linear-algebra

[vast torrent]

T is injective iff ker(T)={0}

[vapid coyote]

this is still going on

or is it different?

[vast torrent]

Iff means if and only if

Different

[wintry steppe]

this is about kernel and range now

[vast torrent]

That's a very important theorem and you should memorize it

[wintry steppe]

i see

[vast torrent]

Lemme see if i can find a khan academy video proving it

[wintry steppe]

i know about injective/surjective/bijective from analysis/calc

[vast torrent]

https://proofwiki.org/wiki/Linear_Transformation_is_Injective_iff_Kernel_Contains_Only_Zero lol i wrote this in 2012

[vapid coyote]

I wouldn't really call this a theorem haha

it's almost true per definition

[vast torrent]

It's a theorem

[vapid coyote]

Technically sure

[vast torrent]

It's like how continuity at 0 is enough to show continuity

[vapid coyote]

I just don't think that you need to memorize this as with enough mathematical understanding you will realize this over and over again

[vast torrent]

Yeah sure but you get that huge t.f.a.e. for invertible operators near the end of lin alg 1

[vapid coyote]

it even shows lipschitz continuity :p

t.f.a.e?

[vast torrent]

The following are equivalent

And then like 20 ways to descrive invertible transformations

[vapid coyote]

I mean

my linear algebra class was probably quite different from yours

so that doesn't tell me anything

at the end of linear algebra 1 we did eigenvalues and that stuff

[vast torrent]

Injective iff surhective iff row reduce to identity iff kernel is trivial iff bijective iff invertible iff full row rank iff full column rank

That one

[vapid coyote]

that was like a couple weeks into the semester

not at the end

[vast torrent]

Whatever!

[vapid coyote]

It's kind of irrelevant what people consider a theorem or not anyway

it's just my personal feeling

that a theorem has to be something non-obvious

but that always dependso n the person anyway

I feel like the math you did two months ago is always easy to you now haha

not always though

[wintry steppe]

ok so like the thing im confused about rn is

i mean, x1 and x2 seem to be 0 and when i apply the transformation to that vector it results yet again in 0 so thats my kernel then?

or like a basis

[vapid coyote]

the kernel is a subvectorspace, so it does have a basis

if the kernel is ${0}$ then its basis would be the empty set

[stoic python]

mophra:

[vast torrent]

Wouldn't its basis be {0}

[vapid coyote]

your kernel is the set of all vectors that map to 0

that can't be a basis since the nullvector is linearly dependent

[vast torrent]

Ah okay

[wintry steppe]

ah i can find a basis by augmenting the matrix with the identity matrix

[vapid coyote]

so e.g. if your linear map is given by f(x,y)=x-y

then your kernel would be spanned by (1,1)

so all vectors of the form (x,x)

[wintry steppe]

right

[vapid coyote]

so if you want you could solve a linear equation system

for the kernel

[wintry steppe]

isnt it enough to get the matrix in REF form for the kernel?

how does that work then? ill have something like

[vast torrent]

Ref won't change the kernel

[wintry steppe]

$ \begin{bmatrix}
3 & -5 \
0 & \frac{14}{3} \
0 & 0
\end{bmatrix}
\begin {bmatrix}
x_1 \
x_2
\end{bmatrix} $

[stoic python]

yami:

[wintry steppe]

now how do i know what my x1 and x2 is

[vast torrent]

To set to 0?

The second row tells you x2=0

Because 14/3 x2=0

[wintry steppe]

oh fuck LOL

[vast torrent]

So the first row simplifies to 3x1+0=0

Which means x1=0 too

[wintry steppe]

yeah ok somehow i got around to x2 = -14/3

lol

[vast torrent]

Bzzzt

[wintry steppe]

okay then, at least i did an hour of research on kernels

ok so then my range is R2?

although given that it maps to R3 that makes me skeptical

[vast torrent]

Yeah R2 isnt a subspace of R3

But

If the range is 2 dimensional

Then the range is "very similar" to R2

Any plane in 3d space is very similar to the xy plane, or the yz and xz planes

[vapid coyote]

the kernel is a subspace of the domain

not the range

[vast torrent]

The image has to sit inside the codomain

That's true for any function

[vapid coyote]

of course

so I guess there is a misunderstanding of dimension of a subspace and dimension of the ambient space?

[vast torrent]

So if the codomain is R^3 the range cant be R^2 but it can be something similarly shaped

Yami said that he got the image as R2 but the codomain is R3 so he was skeptical, as he should be

[vapid coyote]

maybe he just got that one component is always zero or sth.

[vast torrent]

Well he got that the image is 2 dimensional

[vapid coyote]

which makes sense if the map was from R^2 to R^3

[wintry steppe]

well the basis for the range of F is 2dim

[vast torrent]

Yes

[vapid coyote]

you mean the basis has 2 elements in it

[vast torrent]

So the range is a plane

[wintry steppe]

$ S = \left{ \begin{bmatrix}
0 \
1 \
3
\end{bmatrix},
\begin{bmatrix}
-1 \
3 \
-5
\end{bmatrix} \right} $

this is the basis

[vapid coyote]

I would much rather write vectors separated by commas haha, but I'm way too lazy to put that into latex

and only the first one is the basis

interpreted as two vectors

[wintry steppe]

its supposed to be a set i guess

[vapid coyote]

$\left{ \begin{pmatrix}
0
1 \
3
\end{pmatrix},
\begin{pmatrix}
-1
3 \
-5
\end{pmatrix}
\right}

[vast torrent]

\ { \ }

[stoic python]

mophra:
Compile Error! Click the reaction for details. (You may edit your message)

[vapid coyote]

yikes

what does that not work haha

why*

[wintry steppe]

there u go

[vapid coyote]

I want the brackets to be larger

[vast torrent]

