linear-algebra

[gray fable]

hello can someone help teach me about the four fundamental subspaces

[vital swallow]

I don't recognize that term

[gray fable]

null spaces column spaces row spaces

and left null spaces

[vital swallow]

I don't think I'd have anything useful to say on that, so I'll back off

[gray fable]

kk np 🙂

are you in uni?

[pale shell]

Excuse it I have ONE question

When writing a proof of ONE vector property

And you have a theoretical vector like [a1, ... an]

Can you specify that a1...an would just be members of the reals

Or WHAT.

[gray fable]

yea if the vector property you are trying to prove only pertains to the reals

just write that down as your starting assumption/condition

proof: assuming an is a member of the reals

...blah blah

[pale shell]

ALSO

What to assume it for ONE SCALARS

[gray fable]

I'm not sure I understand, are you asking what you should say in the start of your proof when only discussing scalars?

like the scalar "c" in the statement: c[1,2,3,4]?

[pale shell]

YES

whatting to say it

C is member offfffff

Theeee

what.

[gray fable]

reals

[pale shell]

Okay thank it so much

[gray fable]

it can be a complex too, but I don't know if your vector property pertains to complex numbers as well

[pale shell]

Yes but for one you know like a how you say

Regular euh linear algebra course

[gray fable]

yea then reals is fine I guess, but idk what you're trying to prove

[pale shell]

Euhhh

You know

Like eh

The basic propertationals and also subspacity type stuff

[gray fable]

no idea lol

[pale shell]

To make the basic subspace and property of V E C T O R

Little bit B A S I C proofe

[daring solstice]

yo where did cos(theta) go

[vital swallow]

this isn't linear algebra, most likely. but that's just the formula

[daring solstice]

dot product is just vector stuff yea

[vital swallow]

but not all vector stuff is linear algebra

[daring solstice]

yea

urm i can move my question to prealg-algebra

or precalc might be better

[vital swallow]

I was thinking multivar-calc-and-diffeq

[gray fable]

clementine i don't remember exactly

[vital swallow]

temperature is probably the easiest way ?

[gray fable]

but is that formula for perpendicular vectors?

@daring solstice

[daring solstice]

it's dot product (and hence also how to find an angle between two vectors)

[slow scroll]

yo where did cos(theta) go
the whole point of the dot product is that you can compute it without doing any geometry/trig. The fact that you can make the cos(theta) go away is the whole point.

[daring solstice]

oh and don't worry I reread my notes and realized I was just asking a really dumb question

[gray fable]

its not u-v

[daring solstice]

I thought right was an expansion of left

[gray fable]

its mag of u. mag of v

[daring solstice]

but actually it's a definition

[gray fable]

you can derive it too

[daring solstice]

full context makes it more obvious

[gray fable]

just think of magnitude as pythagorean theorem

yea

[daring solstice]

how do u derive u1v1 + u2v2 from just

u need to apply the cosine rule don't u?

[gray fable]

ok so ||v|| is sqrt(x and y components squared)

|v|

think of that as a triangle (the x and y components)

[daring solstice]

yea, the definition for length is from pythagoras afterall

[pale shell]

How to visualize one fourth dimension please.

[quartz compass]

I never tried to visualize 1/4th dimension before

[pale shell]

nu

I mean it not like 1/4

I meaning FOUR DOMENSNION

But thad does raise an interesting question

What is one fraction dimension.

[quartz compass]

there is a way to look at dimensions so that you can get non integer dimension

and it's not too bad to construct fractals of every possible positive real number dimension this way either

[steady fiber]

Hausdorff dimension is slightly different from a linear algebra dimension though

consistent in some ways

but a different concept of dimension

[pale shell]

Wait but anyway

I think that is a bit farther than a regular liner algebra class

[quartz compass]

just have a finite, noninteger number of basis vectors

[pale shell]

How could that be though

[quartz compass]

lol I'm joking

[pale shell]

Oh

Anyway

I wan tto know it

How to visualize one fourth dimensions

[steady fiber]

if we are sticking in a nice linear algebra context, dimensions are natural numbers

[pale shell]

Ok so like in a regular linalg clzs they will always be WHOLE nonegative nombers?

[steady fiber]

yes

a dimension can only be that

[pale shell]

Thanks

Anyway

[gray dust]

tfw mero forgets not everyone can detect his sarcasm

[pale shell]

Do you know any good techniques for visualize four dimensisons

Fourth

HI ROBAKE.

I FOUND OUT WHAT a codomain IS.

!

[gray dust]

oh goody

[pale shell]

Is very like so cool.

[steady fiber]

depends on what you want to visualize in four dimensions

[pale shell]

Hmmm

Welll

[steady fiber]

like a C->C function can be visualized with domain coloring

[pale shell]

I guess basic things like

[steady fiber]

and that is 4 dimensional

[gray dust]

did you get domain/image down yet

[steady fiber]

kind of

[pale shell]

Vectors

And maybe just the space itself

Robakes

I know domain and renge

And codomes

[steady fiber]

actually I guess domain coloring works for any 4D quantity

kind of

but not well

[gray dust]

let's shed that brainlet skin for a bit

call it image not range

[pale shell]

Codomain is just something you say all the possible outputs are right, even if the domain can’t reach them?

[steady fiber]

yes

[pale shell]

O sorry about that

Like the reals for x^2

Well ig to start simple

How to visualize a 4d space

With nothing in it yet

Just like you know

A 2d space

Like how you can just imagine a grid sort of

[steady fiber]

you can project 4D space down onto 3D (or even 2D) space

and visualize that

[pale shell]

hmm

Interesting

How to do

[steady fiber]

you think of a 3D hyperplane in 4D, and find the projection matrix from 4D onto that hyperplane

you can also look at slices of 4D

like let's say you're a 2D creature and cannot understand 3D, but you want to understand a 3D sphere

you can look at the slices of the sphere from top to bottom

and you would find that you get circles increasing then decreasing in size

and the 2D creature can understand circles

same thing happens for a hypersphere

taking slices of that would result in a point increasing to a 3D sphere back to a point

[pale shell]

hmmm

So interest

I honestly like liner aklebos so much

Cool thing like kernels and null space

Also other dimensionals

Coolest math ever experience in life.

What is the hat for 4d czlled?

[gray dust]

in convention we only use letter names for the standard basis up to R^3

[pale shell]

oooh

[gray dust]

for $\bR^n$ we prefer $e_1,\dots,e_n$

[stoic python]

RokettoJanpu:

[pale shell]

So what for MULTIPLE higher dimensi

[steady fiber]

ya, the e_k notation is much more generalizable

[gray dust]

there are only 26 english letters but many more naturals

[steady fiber]

are you sure

[gray dust]

yes i can do a proof by confidence

i know there are. qed

[steady fiber]

I see

[pale shell]

So to doing one of the thing is

FOR 4D UP

[steady fiber]

can always do the excel method of naming with letters

[pale shell]

you wanna getting e4

and e5

AND SO ON.

