linear-algebra

[scarlet ermine]

i'm working in the null space rn

so if a is a 6x8 matrix, the smallest number of dimensions in the null space is 2, because there's 2 free variables i think

[vast torrent]

"Null space Rn" doesn't make sense

[scarlet ermine]

so if there's 1 free variable its 1 dimensional

rn = right now

[vast torrent]

Oh sorry lol

[scarlet ermine]

anyway but if there's only 1 solution does that mean that the null space is 0 dimensional?

like theres 0 free variables

[vast torrent]

Do you know the rank nullity theorem

[scarlet ermine]

i've heard of it but i don't understand it

i tried to read the wiki page but it went way over my head

[vast torrent]

Khan academy is probably better

The matrix version is

Let A be an mxn matrix of numbers

[scarlet ermine]

but anyway so i just want to make sure that i understand, so if the same problem was 'a is a 8x6 matrix, whats the smallest dimension of the null space', would the dimension be 0 because its possible there's only 1 solution?

[vast torrent]

The rank is the dimension of the span of it's columns

[scarlet ermine]

i'm having a hard time visualizing it and a 0 dimension anything doesn't make sense but i don't see any other solution

[vast torrent]

The 0 dimensional space is

{0}

The set with the 0 vector and nothing else

Its basis is the empty set

[scarlet ermine]

i see

[rustic panther]

@slow scroll If $A, B \in M_3 (\mathbb{R}) ; and ; det A = -7, det B = 4$, what is $det (A^{-1} B)$ to 2 decimal places?

(don't tell me the answer just point me in the right direction)

Is $M_3$ here a 3x3 Matrix?

[stoic python]

glee:

[scarlet ermine]

so is a 1 dimensional space both a matrix that contains just 1 solution and a solution that has one free variable?

[slow scroll]

hey i pinged you in #help-3

[vast torrent]

Glee i answered , scroll up

A matrix is not a space

[rose grotto]

how do I find the basis

for the subspace S

ik that basis is a row with a leading 1 right

[gray dust]

you can use the condition in the set to do this really quick

[north sierra]

columns

[gray dust]

if we have that all vectors in S satisfy x-2y+5z=0, they also satisfy x=2y-5z

[north sierra]

would this be related to the null space?

kinda curious now

[gray dust]

na

[north sierra]

oh

cause thought it would be a basis of NulA

since youre solving for x

but how would this be done?

[gray dust]

wth is A

@rose grotto reply when you read this

[north sierra]

oh yeah

idk

really curious how to solve this lol

[rose grotto]

@gray dust there’s multiple things that can be A

like infinite possibilities

Right

For a basis of that

[gray dust]

WTH IS A

[north sierra]

lol

[rose grotto]

What u mean

[north sierra]

yeah i dont know where i got A from

[gray dust]

the q makes no mention of anything with the name A

[rose grotto]

Oh ur not talking to me

[north sierra]

what type of basis are we finding?

[gray dust]

you can use the condition in the set to do this really quick
if we have that all vectors in S satisfy x-2y+5z=0, they also satisfy x=2y-5z
all vectors in S of the form (x,y,z) satisfy x-2y+5z=0, and they also satisfy x=2y-5z bc algebra

[north sierra]

yeah

[rose grotto]

thats the basis?

it gives

2 answers

and says the dimension is 2

[gray dust]

if you're still stuck the idea is this. x,y,z are not totally free variables. they are constrained to take on a particular set of values due to this eqn

[rose grotto]

but why is there only 2 answers

[gray dust]

what you can do is rearrange the eqn

and make one of the variables dependent upon the other two

do you see how i got x-2y+5z=0 to x=2y-5z

[rose grotto]

yeah

[gray dust]

from this we can say that y & z are free vars, and whatever values they take on, x must have a value dictated by x=2y-5z @rose grotto

that's kinda what i'm getting at but not quite

and it's not your q

[north sierra]

sry im curious

[rose grotto]

ty

[vast torrent]

It says find "a" basis, there's infinitely many bases

@rose grotto

[rose grotto]

but why is the dimension 2

and not

3

[dusky epoch]

it's the dimension of S, not of R^3.