You need \ \ for next line

[wintry steppe]

sigh

[vapid coyote]

and you usually do that with \left and \right

for them to automatically align

[vast torrent]

\ left\ { { } \right \ }

[vapid coyote]

$\left{ \begin{pmatrix}
0 \
1 \
3
\end{pmatrix},
\begin{pmatrix}
-1 \
3 \
-5
\end{pmatrix}
\right}

[stoic python]

mophra:
Compile Error! Click the reaction for details. (You may edit your message)

[vapid coyote]

closer

[stoic python]

yami:

[wintry steppe]

r we happy now

[vapid coyote]

not yet

[wintry steppe]

the -

[vapid coyote]

$\left{ \begin{pmatrix}
0 \
1 \
3 \
\end{pmatrix},
\begin{pmatrix}
-1 \
3 \
-5 \
\end{pmatrix}
\right}

[stoic python]

mophra:
Compile Error! Click the reaction for details. (You may edit your message)

[vast torrent]

Then the range is span(S)

[vapid coyote]

why do I continue fucking this up

$\left{ \begin{pmatrix}
0 \
1 \
3 \
\end{pmatrix},
\begin{pmatrix}
-1 \
3 \
-5 \
\end{pmatrix}
\right}

[wintry steppe]

ur missing a \ and a $

[stoic python]

mophra:
Compile Error! Click the reaction for details. (You may edit your message)

[vast torrent]

Two slashes for next line

[wintry steppe]

pmatrix ugly btw

[vapid coyote]

I do that

pmatrix looks way better

$\left{ \begin{pmatrix}
0 \
1 \
3 \
\end{pmatrix},
\begin{pmatrix}
-1 \
3 \
-5 \
\end{pmatrix}
\right}$

[stoic python]

mophra:

[vapid coyote]

$\left{ \begin{pmatrix}
0 \
1 \
3 \
\end{pmatrix},
\begin{pmatrix}
-1 \
3 \
-5 \
\end{pmatrix}
\right}$

[stoic python]

mophra:

[vapid coyote]

there we go

[vast torrent]

The range is all linear combinations of these

[wintry steppe]

those span r2 no?

[vapid coyote]

no they don't

[wintry steppe]

because they have 3 elements?

[vapid coyote]

they span a 2-dimensional subspace

[wintry steppe]

fuck

[vapid coyote]

yes exactly

[wintry steppe]

ok at least i understand that now

[vapid coyote]

In a specific rigorous sense you are correct

this is precisely R^2

but they are not equal as sets

they are isomorphic

which you will probably learn soon

[wintry steppe]

does a subspace always imply that its a lower dimension?

[half ice]

No, for example, every space is a subspace of itself

[wintry steppe]

subset or subspace?

[half ice]

Sorry I meant space lol

[wintry steppe]

that makes sense because R3 has R2 and R1

has is a very weird word to use here but i think u get what i mean

[half ice]

R³ is a subspace of R³

[wintry steppe]

oh

[half ice]

As an example

Nothing complicated. Without talking too much about the axioms, the space itself is a subspace of the space

[vapid coyote]

R2 is not a subspace of R3 in the sense we're using here

neither is R1

[vast torrent]

@wintry steppe it's like how

In the xyz space

The plane z=0 corresponds to all points (x,y,0)

This is very similar to points (x,y) in R2

But not equal

[wintry steppe]

yeah

[wintry steppe]

,tex $ \boldsymbol{T} : M (2 \times 2) \rightarrow M (2\times 2) \ \
\boldsymbol{T}
\begin{bmatrix}
a & b \
c & d
\end{bmatrix}

\begin{bmatrix}
a & 0 \
0 & d
\end{bmatrix} $

[stoic python]

yami:

[wintry steppe]

$ \begin{bmatrix}
0 & 4 \
2 & 0 \
\end{bmatrix} \in ker(T) $

[stoic python]

yami:

[wintry steppe]

is that right

[slow scroll]

sure

[wintry steppe]

then is this true as well?

$ \begin{bmatrix}
3 & 0 \
0 & -3 \
\end{bmatrix} \in im(T) $

[stoic python]

yami:

[wintry steppe]

given that the basis seems to be the r2 identity matrix

[vast torrent]

Uh

Not quite

a can be different than -d

one choice for the basis of your image would be:

if you wanted to have your matrix just have 0s and 1s like the standard basis vectors R^n

[stoic python]

gfauxpas:

$\mathcal B = \left\{ {\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} ,{\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}  } \right\}$
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.11 ...ix} 0 & 0 \ 0 & 1 \end{bmatrix} } \right
}$
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[wintry steppe]

wait why is it 2 2x2 matrices

[vast torrent]

if you wanted a basis with 0s and 1s in it, that's it

well what's the codomain?

[wintry steppe]

oh

[vast torrent]

so what's the dimension of the image(=range) here?

2

because a basis for the image has 2 vectors. the vectors are matrices here, but the dimension is still the number of vectors in a basis

[wintry steppe]

because a basis for the image has 2 vectors
2 because of the dimension, yes?

[vast torrent]

yes that's the definition of dimension

how many vectors are in a basis

of that space

[wintry steppe]

hmm

so are we just providing 2 bases or whats the catch

[vast torrent]

what question are we trying to answer now

[wintry steppe]

that matrix above was in im(T) yes?