[steady fiber]

a, ..., z, aa, ..., zz, aaa, ..., zzz, ...

ez

[gray dust]

as an aside, i've made a petition to name the standard basis for $\bR^4$ as $\brc{\i,\j,\k,\l}$

[stoic python]

RokettoJanpu:

[steady fiber]

why

[pale shell]

no i disagree

I wnana getting ONE W HAT

it being i j k w hat

so when to use e

e for higher than three or WHAT.

[gray dust]

for ANY $\bR^n$ space i can use $e_1,\dots,e_n$

[stoic python]

RokettoJanpu:

[pale shell]

oooooh

anis

But of course four and up because we know first threes

So i hat is e1, j hat is e2, k hat is e3 and SO ON.

[gray dust]

this is a lot of excitement over "new" notation than necessary but sure

[pale shell]

Sorry about that

I get excited

Buttttttttt

The fours dimensions NEVER used in a reg linalg class

or is it..?

[gray dust]

the more linalg you study, the more you realize that "vector space" refers to something more general than a box of pointy arrows

[pale shell]

uhhhhhh

what.

what vector space vs subspace

[gray dust]

that's not what i'm talking about

[pale shell]

o

[gray dust]

is it not true when i say vector, you think of pointy arrow?

[pale shell]

Not me

I think of a little

Uhh

How you say

The little thing with the bracket and coordinates

[quartz compass]

there are more vector spaces other than just "spatial" space

[pale shell]

But got to say it my favorite vector is one vector,

the

ZERO VECTOR!!!!!!

[quartz compass]

lol

[gray dust]

of what space?

[quartz compass]

yeah zero vector is pretty nice

[pale shell]

who does admit it or what

zero vector is like a so good one

is infinite dimension exist.

[quartz compass]

wot

[pale shell]

Infinite dimension in liner algebr

[quartz compass]

oh that was a question

[gray dust]

not in YOUR class

[pale shell]

Oh ok so little bit

Uhh

LATER.

NOT LEARN YET.

So

If you have a 2x2 matrix

Filled with all zeroes

Then apply to 2d for transform it.

Everything is in null space or what

Or maybe little bit wording not good so how you say

The null space of matrix or what

Or the null space of transformed space

[gray dust]

nullspace of that matrix

[pale shell]

oh so then all of space would go to one single point basicallt?

If you wanna doing all vectors in this liner map of course in 2d

[gray dust]

that 0 matrix would map all of R^2 to 0

[pale shell]

Wowww

So interesting

So you just killed all the vectors wh

O live in 2d

[pastel saddle]

Could someone help me with this one? I cant seem to find an operator

[slow scroll]

you need S*S = SS*. There are few times when operators commute like this: when S* is the inverse of S, or whenever S* is some constant multiple of S.

[dry sail]

Hey guys? What does f: P3 -> M2(R) mean

[cyan bluff]

Pls somebody help

[eager locust]

What does f: P3 → M2 (R) mean
It means a function from the set P3, which is probably the set of polynomials (with real coefficients, I suppose) to the set M2 (R), which is probably the set of square 2x2 matrices with real entries.

Hug, a few things:

1. May I show you an example for a line and have you do the rest yourself?
2. I know the name is confusing, but Linear Algebra is a college / university course that deals with functions more complicated than this. I think this goes into #prealg-and-algebra.

[dry sail]

@eager locust Thank you

[eager locust]

Wait, sorry, I forgot something:
P3 is not all the polynomials - but all the polynomials of degree 3 or lower, such as x³ + 3x² + 2x + 1.
I would assume that this function sends each coefficient into its own coordinate in a matrix, or something of the sort.

[shadow drift]

Is it correct, that some span of vectors, may not necessarily contain the (0,0) vector.

[sonic osprey]

no

[shadow drift]

So all spans of some vector contain, the (0,0) vector?

[sonic osprey]

I mean they all contain the zero vector of your vector space yes

[shadow drift]

I assumed that the set of all linear combinations of any set of vectors:
av + bw
Would include the vector (0, 0).

[sonic osprey]

I mean (0,0) isn't even an element of like

R^3

[shadow drift]

Oh right.

So, all "spans" would include the zero vector of the dimension.

So any span on R^3 would include (0,0,0)
And any span on R^2 would include (0,0)
etc?

[sonic osprey]

yes

you should be able to see why

[shadow drift]

Yea

Right now, it seems like spans are the same as subspaces. Just described in a different way.

But I think this is wrong.

[sonic osprey]

I don't think its bad intuition

All spans are subspaces, and all subspaces can be described as some span

[shadow drift]

Right, so for now, since I'm only at the elementary level, I'll probably stick with this intuition then.

[sonic osprey]

I think its good intuition

[shadow drift]

Aight thanks Zophsama!

[long blade]

hello guys

just a quick question

if det(A) = 2 and det(B) = -1

then um whats the det(A^5 * B ^4 * (B^T * A ^6)^-1 * A ^T)

[pallid rampart]

determinant of product is the product of determinant

[long blade]

its the ^-1

[pallid rampart]

determinant of inverse of A is 1 over the determinant A

[long blade]

thats confsuing

[pallid rampart]

that's the inverse

of the matrix B^T * A^6

[long blade]

ohh

kk

i got it

ty

ty

[shadow drift]

So the formulas for dot product of 2d vectors is:
|A| * |B| * cos(theta) = C (where theta is the angle between A and B)
And
A_1 * B_1 + A_2 * B_2 = C

This means that when vector B is some scalar multiplication of A, so A = nB
cos(theta) = 1
Since they have no difference in angles.

So, this means |A| * |B| = A_1 * B_1 + A_2 * B_2

Which can be expressed like so (A_1 = a, A_2 = b, B_1 = c, B_2 = d)

$$\sqrt{a^{2}+b^{2}}\cdot\sqrt{c^{2}+d^{2}}=a\cdot c+b\cdot d$$

[stoic python]

Lang:

[shadow drift]

Where c = n*a, and d = n*b, where n ∈ ℝ

This seems provable using the dot product stuff from before.
But I was wondering if anyone had seen or would have any idea on how they'd go about proving this.

I got up to something like:

$$\sqrt{n^{2}\left(a^{4}+2a^{2}b^{2}+b^{4}\right)}=a^{2}n+b^{2}n$$

[stoic python]

Lang:

[shadow drift]

Nvm figured it out.

[tender stone]

👍

[pale shell]

What does a negative determinant actually mean, geometrically?

[wintry steppe]

How do I convert this into a linear equation

[shadow drift]

Summoner

Summoner name has 6 months protection

• 1 month per 6 levels (with max 30 months)
  I don't know, but I'd turn this into a piecewise.