[rose grotto]

ik

but why is it 2

if thees infinite posibilities

[dusky epoch]

because you've found a basis for it that consists of two elements

and as such, any basis for S will have 2 elements

which is what it means for dim(S) to be 2

[rose grotto]

why 2 elements

[gray dust]

try using this to make a sub:

from this we can say that y & z are free vars, and whatever values they take on, x must have a value dictated by x=2y-5z @rose grotto

[rose grotto]

why not 3 elements

[vast torrent]

what was the original question fruit

[dusky epoch]

there are two elements
one is [2, 1, 0]
the other is [-5, 0, 1]

[north sierra]

original question is here: https://cdn.discordapp.com/attachments/540211747613704221/655947307887493131/Screen_Shot_2019-12-15_at_8.40.07_PM.png

@vast torrent

[rose grotto]

but why can't .I do like 0, 10,5 @dusky epoch

[dusky epoch]

well, [0, 10, 5] isn't in S, for a start.

but you can make another basis for S if you want.

if you really don't like that one

you can make another one

but even then

it'll have

two elements

which is

what it means

for a space

to be of dimension 2

[rose grotto]

I meant like 10,4

0,10,4

@dusky epoch why can't I do that

as one of my elements

then have 3 elements

[dusky epoch]

because if you add [0, 10, 4] to your set

it'll start being linearly dependent

and therefore, not a basis

[rose grotto]

it wont work in the subspace?

like it wont fall under the conditions

is that what u mean

[vast torrent]

do you know what a basis is

[rose grotto]

yeah

LI

column

buut how is that column

LD

[vast torrent]

tell me what a basis is

I've never heard of one before

I know about vectors and stuff

[rose grotto]

oh

[vast torrent]

teach me what a basis is

[rose grotto]

its like

[vast torrent]

(I'm pretending)

[rose grotto]

a column

that has

a leading 1

ya ik

right

[dusky epoch]

[gray dust]

you had me good there fauxpas

(not really)

[vast torrent]

I gave up acting to do math

[dusky epoch]

you don't know what a basis is.

[rose grotto]

isnt it columns

that are Linearly independant

?

[dusky epoch]

jeez

i can't say no because that'll make you come up with even more ridiculousness but i can't say yes either because that's not nearly close enough to a coherent and correct answer

[rose grotto]

;

[timber minnow]

I have something that's half linear half stat, should I post here to move things along lol

[north sierra]

what are you having trouble with mob

[rose grotto]

I dont understand why they chose those 2 basis as the answer

and not something like 0,10,4

[gray dust]

can you re read what i suggested

[rose grotto]

is it cuz

0,10,4 doesn't have a 1 in it

so its not

a basis

[timber minnow]

no?

[half ice]

There's many different possible basis for a space

Especially when it's 2D

[rose grotto]

@gray dust what u said was x=2y-5z

but doesnt 0,10,4

fit that

@half ice ya so why doesn't 0,10,4 work

[half ice]

Well, any possible basis of this space has two vectors

[vast torrent]

you should learn what a basis is

before figuring out how to find one and what isn't one

[half ice]

(5,0,-1) is in the space, but can't be expressed with the span of your basis

[dusky epoch]

0,10,4 doesn't have a 1 in it
so its not
a basis

[timber minnow]

^^ there's your problem

[dusky epoch]

you don't know nor understand what a basis even IS, and any further discussion is not going to fix that

you need to go back to your notes, your textbook, your lecture slides, WHATEVER, and learn the actual definition of a basis, including what type of object a basis IS in the first place, since you seem to be missing even that

[rustic panther]

Definition 4.5. Let V ⊂ Rn
be a linear subspace, a basis of V is a set of vectors v1, v2 · · · , vk ∈
V such that

(i) span{v1, v2 · · · , vk} = V , and
(ii) the vectors v1, v2 · · · , vk ∈ V are linearly independent.
So a basis of V is a set of vectors in V which generate the whole subspace V , but with
the minimal number of vectors necessary

[dusky epoch]

h

davidL this belongs in #probability-statistics

[timber minnow]

kk

wasn't sure bc it's about column space and projection so

[quaint pelican]

These equations can not be solved since the all zero row in coefficient matrix doesn't have 0 correspond in b right(What i mean is Ax=b and here is the augmented matrix (A|b))

[magic slate]

@rose grotto idk if this helps but I'll send.it anyways

since the y and z are just free parameters you can replace them with some s and t

and then you can write <x, y, z> in terms of those two params

[gray dust]

no real need to replace y & z, replacing x with 2y-5z is fine enough

[magic slate]

and you can find that the 2 in the pic are basis vectors

i mean tru

ive just been taught that way

and it makes it a little clearer to me

[rose grotto]

@magic slate I found out how to do it, ty tho

[magic slate]

ok good

[north sierra]

@quaint pelican ??

a and b are in REF form

c isn't

[quaint pelican]

We can make c to REF form by row exchange operation, right ?