[vast torrent]

it was an element of the image of T, yes

[wintry steppe]

ok then its just about showing ker(T) and im(T) by providing a basis for each one of them

[vast torrent]

well if the image is two dimensional you need 2 vectors to form a basis

I found one by considering matrices that only have 0s and 1s in them and just observing that

$a \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} + d {\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} = {\begin{bmatrix} a & 0 \ 0 & d \end{bmatrix} $

[stoic python]

gfauxpas:

$a  \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + d {\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}  = {\begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} $
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.11 ...in{bmatrix} a & 0 \ 0 & d \end{bmatrix} $

I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[wintry steppe]

oh

that makes sense now lol

[vast torrent]

bases arent unique

I just wanted one that was all 0s and 1s

[wintry steppe]

yeah i know that i just didnt know what u actually meant by a can be diff from -d

[wintry steppe]

is there a way to (approxmiately) tell what a transformation matrix will do with my vectors?

[quartz compass]

determinant of a matrix will tell you by how much it'll scale a unit square, well, roughly speaking

[feral mountain]

for 2x2 the columns are where i and j get sent, so you can use that to imagine the transformation

[wintry steppe]

hmm

im having a rather hard time imagining where stuff will go when i have something like

$ \begin{bmatrix}
\tfrac{-\sqrt2}{2} & \tfrac{-\sqrt2}{2} \
\tfrac{\sqrt2}{2} & \tfrac{-\sqrt2}{2}
\end{bmatrix} $

[stoic python]

yami:

[quartz compass]

one way to think of linearity is

if you know where your basis vectors go, then you know where all vectors go

because they're just all scalar multiples of the columns of the transformation matrix added together

maybe it helps to think, suppose you know where a single basis vector v goes to, it ends up at Av

so where do its multiples end up?

A(kv) = k(Av)

they're all just scalar multiples of that

similarly, if you know where some other basis vector say u goes, to Au

then where does u+v go? well of course to Au + Av

you'll also eventually learn about something called eigenvalues and eigenvectors

[wintry steppe]

hmm, i feel like i cant fully grasp what youre saying because im not far enough yet

and yeah ive already heard those terms getting thrown around plenty

[quartz compass]

in some sense this is the natural choice for a specific linear operator because in that basis all your transformations are just scaling

[vast torrent]

the matrix you posted yami is easier to understand if you factor out sqrt(2)/2

since a scalar times a matrix is all of the entries multiplied by the scalar

[wintry steppe]

right, yeah

[quartz compass]

in some sense I'm biased, for that specific matrix the sqrt(2)/2 makes it kind of clear that it's probably some multiple of a 45 degree rotation, then by throwing (1,0) or (0,1) at it you can determine where it actually lands

[wintry steppe]

the x axis acts as a mirror for that matrix

[vast torrent]

also the columns and rows are unit length

[wintry steppe]

or like a shadow if you will

[vast torrent]

that's nice

another reason to keep the sqrt(2)/2 in

[clever cedar]

having trouble solving this

i found that the matrix A is nontrivial

but i dont know how to find two vectors that result in the same transformation

any hints would be appreciated

[dusky epoch]

what do you mean by nontrivial

[clever cedar]

there is more than one solution

when i setup and RREF a homogenous augmented

[dusky epoch]

what do you mean by more than one solution

more than one solution to what

[clever cedar]

to the system of linear equations

[dusky epoch]

what system

[clever cedar]

0 - 3y -6z = 0

0 + 1y + 2z = 0

etc

[dusky epoch]

what system

maybe if you write it in matrix form it'll be clearer to you how to proceed

[clever cedar]

Thats what I have done. I took A and RREF

I saw that the RREF had infinitely many solutions

[dusky epoch]

why did you RREF

matrices don't have solutions

systems of equations do

what was your original system of equations, in matrix form

[clever cedar]

the one presented in the image above denoted as A

[dusky epoch]

i said

system of equations

not matrix

[clever cedar]

i figured it out

just solve the system into RREF

then plug in the free variable

[dusky epoch]

h

[north sierra]

lol

i need help

[dusky epoch]

@north sierra?

[fiery idol]

are orthogonal and perpendicular the same meaning

they are both in straight lines right

[wintry steppe]

umm

[dusky epoch]

?

[brittle orchid]

Are there some good online YouTube playlists for university level linear algebra to supplement reading material?

None of my Mathss university lectures are online 😄

[dusky epoch]

3b1b, essence of linear algebra

[undone garnet]

any idea for this problem?

[dusky epoch]

$A$ is posdef and $B$ is negdef, so $x^TAx$ is always positive and $x^TBx$ is always negative for $x \neq 0$

[stoic python]

Ann:

[undone garnet]

😮

thank you

[royal timber]

If a vector X in base B is

And vector X in base B' is

and S - change of basis matrix from B to B'

Is this relation true?

Shouldn't it be S instead of S^(-1)?

[vast torrent]

seems like a matter of convention more than that of whether it's true as a theorem. let me look it up quick

@royal timber what are the columns of S?

[vast torrent]

I see

so you're looking for solutions to

$\sum c_i \mathbf b_i = \sum c'_i \mathbf {b'}_i$

[stoic python]

gfauxpas:

[vast torrent]

the left is a vector c written in terms of B, and the right is a vector c written in terms of B'

right?

[royal timber]

yes

[vast torrent]

I'm finna factor both sides

let me work out the latex and ill be back

$\begin{bmatrix} \ \mathbf b_1 & \mathbf b_2 & \cdots & \mathbf b_m \ \ \end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ \vdots \ c_n \end{bmatrix} = \begin{bmatrix} \ \mathbf b'_1 & \mathbf b'_2 & \cdots & \mathbf b'_m \ \ \end{bmatrix} \begin{bmatrix} c'_1 \ c'_2 \ \vdots \ c'_n \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

I just factored out the matrix

[royal timber]

And I need to find a matrix that multiplied to the lefthand side gives the right hand side

[vast torrent]

well it's going to be

$\begin{bmatrix} \ \mathbf b'_1 & \mathbf b'_2 & \cdots & \mathbf b'_m \ \ \end{bmatrix}^{-1} $

isn't it?