[static bison]

just need some verification:```
| a 0 0 | | a 0 |
det( | b c d | ) = c * det( | e f | )
| e 0 f |

[quartz compass]

yeah looks good

[static bison]

ty b

[quartz compass]

it might be easier to expand along the first row with a

that way you don't have to keep track of the sign

[wintry steppe]

Yeah, I agree with @quartz compass

[quartz compass]

doing it both ways is also a way to check yourself

[tired garden]

Hi, I have a question

So I've got this matrix

$$
\begin{bmatrix}
3/2 & -1 \
-1/2 & 1/2 \
\end{bmatrix}
\quad
$$

[stoic python]

Illimar:

[tired garden]

The lambda values are

$$\lambda1 = 1-\frac{\sqrt{3}}{2} $$ \ $$\lambda2 = 1+\frac{\sqrt{3}}{2}
$$

[stoic python]

Illimar:

[tired garden]

So I'm trying to find the eigenvectors of that matrix

The first step is to pick one of the eigenvalues, I pick lambda one, and then subtract that from the a and d of the matrix

Here is a calculation for the first element of the matrix

$$ \frac{3}{2} -(1-\frac{\sqrt{3}}{2})
$$

[stoic python]

Illimar:

[tired garden]

I'm stuck here, how do I subtract this eigenvalue from the first element?

[steady fiber]

$\frac{1+\sqrt{3}}{2}$

[stoic python]

PorosInMyAshe:

[tired garden]

Ahh thanks

[dusky epoch]

@tired garden use _ for subscripts

$\lambda_1, \lambda_2, \lambda_{10}$

[stoic python]

Ann:

[tired garden]

@dusky epoch Awesome, thanks

Ok so after subtracting the eigenvalues, I get the following matrix:
$$
\begin{bmatrix}
\frac{1+\sqrt{3}}{2} & -1 \
-1/2 &\frac{-1+\sqrt{3}}{2} \
\end{bmatrix}
\quad
$$

[stoic python]

Illimar:

[tired garden]

So here I have to perform row reduction to find the first and second value of the eigenvector

Correct?

But I don't see an easy way to do that

My understanding is that the first column is the first value of the eigenvector and column 2 is the second value of the eigenvector

So I have to perform operations to get either one of the rows to 0's

Which leaves the remaining row with eigenvector values

[steady fiber]

the row reduction is very simple here

you can just set one of the rows to 0

since the rows must be linearly dependent

and there's only 2 rows

so the rows are just a constant multiple different

[tired garden]

Ok I got it thanks, I'm currently trying to find the right operations that would get one of the rows to 0

[static bison]

I have a question about jordan conical form of a 3x3 matrix with two eigenvectors. ```
0 0
v_1 = 1 and v_2 = 1
0 1

How do I make a jcf using those vectors?

[steady fiber]

so you'd have to find a generalized eigenvector too

so let's say the generalized eigenvector is $v_2^$, then you make a matrix $P = [v_1 | v_2 | v_2^]$, and $J = P^{-1}AP$

[stoic python]

PorosInMyAshe:

[steady fiber]

where A was your original matrix

[static bison]

ah so?

how do find the value of the generalized eigenvector

[steady fiber]

so for an eigenvalue $\lambda_1$, you have an eigenvector $\mathbf{v}_1$ for a matrix $A$, then $(A-\lambda_1 I)\mathbf{v}_2 = \mathbf{v}_1$

[stoic python]

PorosInMyAshe:

[static bison]

btw```
1 0 0
A = 5 1 1
3 0 2

[steady fiber]

solve for v_2

and that is your generalized eigenvector

for lambda_1

and you can keep iterating this to find more and more generalized eigenvectors if you want

another way is to solve for nullspace of $(A-\lambda_1 I)^k$, where you need to find k-1 more generalized eigenvectors

[stoic python]

PorosInMyAshe:

[steady fiber]

and the nullspace will give you the k-1 more eigenvectors (along with your original eigenvectors)

[static bison]

                                0
for (A - lambda_1 I)V_general = 1 , where lambda_1 is equal to 1
                                0

@steady fiber

and I'm solving for v_general

[steady fiber]

ya so find a solution to that

that is not an eigenvector

[static bison]

wait

[steady fiber]

there's infinitely many solutions actually

[static bison]

I'm lost

[steady fiber]

but just find any one

[static bison]

so when I follow that process: ```
0 0 0 1 0 0
(A-lambda_1*I) = 5 0 1 =rref=> 0 0 1 , right?
3 0 1 0 0 0

then,
1 0 0 a a
0 0 1 * b = c , right?
0 0 0 c 0

then we find an find the values for a, b, c
a 0 0
c = 1 , which means a = 0, c = 1, b = x, so v_gen = x , where x = 1
0 0 1
0 0 0
so the form is 1 1 1 where [v_1 | v_2 | v_gen]?
0 1 1

@steady fiber

[steady fiber]

what

\begin{pmatrix}0&0&0\ 5&0&1\ 3&0&1\end{pmatrix}\begin{pmatrix}a\ b\ c\end{pmatrix}=\begin{pmatrix}0\ 1\ 0\end{pmatrix}

[stoic python]

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

[steady fiber]

should be getting this

when you do what I said

don't row reduce or anything

that's too much work

you can notice that since the middle column of the matrix is 0

b can be anything

does not matter

now, for a and c

we just get two equations

5a+c=1
3a+c=0

solve that

find a, c

and you have your generalized eigenvector

[static bison]

are we not suppose to rref or do we just not do because we don't have to?

[steady fiber]

we do not have to do anything like that here

just overly complicates it

there's so many 0s in the matrix

it's already nearly trivial to solve for the vector

you can do it by row reduction too, but you probably made a mistake somewhere

v_gen cannot be (0 1 1) since that's an eigenvector

a generalized eigenvector can never be an eigenvector

[static bison]

so a = 1/2, b = x, and c = -3/2; with x = 1; ```
0 0 1/2
so now rref( 1 1 1 )?
0 1 -3/2

[steady fiber]

you have a and c right

but as I said before

b can be literally anything

it is easier to choose b=0

and so you will get $P = \begin{pmatrix}0&0&\frac{1}{2}\ 1&1&0\ 1&0&-\frac{3}{2}\end{pmatrix}$

[stoic python]

PorosInMyAshe:

[steady fiber]

remember $J = P^{-1}AP$

[stoic python]

PorosInMyAshe:

[steady fiber]

$J=\begin{pmatrix}0&0&\frac{1}{2}\ 1&1&0\ 1&0&-\frac{3}{2}\end{pmatrix}^{-1}\begin{pmatrix}1&0&0\ 5&1&1\ 3&0&2\end{pmatrix}\begin{pmatrix}0&0&\frac{1}{2}\ 1&1&0\ 1&0&-\frac{3}{2}\end{pmatrix}$

[stoic python]

PorosInMyAshe:

[steady fiber]

if you solve for it, it does indeed give you the jcf

I did it with an online calculator to check

[static bison]

ah okay

give me moment

Thanks a lot too

[tired garden]

$$
\begin{bmatrix}
\frac{1+\sqrt{3}}{2} & -1 \
-1/2 &\frac{-1+\sqrt{3}}{2} \
\end{bmatrix}
\quad
$$

[stoic python]

Illimar:

[tired garden]

I multiply the bottom row with $\sqrt{3}$

[stoic python]

Illimar:

[tired garden]

And add the new row to top row

This will result in:

[steady fiber]

why do that

[tired garden]

I need to multiply the rows with scalars and then add them to each other to get one of the rows to 0 right?