[north sierra]

you need to make sure a zero row is below all the other rows

and that all the entries under a pivot entry are 0s

[quaint pelican]

Ok i get it now, but these system of equations is not solvable since they are not consistent, right ?

[north sierra]

it's not augmented so you can't say they aren't solvable

oh it says its augmented

nv

m

yea

none of them are consistent

[quaint pelican]

Ok thanks a lot !!

[north sierra]

np

[quaint pelican]

Can i ask for the books on Linear Algebra which i can use for learning on my own ?

[north sierra]

dm me

i have a really good one

[quaint pelican]

Okay

[wintry steppe]

If I’m trying to prove linear independence of a set of vectors, what exactly am I expecting from my solution when I row reduce my matrix?

So like, if linear dependence is if one of the vectors is a linear combination of the other with at least one of the scalars being nonzero, I know that I either expect a RREF matrix with a unique solution (no free variables) or infinitely many solutions (one free variable

[native lodge]

Ones along the main diagonal

If you rref, you are expecting to get the identity matrix

[wintry steppe]

But that gives a unique solution

Don’t I need my scalars to all be zero for linear independence

[vast torrent]

Amphy Zoro didn't say the matrix was square

[native lodge]

Sad

Say it was square though

You end up with Ix=0

Which then says x=0

So the vector of scalars is the zero vector

[wintry steppe]

Is there a better way of proving linear dependence and independence

Other than putting the vectors in a matrix and solving it

[native lodge]

Not that I know of. rref is the most conclusive way to tell

If it’s dependent, then you know how many dependent vectors there are as well using the rref

[wintry steppe]

I mainly ask this because I use this for when they ask if a matrix is invertible or not and I always have this in my head that if the columns of the matrix are linearly independent then it’s an invertible matrix

[native lodge]

Easiest case that requires little work. If you have more vectors than the dimension of the space you want to span. In other words, you have an mxn matrix where n>m, then surely you have a dependent set of vectors

[vast torrent]

If M is a full rank square matrix with columns a1, ..., an, and the gram schmidt process gives you the orthonormalized e1,...,en, why is the R matrix what it is? <ei,aj> as entries in the upper triangle?

let's say a 3x3

intermission while I work on the latex in bots

$\begin{bmatrix} \vdots & \vdots & \ a_1 & a_2 & a_3 \ \vdots & \vdots & \vdots \ \end{bmatrix} = \begin{bmatrix} \vdots & \vdots & \vdots \ \hat{u_1 } & \hat{u_2} & \hat{u_3} \ \vdots & \vdots & \vdots \ \end{bmatrix} \begin{bmatrix} \langle \hat{u_1}, a_1 \rangle & \langle \hat{u_1}, a_2 \rangle} & \langle \hat{u_1}, a_3 \rangle \ \cdot & \langle \hat{u_2}, a_2 \rangle} & \langle \hat{u_2}, a_3 \rangle} \ \cdot & \cdot & \langle \hat{u_3}, a_3 \rangle} \end{bmatrix}$

{u1,u2,u3} are the Gram Schmidt orthonormalization of {a1,a2,a3}

$\begin{bmatrix} \vdots & \vdots & \vdots \ a_1 & a_2 & a_3 \ \vdots & \vdots & \vdots \ \end{bmatrix} = \begin{bmatrix} \vdots & \vdots & \vdots \ \hat{u_1 } & \hat{u_2} & \hat{u_3} \ \vdots & \vdots & \vdots \ \end{bmatrix} \begin{bmatrix} \langle \hat{u_1}, a_1 \rangle & \langle \hat{u_1}, a_2 \rangle} & \langle \hat{u_1}, a_3 \rangle \ \cdot & \langle \hat{u_2}, a_2 \rangle & \langle \hat{u_2}, a_3 \rangle \ \cdot & \cdot & \langle \hat{u_3}, a_3 \rangle \end{bmatrix}$