[stoic python]

gfauxpas:

[royal timber]

maybe

[vast torrent]

something seems off

[charred stirrup]

I am having trouble understanding literal matrix algebra in the forms of equations

I get that I should be "reading" the transformations right to left

But what does that mean in terms of operations?

For example, should this be read as "Square the matrix B, inverse it, and the multiply it with the tripled transpose of A ? "

[gray dust]

sure, then take the det of whatever that is

[charred stirrup]

do you have any recommendations on practicing the rules?

I do not know why my textbook does not cover any of it

[gray dust]

btw it's called "left-multiplying" $(B^2)\inv$ by $3A^T$

[charred stirrup]

I really got lucky on my last assignment

[stoic python]

RokettoJanpu:

[charred stirrup]

like for both of these, I completely guessed

how does matrix algebra work?

Like dividing or multiplying over a matrix

I dont even have the right term

googling "matrix algebra" gives me algebra with within matrices

not in terms of them

[half ice]

It's just like regular algebra with real numbers, except with a few changes.

• Multiplication is not commutative. AB is not BA. You have to keep the order of the matrices consistent.
• There is no matrix division. Instead, think of multiplying by the inverse

[charred stirrup]

thank you fam I love you

I noticed in the answer I got right

moving matrix "B" and matrix "A" over and turning it into their inverses made them end up in different spots

why is that?

[wind yacht]

https://math.stackexchange.com/questions/2922099/rearranging-matrix-equation-and-solving-isolating-for-x-matrix

This may help walk through one example

[half ice]

AXB = (BA)²

Multiply both sides ON THE LEFT by A' (let ' mean inverse)
A'AXB = A'(BA)²

XB = A'(BA)²

[charred stirrup]

Oh shit

That's a really good intuitive explanation

[half ice]

Then do it again with B' on the right

[charred stirrup]

I can only cancel out A with its inverse by multiplying it exactly in that order?

[half ice]

A'A = AA' = I

And I goes away when multiplied by anything

[thorn egret]

Can i pretend a matrix is just a set of vectors?

[half ice]

No, but the column space of a matrix IS a set of vectors

[thorn egret]

When looking for a basis of a matrix i just need to pretend every column is a vector and find the basis of the space those vectors span

?

[half ice]

You don't find a basis of a matrix. A basis is a thing that sets of vectors have.

[thorn egret]

yeah but if a vector space is defined as a matrix

like

a b 0
0 0 a

[half ice]

Oh, like if the matricies are the vectors in your vector space?

[thorn egret]

Let V = M(2x3) on R, and U =

a b 0
0 0 a

a subset of V

Prove U is a subspace of V and find a basis of U

i know how to prove its a subspace

(a and b are reals)

[half ice]

Hah, okay. So indeed, in THIS CASE, your "vectors" are actually 2×3 matrices

[thorn egret]

so is a b 0 a vector or is a 0 a vector

and b 0 and 0 a

[half ice]

No, those don't exist here

[thorn egret]

oh

so how do i find a basis of U

[half ice]

I'll give you a basis:
1 0 0
0 0 1

0 1 0
0 0 0

[thorn egret]

Ohhhh

So the entire matrix

[half ice]

Because a(the first one) + b(the second one) is your general vector

[thorn egret]

Is one of the vectors

I get it

thanks

[half ice]

Np, tricky question! Feel free to ask if you have anything else

[thorn egret]

thanks

[arctic fox]

just a quick question, if z is a fixed(but unknown) vector in R2, am I allowed to assign random variables to show z like z=[z1,z2]?

the word "fixed" is kinds throwing me off

[clever cedar]

is the image of a linear mapping just all possible combinations of the product Ax (where x is a column vector)

[desert ibex]

How the heck does my teacher get that as a basis?

I'm racking my brain

[clever cedar]

which one

a) or b)

[desert ibex]

a)

[clever cedar]

lemme try it

[desert ibex]

I found the null space basis, I can find the row and column basis

[clever cedar]

i mean the standard basis is [1,0,0,0] ,[0,1,0,0] and [0,0,1,0]

[desert ibex]

so where does the 13 come from?

or the -9?

[clever cedar]

it should be x + y = 9

no

sorry

[desert ibex]

to solve this aren't I supposed to put the vectors in a matrix as columns and then reduce to echelon form?

[clever cedar]

ya

then the ones that are pivot columns

are linearly independant

the column that isnt, is linearly dependant

this is a linearly dependant set

because the third vector is not independant

oh wait i think i know what he did

sec

[desert ibex]

once I get [-2, -1, 0, 1] as my null basis...

[clever cedar]

wdym ur null basis has to have at least on free variable

[desert ibex]

I should just know that v1, v2, and v4 are can be cloumn basis, right?

er...

[clever cedar]

to solve null basis u setup augmentented matrix with a zero cofficent matrix

then solve the system

into RREF

[desert ibex]

yeah I get

| 1   0  2  0 |
| 0  1   1   0 |
| 0  0  0  1 |
| 0  0 0  0 |

[clever cedar]

yh thats right

[desert ibex]

so the null basis is [-2, -1, 0, 1]

[clever cedar]

z = s, w = 0 y = -s, x = -2s

so ya thats right

ur answer is right

[desert ibex]

can I just leave that and it works?

[clever cedar]

i believe so

since ur null rank (aka nullity) is 1

[desert ibex]

gat dangit

[clever cedar]

since u only have 1 variable

[desert ibex]

spent like 2 hours of exam prep so lost

[clever cedar]

ask more questions

[desert ibex]

alright thank you so much for the help

[clever cedar]

np

[desert ibex]

will do!

[clever cedar]

gl

[wintry steppe]

Hi I'm having issues with solving this question

Can someone explain how the process works?