[steady fiber]

as I said before

both rows are linearly dependent

and there's only 2 rows

so they are scalar multiples of each other

you can instantly set one of the rows to 0

you already know they are scalar multiples

you do not have to go through tedious calculations to prove what you know

[tired garden]

Say I set top row as 0 for example

How does that give me an answer

[steady fiber]

are you looking for an eigenvector?

[tired garden]

Yes

I thought that is achieved by "row reduction"

[steady fiber]

$\begin{pmatrix}0&0\ -\frac{1}{2}&\frac{-1+\sqrt{3}}{2}\end{pmatrix}\begin{pmatrix}a\ b\end{pmatrix}=\begin{pmatrix}0\ 0\end{pmatrix}$

so you can do this

[stoic python]

PorosInMyAshe:

[steady fiber]

and you know that a times the value in 2nd row 1st column + b times value in 2nd row 2nd column = 0

and so you can pick any a & b that satisfies that

and you found your eigenvector

$-\frac{a}{2}+b\frac{-1+\sqrt{3}}{2}=0$

[stoic python]

PorosInMyAshe:

[steady fiber]

for example b = 1, a = -1 + sqrt(3)

[static bison]

so P^-1AP is ```
2 0 0
0 1 1
0 0 1

[steady fiber]

yes

[static bison]

noice

thank you a ton

[steady fiber]

you can see that you only have diagonals, and 1s on off diagonals for the bad eigenvalues

[tired garden]

Ok let me digest this

[steady fiber]

so it is jcf

[tired garden]

Ahhhhhh everything clicked now

Thank you so much

[rough pilot]

Any ideas on how to prove this identity?

(cos a + i sen a)^n ) cos (na) + i sen (na)

The idea is not to use induction

[steady fiber]

not a linear algebra question, it's a #real-complex-analysis question

I can answer it there

[rough pilot]

Thanks

[wintry steppe]

Can someone help me with a recursion problem please?

I need to find a geometric sequence for a third order recursion

[quartz compass]

post the problem and what you've tried

[ebon mountain]

Pascal can u add me as a friend

@wintry steppe I have the answer

we have the same question

[shadow drift]

$$\left(ab-cd\right)^{T}=\left(ab\right)^{T}-\left(cd\right)^{T}=ba-dc$$

Is this correct?

[stoic python]

Lang:

[limber sierra]

are a, b, c, d matrices?

[shadow drift]

last part seems off for me

yeah, or vectors

Or maybe it should be

$$b^{T}a-d^{T}c$$

[stoic python]

Lang:

[shadow drift]

The middle part I think is okay, but last bit not sure.

[limber sierra]

well, that doesnt make sense if the matrices arent square matrices

anyway, i have no clue how you got to the last step

perhaps you're thinking of the identity

$(AB)^T = B^TA^T$

[stoic python]

Namington:

[shadow drift]

Oh right

$$\left(ab-cd\right)^{T}=\left(ab\right)^{T}-\left(cd\right)^{T}=b^{T}a^{T}-d^{T}c^{T}$$

[stoic python]

Lang:

[shadow drift]

So this one seems correct

[limber sierra]

that seems fine, yes

[shadow drift]

aight cool thanks

[proud otter]

Can a basis of an eigenspace be found for the matrix M = (A - λI) if M ends up being an inconsistent matrix?

| -12 -24 -12 | 0 |
|   0   0   0 | 0 |
|   0   0   6 | 0 |``` I have a matrix that turned out looking like this.

[half ice]

Systems are inconsistent, matrices aren't

[proud otter]

oh well yeah, sorry 😞

[half ice]

Oh I see what you mean

That makes me think the subbed value of λ is wrong

[proud otter]

Oh I see

I made a mistake

LOL

| -12 -24 -12 | 0 |
|   0   0   0 | 0 |
|   6  12   6 | 0 |```

Was writing too fast, but I forgot the first 2 digits in row 3

That was scary haha

[sick dragon]

span confusing me

[quasi vale]

@sick dragon By span(u,v), they mean that every vector in R^2 can be written as a linear combination of the two vectors u,v.

Or every vector in R^2 can be formed by using these two vectors.

[sick dragon]

It would be true then? Unless there is a case where it's not.

[quasi vale]

Is this a true false question?

[sick dragon]

yeah

[quasi vale]

like a test or what?

[sick dragon]

hw

[quasi vale]

k, what do you mean by a case when it's not?

the key word is that these vectors are not parallel

[sick dragon]

like what if they lie on the same plane

[quasi vale]

in R^2 they will lie on the same plane, the xy-plane

[sick dragon]

oh, i read this as perpendicular

[quasi vale]

but they're not parallel

that's the key word

[sick dragon]

well it says 'not' instead of 'never'

does that matter

[quasi vale]

same thing

[sick dragon]

alright, well im convinced its true then 😄

[cold topaz]

for b), I found <-1, -1, -1, 1>, but Slader says <1 1 1 -1>. Which one us is correct?

[quasi vale]

both are correct

[sick dragon]

what do they mean by contained here?

[cold topaz]

oh

[gray dust]

is $w\in\Span\brc{u,v}$?

[stoic python]

RokettoJanpu:

[sick dragon]

so it's asking me can i manipulate u and v to look like w?

[gray dust]

it's requiring you to get a handle on what span means

[sick dragon]

still not there 😄

[gray dust]

in short, the span of a set of vectors is the set of all possible linear combos of those vectors

[sick dragon]

so it's saying is w a solution to those vectors?

[gray dust]

that's poor wording and forget w for a sec

[sick dragon]

so my intiution says yes but idk how to prove it

not really good w/ the vocab definitions of this class

[gray dust]

if you took the set of all possible linear combos of u & v, what would that look like visually?

[sick dragon]

it could be anything in 2d

assuming you're working with both of them

[gray dust]

do you know what a linear combo is?