$\begin{bmatrix} \vdots & \vdots & \ a_1 & a_2 & a_3 \ \vdots & \vdots & \vdots \ \end{bmatrix} = \begin{bmatrix} \vdots & \vdots & \vdots \ \hat{u_1 } & \hat{u_2} & \hat{u_3} \ \vdots & \vdots & \vdots \ \end{bmatrix} \begin{bmatrix} \langle \hat{u_1}, a_1 \rangle & \langle \hat{u_1}, a_2 \rangle & \langle \hat{u_1}, a_3 \rangle \ \cdot & \langle \hat{u_2}, a_2 \rangle & \langle \hat{u_2}, a_3 \rangle \ \cdot & \cdot & \langle \hat{u_3}, a_3 \rangle \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

here {u_1,u_2,u_3} are the Gram Schmidt orthonormalization of {a1,a2,a3} which are assumed lin. ind

here A = QR, and wikipedia says R is unique is we choose diagonal matrices to be positive. It doesn't say what we need to impose to make it unique if R has nonreal entries

[stoic python]

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

[vast torrent]

Okay i just figured out why R is what it is. My question is still what do we impose to get uniqueness

Wait no i didn't figure it out D:

part of my confusion as to uniqueness is my presumption that the orthonormalization of {a1,a2,a3} is not unique? does GS produce a unique result? D:

[vast torrent]

<@&286206848099549185>

[empty copper]

The GS process requires picking one vector to be simply scaled by its own norm, so I would say that GS produces inherently non-unique orthonormal bases

depending on which vector you pick first

[vast torrent]

well the order of the columns in A determines the order

where A = QR

but the gram schmidt process presumably isnt the only way to get orthogonalization. and WP says R is unique, not that Q is, on second reading

[carmine terrace]

does anyone konw how to do this problem?

https://i.imgur.com/ttn9VFX.png

i can't find it in my notes

[gleaming topaz]

Cant you just do orthogonal projection?

[carmine terrace]

@gleaming topaz is that all it is?

so V onto T or the other way around?

[leaden ermine]

Hi, quick question. If a matrix has orthonormal columns, and another matrix acts on it, will the resulting matrix have orthonormal columns as well(Aside from that other matrix being the zero matrix), regardless of what the other matrix is? I am having a hard time showing that, if it is true.

[vast torrent]

hint: if M is orthonormal, is 5M orthonormal?

or -2M or whatever

any non unit scalar

[leaden ermine]

Hi, Yes

However a matrix would apply different numbers to each entry. So if I had a matrix [[1,0],[0,1]] and a matrix [1,1,1,1] acted on it, the output would be [[1,1],[1,1]]

where my original matrix was orthogonal, but my new one is just a matrix of all 1's, which isn't orthogonal

[vast torrent]

if the determinant isn't +-1, the matrix isn't orthonormal

[leaden ermine]

Ah, you are right

Totally forgot the normal part

[vast torrent]

still worth asking your question about orthogonal matrices

[leaden ermine]

Well this is in the context of primary component analysis

where I am taking a data matrix, A. Performing SVD. I then have a matrix V(Transpose), which are the orthonormal eigenvectors of A_Transpose x A, right?

So that matrix V(Transpose), is an orthogonal matrix, and I will be taking the first k eigenvectors of it

My dataset will then act on that matrix to produce a smaller data matrix. But the question is, does the new data matrix have orthogonal columns

[vast torrent]

not sure

[wintry steppe]

how do i

do c)

[vast torrent]

Why is your basis of vectors such large entries

[vast torrent]

You can scale them all by 1/5 for example

[wintry steppe]

it's just in regards to the word problem

I'm just having trouble understanding

what exactly it is I'm supposed to do

[vast torrent]

Oh

[wintry steppe]

what does this represent

[gleaming topaz]

a in terms of that basis

So basically what you need to multiply each basis vector by in b) to get the vector a

[wintry steppe]

Anyone have a text or some resource to help make linear algebra less abstract?

Although I do know a few applications where I can use linear algebra in

But for the most part, this semester’s Linear Algebra 1 course has been very abstract

For one in Calaculus I had an easier time connecting theory to application but for linear I can’t do that as often apart from some of the more subtle and easier things

[magic slate]

3blue1brown

his video series on the essence of linear algebra

it's intended as a companion for a proper course but he gives some phenomenal intuition on a vast amount of linear algebra topics

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[wintry steppe]

@gleaming topaz

sorry what exactly do you mean

[gleaming topaz]

Well you know what basis's are right?

[wintry steppe]

isn't it like a combination of vectors

that can create

any other vector in that given space

[vast torrent]

uniquely

@wintry steppe

[wintry steppe]

wdym uniquely

a combination of unique vectors?