[charred stirrup]

@wintry steppe you make a matrix and row reduce into into RREF

you will have some elements of k in the RREF, and then you will simply make k the value whatever you want to make the vectors independent of eachother

in other words, they will span r4

my solution was B = 9i , but I don't think they've introduced the concept of complex numbers yet, where did I go wrong?

[gray dust]

show work?

[gray dust]

where were you told A and B are 4x4 matrices?

[charred stirrup]

sorry I should have included that

[gray dust]

work looks fine. complex det is allowed if B has complex entries

btw det(B)=-9i also works

[charred stirrup]

Oh ok thank you

[clever cedar]

can someone please help me understand this

[half ice]

If every element in the codomain is mapped to

The definition is saying "if there's some element that maps to any element of your choice"

[clever cedar]

sorry for late response

oh

so it just says that for a given x_2 vector whether there is a x_1 vector that maps to it?

[clever cedar]

@half ice

[clear otter]

@clever cedar yes

I think of onto as "I can hit everything in the codomain"

[clever cedar]

so the domain and codomain have to be the same size?

[clear otter]

No

Let me give you an example

[clever cedar]

thank you

[clear otter]

Consider the map that maps from R^2 to R ^1 given by (x,y) to (x)

Apologies I don't want to type LaTeX because I'm on a garbage Motorola phone.

[clever cedar]

no its fine

dw

[clear otter]

That map is surjective because I can pick any vector in R^1 and can get it from a vector in R^2 and that map.

Er, onto

[clever cedar]

in my head it seems like thats always possible for any case

[clear otter]

Okay, consider this other map then

From R^2 to R^3 given by (x,y) to (x,y,1)

Is there any valid input that would give you the vector (1,2,3)?

[clever cedar]

oh shit i see what u mean

[clear otter]

:)

So your domain needs to be at least the same size as your codomain in order to be onto

[clever cedar]

damn tyvm

ok good to know i really appreciate it

[wintry steppe]

how do i get the t vectors

[toxic pendant]

If I'm given two vectors and told to find the equation of a plane and no other information, how do I find d?

[clever cedar]

@wintry steppe

do u still need help?

@toxic pendant u still need help?

[toxic pendant]

Sure

[clever cedar]

the equation of a plane is

n (dot product) PoP

where N is a normal vector

[wintry steppe]

yes

[clever cedar]

and PoP is a vector

Kryptic

Row reduce the matrix

[wintry steppe]

first i reduce the matrix right

[clever cedar]

to RREF form

yes

[wintry steppe]

then what

[clever cedar]

then solve it for a homogenous system

such that all the equations = 0

so setup an augmented matrix

for example if u get

x + 5y = 0

isolate for x

then y

et

etc

[wintry steppe]

ok sec

[toxic pendant]

So as far as I know, the equation of the plane is ax+by+cz-d=0

[clever cedar]

ill give u an example bait

one second

[toxic pendant]

Normally I'd find the normal vector by using cross product

[clever cedar]

ok

u do this

[toxic pendant]

But I'm unsure/forgotten how to get d

[clever cedar]

n (dot product) ([x,y,z] - vector2)

for example

if vector 2 = [1,2,3]

it simplifies to

n (dot product) [x - 1, y -2, z -3])

then u find the dot product of that

and rearrange for the constant is on one side

[toxic pendant]

Wait so you can sub the vector into the equation?

[clever cedar]

ill show u a picture

[toxic pendant]

What you wrote is basically the precursor for the equation of the plane

[toxic pendant]

That's a point though

[clever cedar]

no its the plane

[toxic pendant]

I'm looking at 4.45 right

[clever cedar]

the whole thing

look at all the steps

it shows u how to solve it

[toxic pendant]

Perhaps I'm missing something, but this example is using a known point on the plane

[clever cedar]

what does ur example have

[toxic pendant]

I'm talking about using two vectors

[clever cedar]

a point can be considered a poisiton vector

position*

and one of ur vectors has to be orthogonal to the other vector

otherwise its not possible

[toxic pendant]

Don't you just find the vector orthogonal to both of them

Why do the two vectors need to be orthogonal with each other

[clever cedar]

i mean u could

then ud use that point to solve it

[toxic pendant]

So what I'm getting here is that I could just find the normal vector and then use either vector to find the equation of the plane?

[clever cedar]

exactly

[toxic pendant]

I suppose if you multiplied the other vector by 0 and one vector by 1, the point produced would be the vector

[clever cedar]

yh

itd be a linear combination of the vectors

just a random vector tho

[toxic pendant]

I'm not so sure of this solution though

[clever cedar]

post the full problem

hard for me to gauge

[toxic pendant]

If you multiplied both vectors by 0 it'd still be a linear combination

But not all planes pass through the origin

[wintry steppe]

@clever cedar ok so the reduced matrix is
1 -1 0
0 0 1
0 0 0

x-y=0

z=0

[clever cedar]

ok so it has infinite solutions

as a result it is not one to one

its only one to one if it has one solution

now

wait what are the solutions

x = y

z = 0

y = s

[wintry steppe]

s is the free variable

?

[clever cedar]

yh

[wintry steppe]

ok

[clever cedar]

ok so for the function

where it says

T([vector])

plug in

T[1, 0, 1]

and also to understand why do this

1[vector-row-1] + 0[vector-row-2] + 1[vector-row-3]

not row columns

mb

and solve it

tell me what u get

[wintry steppe]

where do i plug in

in the original matrix?

[clever cedar]

let me write it down

ill take pic

to show u

give me 3mins

[wintry steppe]

like xyz values in the original matrix

[clever cedar]

no

the RREF matrix

[wintry steppe]

ok wait

i dont think im doing this right

[clever cedar]

theres the full solution

[wintry steppe]

how do you get these values

[clever cedar]

i let the free variable be 1

and plug it into the answer to the system

the solution to the system

is x=s, y = 0, z = s

and i picked a random s value

s = 1, so x = 1, y = 0, z = 1

make sensE?