[sick dragon]

no

maybe

do you just mean scalar multiplication

[gray dust]

nope

a linear combo of vectors $u,v$ is $c_1u+c_2v$ where $c_1,c_2$ are scalars

[stoic python]

RokettoJanpu:

[sick dragon]

yeah this is just the reduced row shit ive been doing isnt it

multiply then add the vector

[gray dust]

i never mentioned row reduction

[sick dragon]

any row operation

its doing the same thing?

[gray dust]

if your vectors are in R^n then vector addition is componentwise addition, and scalar multiplication is componentwise multiplication

[alpine echo]

Sorry for interruption, I'll just post a question here, please take a look once the convo is over

Given AX=B, a 3x3 system, what's the exact relationship between the two interpretations;

the solution is the point (line, plane, ...) where the three planes intersect
and
the solution is that vector(s) that lands on B after the transform A
Sometimes thinking about matrices as transform is a bit difficult compared to the other interpretation (in elimination for example).
Taking A^T seems to interchange these two "interpretation"

[gray dust]

if you're still there if we take scalars to be in $\bR$, we define the span of $\brc{u,v}$ as
$$\Span\brc{u,v}=\brc{c_1u+c_2v:c_1,c_2\in\bR}$$

[stoic python]

RokettoJanpu:

[gray dust]

THAT is what i mean by the set of all possible linear combos of u,v

all you have to do is reason out visually what that span looks like and whether w lies within it

[shadow drift]

@alpine echo I would try thinking of it in a 2x2 system, since it'll still be the same

[shadow drift]

span

[unborn trail]

hey im struggling with a linear systems question

is anyone here able to help me with it

[shy atlas]

[golden cloak]

I have a question: what is a dimension of a vector space: $\mathbb{R}$ over a field $\mathbb{Q}$?

[stoic python]

cybergnostic:

[golden cloak]

What is its basis and dimension?

and more, what about, say, Q^2 over R?

[alpine echo]

@alpine echo I would try thinking of it in a 2x2 system, since it'll still be the same
@shadow drift I don't get you ?

And may I know what software it is?

Grapher?

[dusky epoch]

@golden cloak uncountably infinite

[golden cloak]

hi @dusky epoch we meet again 😗

[dusky epoch]

and you can't really write down a basis for R over Q bc it's one of those things that requires the axiom of choice to show its existence

Q^2 is not an R-vector space at all

[golden cloak]

so its completely other field of study?

[dusky epoch]

what

[golden cloak]

non-real vect spaces

[dusky epoch]

...

no i just said it doesn't make sense to consider Q^2 as a vector space over R

[golden cloak]

why not, it satisfies the axioms?

[dusky epoch]

no it doesn't

[golden cloak]

ah

you're right

[dusky epoch]

how are you gonna scale (1/2, 1/2) by pi

[golden cloak]

its not closed

ok ok

[dusky epoch]

i mean

in a broader sense, it's simply impossible to equip Q^2 - or any countable set with more than one element, really - with the structure of a real vector space

if you want, i can prove it

[golden cloak]

no, its perfectly clear. You can if you like - it would be nice just to write some proof 🙂

[dusky epoch]

let V be a nontrivial real vector space. my claim is that V must be uncountable

i prove this by demonstrating that V has a subset whose cardinality is the same as that of R

since V is nontrivial, it contains a vector which is not equal to the zero vector. take one such vector and call it x

now define a map f: R -> V by f(t) = tx for all real t

i claim that f is injective. do you want a more detailed proof of this

[golden cloak]

all clear

[dusky epoch]

f(R) is a subset of V, and since f is injective, R is in bijection with f(R)

therefore V has an uncountable subset and hence must itself be uncountable

[golden cloak]

nice, ty

[pale shell]

Any applications of linear algebra in hs physics or what.

[dusky epoch]

probably none, beyond maybe simplifying some vector calculations and solutions of systems of equations just a little bit

[pale shell]

Hmm

Also

I am kind of confused

Because when I reduce a system of equations with some row reduction

How to know when to stop

[alpine echo]

Given AX=B, a 3x3 system, what's the exact relationship between the two interpretations;
and
Sometimes thinking about matrices as transform is a bit difficult compared to the other interpretation (in elimination for example).
Taking A^T seems to interchange these two "interpretation"
@dusky epoch any thoughts?

[pale shell]

Look think it about the left column is the new î and the right is ONE new j

And same for any dimension basically

[alpine echo]

I understand the two interpretations individuially ... but the relation between the two

in a intuitive manner

[dusky epoch]

@pale shell re. row reduction
you stop either when your matrix is in row echelon form or reduced row echelon form depending on how much extra algebra you want to do

[pale shell]

okay thank it so much ann

[alpine echo]

Is there a geometric intuition for... why the row space and column space have the same dimension?

[dusky epoch]

mm

none that i can think of immediately

hold on

[pale shell]

Hmm

Good question

[alpine echo]

The two ... seem to be... unrellated (I'm not able to connect the two)

[dusky epoch]

earlten, with all due respect, you probably aren't the most qualified person around here to answer this

[pale shell]

So I guess the rank would also describe the row space..

Agreed

Im just tagging along to learn more

[dusky epoch]

i can't think of anything geometric rn

[pale shell]

So the definition of rank is dimension of either row or column space?

[alpine echo]

Rank (dimension of C(A) and R(A))
Transform interpretation ~ the dimensions the independent vectors (columns) of A can span.
Planes interpretation ~ the least independent set of planes required to attain the same system

i can't think of anything geometric rn
@dusky epoch Aight cool.

[pale shell]

so basically yes...?

[alpine echo]

yup

[pale shell]

Interesting

[sonic osprey]

I think the best way to think about this is that

The row space is the orthogonal complement to the null space or kernel

[dusky epoch]

i don't remember the details of the prof that row rank = col rank rn

[pale shell]

Uhh its at one right angle?

Oh no

So dot product

Equals

ZERO

For one kernel and one row space

[alpine echo]

The row space is the orthogonal complement to the null space or kernel
@sonic osprey what does orthogonal complement mean? I understand why column space and the null space are orthogonal

[sonic osprey]

But that's not true

[alpine echo]

Ouch

[sonic osprey]

The column space and null space live in potentially different spaces

When your matrix isn't square

[alpine echo]

Ahh right.

Darn... i'm so confused rn

[sonic osprey]

The row space and null space will always be in the domain

[dusky epoch]

zoph this may be a bit too abstract for aravindh's taste

also you mean col space probably

[sonic osprey]

No I mean row space?