[dire delta]

Is there a generic name for the symetry present in tensors where you can arbitrarily arrange the diagonal

[tidal kettle]

@wintry steppe by uniquely we mean theres only one set of weights that describes each vector in the space

say we took out basis vectors to be the colums of a matrix a

the linear transformation described by A maps each vector x to one and only one unique b

so for any b in the span of our basis vectors, there is only one unique solution x

[wintry steppe]

So i was watching this video about abstract linear spaces

or vector spaces rather

https://www.youtube.com/watch?v=TgKwz5Ikpc8

and he was talking about how derivatives can be represented as matrix transformations

so would the taylor series expansion of e^x be the eigenvector of that derivative matrix?

i don't know a lot about linear algebra

i just wanna know if my intuition is right

[vast torrent]

If e^x is an eigenfunction then it's an eigenfunction

e^x and the power series of e^x are the same object

[wintry steppe]

well in the video, the coefficients of the polynomial are in a vector

[vast torrent]

Power series are not polynomials

[wintry steppe]

but in the video the matrix is infinitely large

[vast torrent]

{1,x,x²,x³,...} does not generate e^x in any linear combination of them

Which time?

[wintry steppe]

9:00

[vast torrent]

Ill click

Yes this is talking about polynomials

{1,x,x²,x³,...} is a basis for the space of polynomials in x

e^x is not in the span of {1,x,x²,x³,...}

[wintry steppe]

ok

also a second question

What do cross product and dot product mean in a physics perspective?

[vast torrent]

However in the space of, say, differentiable functions

[wintry steppe]

oh keep going

[vast torrent]

e^x is an eigenfunction of d/dx

With eigenvalue 1

[wintry steppe]

ok that makes sense

[vast torrent]

But we don't really know what a basis of "the space of differentiable functions" looks like

So we can't use matrices for that space

[wintry steppe]

which is why i assume its an abstract space

[vast torrent]

Well we know all about it

But not what its basis looks like

And without knowing a basis, you cant use matrices. But its still a vector space

[wintry steppe]

thats interesting

[vast torrent]

The axiom of choice implies that every vector space has a basis

But in general we don't know what they look like for infinite dimensional spaces

[wintry steppe]

sorry but what is the axiom of choice

[vast torrent]

Uh its a proof rule

[magic slate]

thats cause you need infinite basis vectors to represent an infinite dimensional vector space right?

[vast torrent]

To put it mildly

An axiom is

A statement you accept as true without proof, which you use to prove theorems

Like "every statement is true or false"

[wintry steppe]

yeah

[vast torrent]

Or "if a set exists, so does its power set"

Things like that

So you can use axioms to prove every vector space has a basis

Without telling you what the bases actually are

The axiom of choice is

Given any collection of nonempty sets

There's a function that maps each set to an element inside the set

The function "chooses" an element in each set

[wintry steppe]

so we just assume that's true?

also that would be like "Find the max number" right?

[vast torrent]

Yeah, mathematicians have proven you can't derive it from the usual set theory axioms

If you had a set of infinitely many pairs of shoes

You could create a map that sends every pair of shoes to the right foot shoe in each one

If you had a set of infinitely many pairs of socks

You would need the a.o.c. to pick a sock from every pair

We assume that picking a sock from each pair is a well-defined process that we are allowed to do

[wintry steppe]

and that would identify each pair?

i thought it would pick one element from the set

oh

nm

[vast torrent]

No, a choice function would send each set {left sock, right sock} to right sock

[wintry steppe]

right, its a infinite set of sets

[vast torrent]

Even though there's no difference between the right and left socks

It's that level of abstraction you need to prove vector spaces have bases

Without knowing what the bases are

[wintry steppe]

So wouldn't that assume that each set is ordered in one way or another

like if you didn't know what right and left meant, then you couldn't do the mapping

[vast torrent]

You probably don't appreciate what a good question that is :)

The choice function is very very closely related to the concept of "putting sets in order"

Of assuming every set can have a nice order assigned to it in a certain way

If you want to learn more about this

The first chapter of Topology by Munkres explains the axiom of choice

[wintry steppe]

thanks a lot

[vast torrent]

Np

[wintry steppe]

if you dont mind can i ask another question

[vast torrent]

And the relationship between ordering and choosing

I, in fact, do not mind

[wintry steppe]

nice

so what is the physics representation of the dot product and the cross product

im understand the calculations

[vast torrent]