[wintry steppe]

should this not equal
2
1
0

[clever cedar]

ya

that means u made a mistake RREF

give me the original matrix

[wintry steppe]

or it would be
1
1
0

[clever cedar]

yh

lol

my mistake

plug those in instead

and it'll equal [0,0,0]

[wintry steppe]

1 1 0 instead of 1 0 1 ?

o i see

x =1 y =1 z = 0

[clever cedar]

yh

now make s = 2

so it'll be

2 2 0

which also equals [0,0,0]

[wintry steppe]

why s =2

[clever cedar]

because its a free variable

ok lets go back

a matrix is said to be one to one, if there is only ONE vector (lets call it x) where T(x) = V, where v is another vector.

If we can find a vector T(p) = V, then

the transformation function ISNT 1-1

because we found two vectors, that result in the zero vector

your given vector A isn't 1-1

[wintry steppe]

oh i see

so its
1 0
1 = 0
1 0

and
2 0
2 = 0
0 0

[clever cedar]

in the first one

the z should be 0

but yes

[wintry steppe]

ok and for onto?

[clever cedar]

1      0
1      0
0     0

2    0
2    0
0    0

wtf

formatting

[wintry steppe]

yup i got it

[clever cedar]

onto is harder to explain

pretty much

row reduce and see how many rows have pivot columns

since there are infinite solutions

the matrix isnt onto

[wintry steppe]

pivot colums = rank ?

[clever cedar]

this is because the function A(x) doesn't result in any vector V

yh

in other words

there needs to be one solution

wait sec

let me

think

you know how it goes T: R^n -> R^m

[wintry steppe]

yes

[clever cedar]

if n < m

then it can not be onto

i know that for sure

[clever cedar]

fuck good textbook

ill rephrase it

what its saying is that if T(x) = B is possible for any vector B

given the constraints

since your matrix A is 3 x 3

natrually, the vector X will be 3 x 1 (because of the laws of matrix multipling)

and the resulting vector T(x) will be 3 x 1 as well

(also because of the laws of matrix multiplication)

as a result i think it sonto

its onto**

because there is a value of x that can result in any T

i mean in any B

[wintry steppe]

lemme check

[clever cedar]

yh sorry i really dont understand onto well

lemme think

what its saying is that

[-9, -10, -3] is a vector that can never be acieved from the transformaiton T(x)

in other words there is no value of x that gets [-9,-10,-3] (which is an arbitrary vector they chose to prove their point it has no signifgane other that it is impossible to get)

i think because u had infinite solutions

it wasn't onto

it has to have only one solution

[wintry steppe]

o so you just pick any random vector

[clever cedar]

yh

here

the rank of the matrix

has to equal the number of rows

it was not onto

because the rank of ur RREF was 2, while it had 3 rows

[wintry steppe]

i see

for r4->r4

how do i choose the free variable

[clever cedar]

whichever column results in 0 = 0

is the free variable

[wintry steppe]

but what do i set it to

i did xyzw

[clever cedar]

w = s

here

x, y, z, w

and ur w column has the free variable

set it to whatever u want

s = 100000 if u want

itll still result in [0,0,0]

[wintry steppe]

is this right

[clever cedar]

yh

just plug it in

and see if u get [0,0,0[

[wintry steppe]

ok yes its all 0

[wintry steppe]

did a few on my own i think i get it now 😌 thank you

[clever cedar]

np

can anyone help me figure out difference between surjective and image

if a vector is in the image of the transformation is it surjective?

[dusky epoch]

"surjective" refers to a property of functions
"image" refers to a set of points

if a vector is in the image of the transformation is it surjective?
no, just because one single vector happens to be in the image of a transformation does not at all mean the transformation is surjective

[clever cedar]

Oh so we say a transformation function is surjective not that a vector is surjective

[dusky epoch]

"transformation function"?

[clever cedar]

linear map I think is what i meant to say

[dusky epoch]

yes, "a vector is surjective" is just nonsense

[clever cedar]

I see, a vector X may be in the image of the transformation, but that doesn't necessarily mean the transformation is surjective because there could be a vector P that isn't in the image

If i'm not mistaken

[dusky epoch]

clunky but correct

[clever cedar]

Thank you, I'll make sure to review parlance

[ebon gate]

Is it true that if you have two vectors, $\mathbf{a}$ and $\mathbf{b}$ which lie in the plane $\Pi$, and a point that not on the plane with position vector $\mathbf{p}$, then the shortest distance from the point to $\Pi$ is given by $|proj_{\mathbf{a} \cross\mathbf{b}}\mathbf{p}|$?

[stoic python]

DoomieDaJuan:

[clever cedar]

one of those vectors have to be normal

for that statement to be true

[ebon gate]

a x b gives the normal doesn't it?

[clever cedar]

the coefficents of the equation of the plane does

im not sure about a x b

3x + 2y + 4z = 9

the normal vector is [3,2,4]

[charred stirrup]

why does the calculator elect to simplify the matrix first?

it seems like expanding along row 2 will be really simple and easy to do, but I get a completely wrong answer

[clever cedar]

is it finding det

[charred stirrup]

yeah

[clever cedar]

why is multiplying by 2

[charred stirrup]

thats what Im saying

[clever cedar]

it usually does proper laplace

fix ur 2

looks sketchy

[charred stirrup]

lol entering any number shifts it to the left for some reason

[clever cedar]

i dont get the steps in ur first screenshot

[charred stirrup]

yeah thats what im not getting

why did it also simplify the matrix?