[alpine echo]

I think the "geometrical meaning" (if there exists one) of transpose would shed some light on this. I'm getting into terms with LA, that not everything can be plotted and seen 😦

Grant (3b1b) spoke about it... but I didn't understand

https://www.reddit.com/r/3Blue1Brown/comments/c2z0qa/transposed_matrix_meaning/ernjqet?utm_source=share&utm_medium=web2x

[sonic osprey]

This is just a more complicated way to say what I said

[dusky epoch]

there is no viscerally geometric way to talk about dual spaces

[pastel saddle]

I'm not really sure how to find the characteristic polynomial for this. I've only done it for matrices

[dusky epoch]

the characteristic polynomial of a transformation can be computed by fixing some basis of the space it acts on, writing down the matrix of the transformation in that basis, and computing the charpoly of that matrix

so you might consider going with something simple like {1, t, t^2, t^3} for your basis

[pastel saddle]

Thank you!

[alpine echo]

The row space is the orthogonal complement to the null space or kernel
@sonic osprey What does this mean?

[pastel saddle]

How would I go about finding the minimal polynomial from here?

[eager burrow]

Since the minimal polynomial has the same roots as the characteristic polynomial, there are now exactly two possibilities for what the min. polynomial can be: the characteristic polynomial itself, or the characteristic polynomial with one factor (t+2) removed

so that restricts your possibilities already

Also: a matrix is diagonalizable if and only if its minimal polynomial factors into distinct linear factors

just some thoughts, might not lead to the most efficient way of solving this

[wintry steppe]

Anyone know what this notation mean

Where B B’ are basis

[shadow drift]

$$\left[T\right]{B}^{B{1}}$$
$$\left[T\right]_{B}^{B'}$$

[stoic python]

Lang:

[eager burrow]

The matrix representing the linear map T with respect to the basis B in the domain, and basis B' in the codomain

[ivory tulip]

Just need help understanding how to solve this type of question.

[stuck tide]

yo @ivory tulip

it would be just the coefficients

so first column wouldbe -7 3 -3

[ivory tulip]

Is it that simple, am I overthinking it?

[half ice]

Because it can be written
[-7 -3 -1] [x1]
[3 0 -6] [x2]
[-3 1 0] [x3]
If you actually perform the matrix multiplication to combine these two, you'll get f(x) back

[ivory tulip]

Thanks for the help @stuck tide & @half ice

[pastel saddle]

I think the minimal polynomial would be (t - 7)^5 since dim[Im(T-7I)^5] = 0. But not too sure where to go from there

[lean spoke]

i need to find x to the power 5 by method of xAx^-1

I found the x vector but dont know how to write x^5 unless i do it as multiply x with itself 5 times

do you know a different way

[ocean sequoia]

The answer is yes right because it satifies all the properities of a vector space right?

[limber sierra]

does it?

hint: \infty - \infty - \infty

[ocean sequoia]

what does \ that mean?

i know it has the 0 vector, it definitely has the additive identity, its communitive, it appears to be closed under addition

[gray dust]

some of us have a habit of typing latex outside mathmode

$\infty-\infty-\infty$

[stoic python]

RokettoJanpu:

[ocean sequoia]

well I know that infinity minus infinity is zero right? because i know infinity + (-infinity) is 0

[limber sierra]

it has the 0 vector, it definitely has the additive identity, its communitive, it appears to be closed under addition

there's only a few properties left to check

try them on the above

[ocean sequoia]

lets see i know there is an additive inverse

[limber sierra]

why do you think i wrote an expression with 3 terms

i want you to check additive associativity

[ocean sequoia]

ohhh

[limber sierra]

does $(\infty + (-\infty)) + (-\infty) = \infty + ((-\infty) + (-\infty))$?

[ocean sequoia]

right because infinity + - infinity = 0 - infinity = - infinity

and

[stoic python]

Namington:

[ocean sequoia]

• infinity - infinity + infinity is 0

hence they arent equal

[limber sierra]

there we go

so this doesn't have associativity, hence it isnt a vector space

[ocean sequoia]

there for associativity doesnt apply

Thanks!

[limber sierra]

in general, if you have a + a = a

and a is not zero

you should be really suspicious of associativity

[ocean sequoia]

thanks im make a note of that

[dusky epoch]

also. "communitive"

it's commutative y'all

commute

[ocean sequoia]

hahahhaha

[limber sierra]

to slightly elaborate on the last thing i said

suppose a + a = a

then a + b = (a + a) + b = a + (a + b)

so, by cancellation, b = a + b

this means that, if you have associativity and cancellation, a + a = a implies a = 0

i mean, obviously you can also cancel directly from the equation a + a = a

but some people prefer a proof using the "properties" of additivity

[not that it actually makes a difference]

[ocean sequoia]

thank you so much

how do we know that (cf)0 = c * f(0)

[eager burrow]

that's generally just how you define scalar multiplication for functions

[limber sierra]

is f a linear function?

oh wait, S is a space of functions

ok yeah thats just how scalar multiplication is defined

[ocean sequoia]

oh ok cool

[ocean sequoia]

oops i also meant to link this

but arent we trying to prove it is a linear function so didnt we just give a definition there that we wanted?it just seems kinda circular i guess... cause we need to show that its closed under scalar multiplication but im not sure how that definition did cause its just a definition if that makes sense?

or when we say its a function does that mean its a linear one?

cause couldnt f(x) be a non-linear function?

[half ice]

We're not trying to prove it's a linear function

In fact many such f won't be

We're trying to show that the functions can be the vectors here

[ocean sequoia]

(cf)0 = c * f(0) but thats not true for all functions right?

[half ice]

Yes it is true, as that's how we're defining (cf)(x)

The function (cf)(x) is the same as c(f(x))

Note that this is not saying f(cx) = c(f(x)) which is part of the definition of a linear function

[ocean sequoia]

ok so the reason that applies is because we aresaying its true

[half ice]

Ya ya. You'll see this is a common thing to define

Because turning things into vector spaces are bae

[ocean sequoia]

hahahah ok im good with that

thanks!

i like the name by the way

[half ice]

Thx I work hard on it every day

And feel free to ask if you need any other help with it!

[ocean sequoia]

thanks so much!

[golden magnet]

So right now i'm trying to rotate an object, it's always at the origin. I want to emulate a trackball. So, from my understanding one way I can do this is to obtain a vector from my mouse drag and then convert that to polar coordinates.

but i'm not sure how to proceed from there, i.e. how to actually do the rotation

[ocean sequoia]

why would you need to do polar coordinates? Cause if we start at the origin couldn't we just solve it as a system of equations? [x1,y2] + [x3,y4] = [x5,y6]?

though if you want to do a rotation generally is just a matrix like

however if we are starting at the origin we cant rotate in a way that i know of because [0,0] * A = [0,0]

someoen who knows more might be able to correct me tho

[runic pewter]

can someone explain what an inner product space is? I am really confused about the whole concept.

[half ice]

It's a vector space that has an inner product

And an inner product is something that acts like a dot product. It takes two vectors, gives back a scalar

[runic pewter]

yea it has an associated norm or something, i remember. But what's the difference? What's all the axioms about symmetry and linearity?