The problem btw with learning about aoc

[wintry steppe]

but i don't get the intuition behind it

[vast torrent]

Is that you get confused whenever you see a news article mentioning AOC

[wintry steppe]

lolol nice

[vast torrent]

Oh I'm not good with those kinds of questions

Ask a physicist

[wintry steppe]

ok

[vast torrent]

Though graphically

The dot product is related to the angle between vectors

a.b=|a||b| cos(theta)

[wintry steppe]

yeah

[steady fiber]

a good physics understanding of the cross product comes from torque

like when you are rotating a wheel

[wintry steppe]

yeah thats what i was wondering

[steady fiber]

the torque is the cross product of the radius and the force

[wintry steppe]

i understand that the torque is perpendicular

but how do you know that the magnitude is the area between the radius vector and the force vector

[vast torrent]

The direction of axb is perpendicular to both a and b and its direction is given by the the right hand rule

The magnitude of axb is the area of the parallelogram with sides a and b

[steady fiber]

the torque is the rate of change of angular momentum

like by definition

[wintry steppe]

yeah

[steady fiber]

$\tau=\frac{dL}{dt}$

[stoic python]

PorosInMyAshe:

[steady fiber]

and $\frac{dL}{dt}=r\times \frac{dp}{dt}+\frac{dr}{dt}\times p$

[stoic python]

PorosInMyAshe:

[steady fiber]

where p is the momentum

the 2nd term is obviously 0

since dr/dt = v, and v x (mv)

is 0

they're parallel vectors

dp/dt is F

and that's how you arrive at the cross product

L = r x p btw

that's why the derivative is that

[wintry steppe]

so when you use the chain rule here

[vast torrent]

Product rule

[wintry steppe]

sorry product rule

is cross product analogues to normal multiplication

[steady fiber]

kind of, yes

[vast torrent]

The product rule looks the same

[steady fiber]

but the order has to be the same

as in the r term is also in front

since it's r x p

not p x r

[vast torrent]

Dot product product rule also looks the same

[steady fiber]

not commutative here

[wintry steppe]

ok cool

that makes much more sense

[steel mist]

anyone online

[wooden pike]

No

[viral mason]

I was able to finish Lin Alg with an A (85) thanks for all your help @gray dust @half ice @dusky epoch @vast torrent

[dusky epoch]

congrats

[viral mason]

thank you

[half ice]

Yay! Glad.

[viral mason]

:)

[gleaming topaz]

Why is the magnitude of the vector product the area of the paralellogram spanned by the 2 vectors?

[cursive narwhal]

Consider the area of the triangle formed by the 2 vectors, $\overrightarrow{a}$ & $\overrightarrow{b}$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

So, the initial points of the two vectors have been joined up and we've connected their terminal points using a straight line.

Now, suppose that the two vectors were directed such that there was an angle of $\theta$ between them.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Now, this is a geometric theorem but we can say that the Area of the Triangle, given by A, is:

$A = \frac{1}{2} \cdot |\overrightarrow{a}||\overrightarrow{b}|\sin(\theta)$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Essentially, we're multiplying the lengths of the two sides of the triangle and the sine of the angle between them. That's something from basic geometry and we've just applied the idea here

Now, of course, you know that the two vectors will form a parallelogram if i draw another 2 vectors that are just parallel translations of the original 2. That second triangle, then, has to be the same as our original triangle. Hence, it's area is also A, given by the formula above. Thus, you have the area of the parallelogram, B, given as follows:

$B = 2A = |\overrightarrow{a}||\overrightarrow{b}|\sin(\theta)$

That's nothing but the magnitude of the cross product of the two vectors.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

If you can't visualize what i'm saying above, try to draw it out. It'll make more sense if you can use my explanation above as a guide but derive most of it yourself. That will, undoubtedly, make it easier to remember.

[gleaming topaz]

@cursive narwhal Thank you, will read it soon and see if I understand. Appreciate the help

@cursive narwhal Not sure if you can simplify it further but I followed to the part where the magnitude of a multiplied by magnitude of b and the angle sin between them is the area of the paralellogram but why is this the same as axb?

I mean magnitude of axb

[cursive narwhal]

Eh there's a way to derive it from the definition of a cross product

Like, the cross product of two vectors is defined by the cross products of the unit basis vectors. We've defined the cross products of any two unit vectors to produce a particular vector using the right-hand rule. I'm pretty sure you can use that to derive the magnitude of the cross product

[gleaming topaz]

Alright, thank you once again.