[clever cedar]

sec i wanna try

[charred stirrup]

ahh so frustrating thought i finally learned how to expand

[clever cedar]

oh

i get it

haha

it skipped the steps

ok watch

https://cdn.discordapp.com/attachments/540211747613704221/647678116134780948/unknown.png

u choose row 2

in row 2 all of the fields except 1

is 0

so the first =

is just finding det of the only 2

and it ignores the zeroes

[charred stirrup]

oh my god

so i did it right but i cant use the calculator

[clever cedar]

haha happens

it should be explicit

[charred stirrup]

how should I punch it in?

[clever cedar]

then for the next step

its 4

because it takes the 2 from the outside

and expands 2(2)

2(6)

etc

this calc kinda sucks to show steps

find another one

[charred stirrup]

where did I goof here?

same matrix

[clever cedar]

the 2

is in position (2,2)

u crossed out column 1

when should be column 2

my recommendation is just expand on the first column

easiest

[charred stirrup]

oh my god

i am blind

thank you

[clever cedar]

lolol

np

[charred stirrup]

wont lie I am kind of confused now because of the whole crossing out thing

what does it mean to actually make the cross?

For me I see the cross, I say a ij

is on the x

[clever cedar]

see how u have |4 - 1, -2 4|

for the first cofactor

[gray dust]

take whatever entries that weren't crossed out and make a new matrix out of that

then focus taking det of that

[charred stirrup]

yup sub matrix

but my a ij is where my x marks ?

[gray dust]

yeah it's where the horizontal and vertical lines meet

at the i'th row and j'th column

[charred stirrup]

but will I ever use that number?

[clever cedar]

yeah

[gray dust]

no it gets crossed out when taking the minor

[clever cedar]

its what u use to multiply

by (-1)

[gray dust]

when you take the minor, just multiply the result by the entry

and don't forget whether to multiply again by -1 or 1 depending on what i and j are

[charred stirrup]

omg

so a ij is the scalar that applies to every single ijth minor?

[clever cedar]

yeah

[gray dust]

multiply the result by $(-1)^{i+j}$ if that helps

[stoic python]

RokettoJanpu:

[charred stirrup]

no wonder by crossing the 0 I will get nothing

Thank you so much

[clever cedar]

$ijth-entry(-1)^{i+j}$

[charred stirrup]

think I can finish the assignment

[gray dust]

please don't mislead, mike

[clever cedar]

???

[charred stirrup]

nah both u and mike were godly today

[clever cedar]

u take the ijth entry

[charred stirrup]

thank you alot

seriously

[clever cedar]

and multiply it by (-1)

wtf r u on about

[charred stirrup]

(-1) ^ ( i + j) yeah?

[gray dust]

multiply the smaller matrix determinant by $(-1)^{i+j}$ if that helps

[stoic python]

RokettoJanpu:

[clever cedar]

hes doing laplace expansion not finding cofactor

but ok

[charred stirrup]

smaller matrix determinant you mean minor?

[gray dust]

yes, but i wanted to avoid potential confusion with terminology

[charred stirrup]

and uhh I dont know if it is difference to u guys, but the lab is just called "determinants through cofactor expansion"

half getting the title meaning haha

[clever cedar]

thats fine, they specified because u can find determinants other way

i.e. upper triangular form

[charred stirrup]

why does the matrix we take of each element in my selected row different?

[gray dust]

you cross out different entries each time

[charred stirrup]

im big depress

thought we were using the same ones each time for some dumb reason

[gray dust]

tip: find the row/column with the most zeros as that will save some time. if you can't, you're outta luck and gotta iterate painfully over an entire row/column

[charred stirrup]

elementary row operations are allowed right?

to make some 0s

[gray dust]

row ops make it a whole new matrix whose det is not guaranteed to be equal to the original's det

[charred stirrup]

😦

[uncut forge]

Certain row operations are allowed

For example adding a multiple of one row to another, won't change the determinants value

Use it to get lots of 0s

Switching two rows will change the sign though

[gray dust]

this is true, but i believe him to be better off knowing that row ops in general don't preserve the det, until some later class where he may learn to compute dets by performing row ops to get a simpler matrix

[uncut forge]

And you can't multiply a row by a constant without altering the determinant, you can however factor out something from a row outside of the determinant, one row at a time

Calculating determinant without row operations seems like a massive pain,

[gray dust]

which is why, fingers crossed, i hope the matrices he deals with have at least 1 or 2 zeros in some col/row

[uncut forge]

Hmm if he hasn't learnt row ops with determinant then I guess exercises are solvable without

And I guess some teachers gives 0 points for solving an exercise with methods not yet taught

[gray dust]

sadly

[uncut forge]

As long as you have 3x3 matrices and one zero it's not too bad.

Yeah

[gray dust]

@charred stirrup i up-arrowed the messages you should read in case you want to learn how row ops can help you find dets

[blissful vault]

just need to make sure

if det(a) = 2

det(a^2) = (2)^2

is that right?

my logic is that det(a^2) = |a| |a| = (2)(2)

[dusky epoch]

yes det is multiplicative

[wintry steppe]

Hi I'm having issues with solving this question

Can someone explain how the process works?

[meager tinsel]

oof
I forgot how tensor contraction works
Can some1 help me pitcure it in my head properly?

[wintry steppe]

@broken hawk sorry for the ping. I'm trying to find a person to guide me through this concept

[broken hawk]

that’s not how this server works

[wintry steppe]

If you're busy, I understand

[broken hawk]

please follow the rules

#❓how-to-get-help

I am not a helper
pinging random people has a high likelyhood of annoying them

[sour spear]

@wintry steppe The set spans when all of the vector are linearly independent

so for that, make a matrix of the vectors in that set and set them all equal to 0

and then solve as you normally would, just with the k in there as a variable

[wintry steppe]

@sour spear

[sour spear]

Your r3 - r2

You didnt subtract 1 in the 4th column

@wintry steppe

[wintry steppe]

it becomes -1 i think?