[runic pewter]

anyone?

[sonic osprey]

if you have an inner product<>, then ||x||^2 = <x,x> will satisfy the properties of a norm

@runic pewter

So inner products always give you a norm, but sometimes the other way around doesn't work

There are examples of norms on spaces, where you can't construct an inner product that gives you that norm

[runic pewter]

okay i think i understand. I probably need to know more about different types of normed spaces to get a clearer understanding. Thank you for the help

[sonic osprey]

basically inner product spaces are always normed spaces

but the other way around isn't true

[half ice]

Difference between what and what? Difference between a space with a norm and a space without a norm?

The axioms just tell you what an inner product IS

[lavish pendant]

ii) f : R2 → R, f(x, y) = x + y

find if the function is bijective if it is find the inverse else find the image

have no idea what to do

[spare crystal]

well how do you determine if its bijective or not @lavish pendant

[lavish pendant]

it has to be injective and surjective

but i do not understand well the part on how to work with R2 and R

[spare crystal]

how can you show that it is injective

or show that it isnt injective

[lavish pendant]

it is injective if every element of the domain has 1 element of the image

[spare crystal]

thats a pretty vague way of defining injective 🤨

a better definition is that no two different values x and y map to the same value

or in other words no two distinct x and y satisfy f(x) = f(y)

if you want to show that something is injective you need to show this is true

if you want to disprove it, all you have to do is find two distinct x and y where f(x) = f(y)

x and y are vectors by the way, not the x and y in the example you had in your question

in this case, its pretty easy to find an example where two different vectors map to the same value

[lavish pendant]

ohh i think i understood

[spare crystal]

also, as a general rule of thumb, given T : Rm -> Rn
if m > n: its not possible to be injective
if n < m: its not possible to be surjective

you can think about why but that means the only invertible linear transformations are from Rm to Rm

[lavish pendant]

thanks for the tips

[spare crystal]

np

[lavish pendant]

❤️

[spare crystal]

Can someone explain why redundant columns in a matrix A (columns that are a linear combination of those before it) correspond to redundant columns in rref(A)? I feel like the answer here https://math.stackexchange.com/questions/489712/relationship-between-the-column-space-of-a-matrix-a-and-its-non-free-pivot-c doesn't actually provide a full explanation ://

[slow scroll]

it depends on what A is

[bleak stone]

nvm dont worry about it

[spare crystal]

if anyone has an answer pls ping me ^

[agile gyro]

rank(A) = # indpt rows = # indpt columns

because rank(A) = rank(A^T)

and row operations preserve rank of a matrix

rank(A) = # indpt columns so n - rank(A) = # redundant columns for a mxn matrix A

basically its because row reducing the matrix doesnt change the rank

[spare crystal]

i see that, but why are the redundant columns the same columns in both matrices

like if columns 2, 4, and 5 are redundant in A, the same columns in rref(A) are redundant

i.e. if you want to write out a basis of im(A) you would use this fact

[agile gyro]

because row reducing doesnt swap the columns

[spare crystal]

idk that just doesnt really seem like a satisfying explanation to me ://

[slow scroll]

hold on

[stoic python]

kxrider:

[slow scroll]

@spare crystal

[spare crystal]

ahh okay thank you for writing this

[slow scroll]

npnp

[spare crystal]

thank you this makes sense! gonna have to read it a few more times to fully digest it but I think i got it :))

[sharp merlin]

is anyone here good with matrices

[slow scroll]

[sharp merlin]

I need help understanding 6,7,10 please

[slow scroll]

are these T/F questions?

[sharp merlin]

Yes

sorry for to include that

Im guessing 6 is false because the left side is supposed to be a matrix as well right

[slow scroll]

that is just notation. its the same thing as saying f(x) = Ax

it reads "x maps to Ax"

[sharp merlin]

it should be A(x) -> cA(x)

shouldnt it be like that

[slow scroll]

nope. you plug in x, and you get out Ax. that is all its saying

[sharp merlin]

oh

[slow scroll]

i can't really think of a simple explanation for 6 though. Are you familiar with the fact that the column space of a matrix is a subspace of the vector space in the codomain?

actually wait no

[limber sierra]

this is a fairly important theorem, i kinda doubt that this would be a student's first introduction to it

[well, specifically the fact that it's an iff is important]

[sharp merlin]

yes @slow scroll

[slow scroll]

lets say f(x) = Ax. so for 6, we know that a matrix is defined by what it does to a basis. i.e. the first basis vector gets mapped to the first column, the second basis vector gets mapped to the second column, and so on...

therefore, f(e_1) = [first column of A]
f(e_2) = [second column of A]
...
and so [f(e_1) f(e_2) .... f(e_n)]
is you matrix A. not really sure if there is an easier explanation than that.

does that make sense?

[sharp merlin]

Yeah ty

For number 7, I get it that it would be a linearly independent

That means that there is a free variable

bc only two pivot points

[slow scroll]

columns are linearly dependent*

[sharp merlin]

but idk what Ax = 0 is supposed to mean

[slow scroll]

null space?

[limber sierra]

Ax = 0 is an equation multiplying a vector x by a matrix A

to get... zero

[sharp merlin]

oh

[limber sierra]

the solutions are the set of vectors x such that Ax = 0

such a vector corresponds to a solution of the linear homogenous system that A represents

[sharp merlin]

Ax = 0 is only true when is it linearly independent right

so that means the answer is false

[limber sierra]

not sure what you mean

take every row of A to be the row (1 1 1)

[sharp merlin]

Ax = 0 can only be 0 if c1, c2, c3 is 0 aka linearly independent

[limber sierra]

no?

consider $\begin{pmatrix}1&1&0\1&1&0\1&1&0\end{pmatrix}\begin{pmatrix}1\-1\0\end{pmatrix}$

[stoic python]

Namington:

[sharp merlin]

that has no pivot points

so it is not linearly independent

[limber sierra]

this example can be modified as needs require

$\begin{pmatrix}1&1&0\0&1&1\0&0&0\end{pmatrix}\begin{pmatrix}1\-1\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

if you have any theorems/results on the solutions of homogenous systems, you might want to review those

[sharp merlin]

no

what I meant to say

It is only a trivial solution when its a 3x3 matrix with 3 pivot points

@limber sierra

[limber sierra]

oh, i see what you mean

alright yeah

if the three rows are linearly independent, then the only solution is the 0 vector, as you said

sorry, misunderstood

i thought by "c1 c2 c3" you meant the entries in the vector x

my bad

[sharp merlin]

👍

Can someone explain 10

[limber sierra]

given a set of vectors {v_1, v_2, ... v_n, 0}

is this set ever linearly independent?

you might want to review your definition of "linearly independent"

[sharp merlin]

no I know what that means

its just the notation that messes me up

so an example of this would be

right

and this will never be linear indepedent

bc not possible for pivot in every column

right

[limber sierra]

sure, yeah

using the more "traditional" definition

a set of vectors is linearly dependent if $c_1v_1 + c_2v_2 + \dots + c_kv_k = \vec{0} \implies c_1 = c_2 = \dots = c_k = 0$

[stoic python]

Namington:

[limber sierra]

but if one of those vectors is the zero vector

$c_1v_2 + c_2v_2 + \dots + c_k0$

[stoic python]

Namington:

[limber sierra]

then the choice of value for, say, c_k will never affect the resulting sum

and so c_k isn't "forced" to be 0

it can be whatever

so this set can't be linearly independent

so yeah, 10. is true

[sharp merlin]

ok cool

also for number 9

can someone verify if I did it properly please

last question

@limber sierra can u explain this plz

[vale arrow]

If it is linear it has to verify t(ax)=at(x)

Is that true here?

