains 0 follows from the other properties btw as long as you know its nonempty

[spare widget]

yeah non-empty is necessary for that to follow

it's equiv either way

[rotund verge]

Ah so it depends

I mean

Probability matrices

Like the markovs

[spare widget]

doubly stochastic?

[rotund verge]

Ahm never heard of that term

[spare widget]

https://en.m.wikipedia.org/wiki/Stochastic_matrix

It's not a vector space

the 0 matrix is not a stochastic matrix

[rotund verge]

Ahhh

Okieee i see

Yeh theres no probability such as null matrix

Thankk uu

This is what i mean btw by scalar

The values in the proncipal diagonal are the same

And they are both lower and upper triangular matrices

[spare widget]

Is that a question?

I am not sure

[rotund verge]

It is still question

[spare widget]

Then I didn't understand what the question is

Do triangular matrices form a vector space? Yes. Do diagonal matrices with the same diagonal form a vector space? Also yes, and so on.

[rotund verge]

Ah okie

I get it now

[spare widget]

There are some basic methods you can use:

1. Subspace: if you have a vector space V, and a subset U of V, then U with the same operations is a vector space if it is closed under addition and scalar multiplication and contains the zero element.
2. Isomorphism: find a bijective map between your subset and a vector space
3. Axioms: show that the axioms for a vector space hold

[rotund verge]

Ahh yeah there ar emany axioms🤧

Last question

How would i know if a matrix is decomposable to L and U

[spare widget]

https://en.m.wikipedia.org/wiki/LU_decomposition

See existence and uniqueness

[rotund verge]

Ahh okie so theres no pattern

U mist have to solve to know

[vocal dock]

is A trivially a symmetric matrix?

or do I have to show that A^-1 exists?

[stable urchin]

does the same go for rref(A)?

[slow scroll]

What does Row(A) mean?

[stable urchin]

rowspace of A

the subspace spanning all the rows of A as vectors

[slow scroll]

Ah okay. Anyway, the answer is yes

The intuition is that the row space is closed under applying row operations, so as long as the nonzero rows you get in the end are linearly independent, those rows will form a basis for the row space

[stable urchin]

That makes sense

thanks for the clarification

[slow scroll]

Npnp

[winter harbor]

No, it is not necessarily true that $A$ is invertible. $A = 0$ satisfies this condition, for instance.

[stoic python]

MisterSystem

[winter harbor]

Hmmm

I have something in mind tho

[vocal dock]

yea i've been thinking about that

thank you 🙂

[stoic python]

MisterSystem

[spare widget]

There are some other fun solutions here: https://math.stackexchange.com/questions/571583/if-matrix-such-aat-a2-then-a-is-symmetric/571602

[winter harbor]

There's one using symmetric + antisymmetric decomposition too

The second is nice, I was thinking about proving that A and A^t commute too

[stoic python]

MisterSystem

[wintry steppe]

let's say you have V and W where dim(V) = n and dim(W) = n + 1 you can have a linear transformation T : V --> W right?

[winter harbor]

yeah, between any two vector spaces you always have a linear transformation between them, just set T(v) = 0 for all v in V.

I guess you are asking about something more specifically

For instance

We can't have a surjective linear transfomration T : V -> W since dim(W) > dim(V)

(As a corollary of rank-nullity)

We can always construct an injective linear map T : V -> W tho

just pick a basis {e_1, ..., e_n} for V and a basis {b_1, ..., b_n+1} for W

and set T(e_1) = b_1, ..., T(e_n) = b_n

So constructing an injective linear transformation is always possible

[grave oasis]

I just started learning some linear algebra and 3Blue1Brown’s videos are really helpful for visualizing things. My question is about pseudoinverses. I just learned about them and am wondering if there’s a simple geometric explanation of them (for noninvertible matrices), like how the inverse of an invertible matrix is “undoing” the transformation of the original matrix. Is the “geometric construction” on here https://en.m.wikipedia.org/wiki/Moore–Penrose_inverse the best explanation of it? Or does it have something to do with least squares?

It’s also entirely possible I don’t know enough to fully understand them. Or there is no “simple” geometric interpretation

[winter harbor]

Are you familiar with the method of least squares?

[grave oasis]

I took a stats class a while back, so vaguely. I was reading about it now

[winter harbor]

The Moore Penrose Inverse of a matrix naturally appears when one is trying to find the best fit solution to a linear equation via the least squares method.

Indeed, it's almost like the Moore Penrose Inverse is just the ''matrix point of view'' of the least squares method.

[grave oasis]

Thank you! I am watching the Khan Academy video on least squares method now