@sour spear

[sour spear]

The third row should be 0 0 1 -1 after that step, yes.

[wintry steppe]

okay

Is this okaynow? @sour spear

[sour spear]

Looks fine @wintry steppe

[wintry steppe]

how would i know k/=/7

[sour spear]

continue reducing the matrix into RREF

[clever cedar]

what does it mean for two subspaces to be isomorphic?

i understand isomorphism for linear map

but not for two subspaces

[half ice]

Two spaces are isomorphic if there's an isomorphism between them

[clever cedar]

hm

so

if x is a vector in R^n and T: R^m -> R^n is ismorphism

and if v is a vector also in R^n and T: R^m -> R^n is ismorphism

those two are isomorphic?

[half ice]

If there's an isomorphism R^m → R^n
Then R^m and R^n are isomorphic

(But there isn't one unless m = n, but that's beside the question)

[clever cedar]

oh so an isomorphism is a way of describing that domain and codomain

are the same

[half ice]

Exactly. There's a mapping between them that preserves EVERYTHING, so domain and codomain must be the same object, so to speak

[clever cedar]

oh wow cool

this may beyond the scope

but does that mean they're 'switchable'

so if T(x) = n

can T(n) = x?

[half ice]

If T(x) = n, then there exists some other linear function f such that f(n) = x

T doesn't necessarily undo itself, but it DOES have an inverse

[clever cedar]

Wow thank you 🙂

I had flu for a few days so I appreciate your help.

<#

❤️

[half ice]

Np at all, glad you got the concept. I know this one is weird for a lin alg course

[clever cedar]

Yeah its interesting because I've learned how functions work before but this takes it one step further

So thanks again

[clever cedar]

i have no idea what is happening

in the second step

[half ice]

Looks like they're expressing (0,1) as a linear combination of the vectors in B2

@clever cedar

[clever cedar]

yeah i think they solve the system for the first vector in b1

but they flip it so it becomes [0,1]

because [a,b] = [b,a]

[wintry steppe]

@clever cedar lyryx textbook?

I recognize it

[clever cedar]

ya

[vapid coyote]

I wouldn't formulate it as saying that domain and co-domain are the same. The respective objects with their respective structures are the same.

There is some cool formalization of this in category theory though.

Like a function from a ring to another ring might be an isomorphism for the additive groups but not a ring-isomorphism

then they are the same as groups but not the same as rings

[wintry steppe]

Hi sorry for the intrusion

I'm trying to understand the concept of a vector in an image of A

[clever cedar]

if a vector is in the image of A

it means that through linear mapping of the defined transformation

you can find the result of that vector

for example

if the vector x is in the image of A

then A(v) = x

is a possible solution

@wintry steppe

[wintry steppe]

so i have to assume the numbers for b1

[clever cedar]

in ur problem

do this

row-reduce A, with the vector b1 as an augment to the matrix

so solve the system for b1

if there is a solution then its in the image

if theres no solution, then theres no solution

make sense?

[wintry steppe]

bring it to RREF?

[clever cedar]

what does it mean for a vector to be in the image of a transformation?

if u can answer that u can do the problem

[wintry steppe]

a vector exists such that Ax = b

[clever cedar]

x + y -2z = 2
-2x -y +10z = 14
2x + 2y -3z = 2
-x -5y -7z = -26

thats the system u have to solve for

the left hand side of the equations

is the matrix A

the right hand side is vector b_1

if after u RREF that system u get a solution

[wintry steppe]

sh

ah

okay

[clever cedar]

then by proxy theres a solution

[undone garnet]

any idea for this prob?

[clear otter]

@undone garnet what have you tried?

[wintry steppe]

systems of equations...

[undone garnet]

@clear otter

using A^2 = tr(A).A - det(A).I

but still get nothing

[undone garnet]

@wintry steppe you mean solving 4 systemequation from A^2B = AB^2?

[floral prairie]

how does dinosaur math differ from normal math?

[dusky epoch]

1. wrong channel
2. it does not

[floral prairie]

ty ok

[dusky epoch]

this is #prealg-and-algebra

"linear algebra" doesn't mean what you might've thought it means

[edgy ridge]

If this channel is clear, I've a question. Let us say we have a matrix, A It is 3x3. It has eigenvalues -2, 1, and 3. How would I find...

The determinant det(2A)
The trace tr(A^2)
And the rank (A+2I), where I is the identity matrix?

I believe that the determinant is 2^3 * det(A), which is the product of the eigenvalues, but how would I find the trace and rank of the above questions?

Is there a way to obtain a unique matrix with those eigenvalues that I can use to calculate it?

[undone garnet]

trace is sum of eigenvalue

[edgy ridge]

Yep, but the square of that matrix does not necessarily square the trace, yeah?

[undone garnet]

😄

if x is an eigenvalue of A

then x^2 is an eigenvalue of A^2

we can easily prove that

Av = xv

A^2.v = A.Av = A.xv = x.Av = x.xv = x^2.v

A has 3 distinct eigenvalues, so A is diagonalizable

A = PDP^(-1)

of course, I = PIP^(-1)

so

A+2I = PDP^(-1) + 2PIP^(-1) = P(D+2I)P^(-1)

rank(A+2I) = rank(D+2I)

[edgy ridge]

I'm still looking into diagonalization, but it seems interesting enough.

[undone garnet]

D is diag(-2, 1, 3)

so

rank(D+2I) = rank(diag(0, 1, 3)) = 2

[edgy ridge]

Erm, forgive me for sounding dumb, but what do these equations mean?

A = PDP^(-1)
I = PIP^(-1)

What is P and D?

I'll assume I is the identity matrix.

[undone garnet]

A is diagonalizable

so A = PDP^(-1)

D is a diagnal matrix