[sharp merlin]

so i multiply a constant with the left side and see if its the same when i muliply on the right side

nvm

wait it doesnt work when u do t(u+v) check right

@vale arrow

[vale arrow]

Doesn't work for t(u+v) neither for t(α.u)

So choose one counter example

@sharp merlin

[sharp merlin]

ok

[hollow finch]

say i have a product of matrices and a column vector ABCv and some of the entries in B are functions (but only B). is taking the derivative of the product ABCv the same as the product AB'Cv where the entries of B' are the derivatives of the entries of B

[ripe hemlock]

anyone know where to begin this?

[quasi vale]

Have you done some working?

[ripe hemlock]

yeah i tried to get it do Row echelon form

but i kept on getting the set of scalars a_1,a_2,a_3 were zero

which means that its always linear independent no?

@quasi vale

[quasi vale]

show the working

[ripe hemlock]

It's very very rough but aight

[quasi vale]

Can't understand much, but the thing is during your row operations you'll have to divide by some expression in k.

[ripe hemlock]

and then?

[quasi vale]

For that division to be possible, that expression cannot equal 0.

which will tell you what value of k is not allowed

start again with better writing

[ripe hemlock]

aight calm thank you

Still getting stuck

[ripe hemlock]

@quasi vale

[quasi vale]

yeah im doing it myself 1 sec

[ripe hemlock]

i think i found another way to do it, ill let you know if it works but basically ill just get the angle between the two vectors and set it to 0

[half ice]

How do you relate the angle between vectors to linear dependence?

[dusky epoch]

[ripe hemlock]

well if its linearly dependent it means it is scalable

so the angle between the two vectors would be 0

[quasi vale]

well im done with my working let me know if you wanna see

@ripe hemlock If you do it the angle way, you'll have to dot product of u,v , u,w , v,w and if those aren't 0, then take linear combinations of u,v and dot with w or combinations or w,v and dot with u, etc..

I think..

[ripe hemlock]

ill do it in the morning probably

[quasi vale]

I'll send you the pics

you can look at them

[ripe hemlock]

but id like to see your solution pls

thank you

[ripe hemlock]

thank you!

[quasi vale]

np

[wintry steppe]

hello

[orchid osprey]

Hello, im in PreCal (MAC1140) and i posted a question in precal/cal but they told me it look more like a linear algebra, is there anyone who can help?

[limber sierra]

have you tried drawing the matrix and filling in what the values must be?

[orchid osprey]

i don't exactly remember how to do that, i looked through the text book, but it doesnt say anything from a given value (ex: a_21=4) to a_11

[rotund parrot]

column entries have a common ratio of 2 between them, so fill in what you can fill in from the two given values

[limber sierra]

if i told you that 4 was the second term of a geometric sequence with commonr atio 2

could you fill out the other terms of that sequence?

similarly, if i told you that 20 was the third term of a geometric sequence with common ratio 2, can you fill out the other terms?

these correspond to two of the columns

[orchid osprey]

Im sorry, im completely lost, ever since online teaching started I have trouble paying attention to the lectures :/

[limber sierra]

ignore the matrix stuff

for now

do you know what "geometric sequence" means? what "common ratio 2" means?

[orchid osprey]

i dont know what common ratio 2 means, but i've done some geometric sequences

[limber sierra]

er

the common ratio is just the ratio between terms in a geometric sequence

for example, in the sequence 2, 6, 18, 54

the common ratio would be 3

[orchid osprey]

Oh i see

[limber sierra]

we know that the columns form geometric sequences

so for example, the first column (a_11, a_21, ... a_n1) forms a geometric sequence

with common ratio 2

and we know a_21 = 4

can you find the rest of the sequence based off that?

[orchid osprey]

oh, so would a_11 be 2?

[limber sierra]

right

[orchid osprey]

oh ok that makes a lot more sense

[limber sierra]

this gives you the sequence for your first column

you can use the fact that a_34 = 20 to get a similar thing for the fourth column

and once you've done that, because the rows are arithmetic sequences, which you now know 2 terms of

you can fill out each row

this will probably "hint" at what your general expression for a_ij should be

[orchid osprey]

ohh thank you for clearing this up for me

would a_14 = 5, a_24 = 10, and a_34= 20?

[limber sierra]

yes, and this pattern continues for however large the matrix is

[orchid osprey]

how would i find let say a_12, or a_13 from all of this?

would it be a_12 = 3, a_13 = 4?

[pliant harbor]

Let S be the set of analytic functions f such that {z, f(z), f(f(z))} is linearly dependent over C. Prove that S is closed under addition.

[dusky epoch]

good luck trying to prove a false statement

$iz^2 \in S$ and $-iz^2 \in S$ but $0 \notin S$

[stoic python]

Ann:

[dusky epoch]

@pliant harbor

[sonic osprey]

it says dependent

[dusky epoch]

oh my bad

i can't read lmfao

brb committing sudoku

[swift plinth]

Hey Guys! hope everyone is doing well!
I have 2 problems that look really simple and trivial but I can't find an answer:

1. Proof: Given a straight line g and a point P outside. Show it:
   The shortest of all lines PQ with Q ∈ g is the one that is perpendicular to g.

2. Proof:
   Show that a line g with a circle can have a maximum of two points in common.

It's obvious but how do I proof it?

[pliant harbor]

1. Let the perpendicular to g through P intersect g at R. Use Pythagorean theorem on PRQ.

2. We may scale and rotate so that the circle is x^2+y^2=1 and the line is y=x+c. We have a quadratic for x, which has at most 2 solutions.

[sonic rune]

quick question regarding sets, if A is a subset of B, and A and B have the same size, are A and B equal?

[pliant harbor]

Yes for finite sets, no for infinite.

[sonic rune]

cool, thanks

[pliant harbor]

Wrong chat by the way.

#proofs-and-logic or #discrete-math

[sonic rune]

my bad my bad