[wind hound]

Hey, I have question, that I have the answer to, I want others to solve it as well
Question:
Take any matrix of any order, find an algorithm to extract any of the elements in the matrix. Allowed operations are Multiplication by other matrices, addition with other matrices and determinant

the matrices used in the algorithm must only depend on the order of the original matrix and the position of the element you're trying to extract

[sonic osprey]

what are the elements of the matrix

[wind hound]

real numbers

[sonic osprey]

Then it's impossible

[wind hound]

why?

[sonic osprey]

Because non-computable numbers exist

[wind hound]

what do you mean?

[sonic osprey]

just google non-computable number

[wind hound]

ok

So?

as I saw, non computable number are those for whom we can't develop an algorithm

for a number n, we can't figure out n decimal places

@sonic osprey ^

right?

@sonic osprey ??

But they are still real numbers, and follow properties of real number...

If you wish to take non computable number in the matrix, let's say, 'Mu' leave it as Mu in the matrix

who told you to find exact value of Mu

[sonic osprey]

Is that not what "extracting any of the elements in the matrix" means

[wind hound]

Ok ok

look

if the 1st row of matrix is [1, Mu, 22]

and you wanted to extract Mu

then after the algorithm, you'll get Mu

no matrix

that's my question

[sonic osprey]

"as I saw, non computable number are those for whom we can't develop an algorithm"

[wind hound]

Is it ok now?

to find the exact value- yes

[sonic osprey]

So how are you going to find mu

if you're not finding the exact value

[wind hound]

Leave it a 'Mu'

[sonic osprey]

That makes no sense

[wind hound]

This assumes you had the value of Mu in the first place

[sonic osprey]

For example

If the number there was pi

[wind hound]

Ok

[sonic osprey]

And you computed the first couple digits

Pi is a computable number

[wind hound]

Yes

[sonic osprey]

But if you computed the first few digits and got

3.141592

You couldn't tell if it was pi

or just 355/113

And even if you kept computing more digits

You still wouldn't be able to tell if it was acutally pi

or some fraction that's close to pi

[wind hound]

Yes

[sonic osprey]

Or even some other irrational number that's close to pi

[wind hound]

pi * 1 = pi ??

is it right?

[sonic osprey]

sure?

[wind hound]

that's what i'm saying

Nowhere you multiplied every decimal place with 1

but you got pi

right?

[sonic osprey]

okay?

Besides

No actual computer can hold real value numbers

[vast torrent]

What

[wind hound]

Yep, but your abstract mind can

and, algorithms are not only for computers

but also for your mind

[vast torrent]

"No actual computer can hold real value numbers" how is that true

[empty copper]

tfw I can't store the real number 0 in my computer

[wind hound]

"floating points"

All I'm saying is the question is possibl

possible*

[vast torrent]

CAS can't store irrational numbers ? Like "the root of x²-2"?

[sonic osprey]

I mean, it's really not storing them, calculations with them are just algebraic manipulation

[vast torrent]

Algorithms that can't even theoretically be implemented aren't algorithms? How would that even work

[sonic osprey]

And you're right, by store I mean actually store the digits of

[vast torrent]

unless its a rational number with a denominator a power of 2

[sonic osprey]

denominator only divisible by 2 and 5

is what you want there

[vast torrent]

Really? It can't store 1/8 in binary?

[sonic osprey]

Uh I was thinking decimal, but

8 is only divisible by 2 and 5 yes

[vast torrent]

Decimal isn't a thing in float, i was told

Maybe told wrongly

[wind hound]

Guys I wasn't talking about computer algorithm

[vast torrent]

8 would be 0.001 BIN

[sonic osprey]

I mean, usually when people say algorithm, they mean turing computable algorithm but uh

is your matrix square

[vast torrent]

@wind hound there are no mind algorithms that aren't for computers, how would it even be an algorithm

[sonic osprey]

I think I misunderstood your question earlier

[wind hound]

doesn't matter

Yep

[sonic osprey]

His idea is that he has a matrix with all the values already in it

[wind hound]

Yes

[sonic osprey]

But you just don't know what the values in the matrix are

[wind hound]

There you go

[sonic osprey]

So that's why you can do the things

my bad

[vast torrent]

Like the data are encrypted?

[wind hound]

It's okk

[sonic osprey]

so is it square

[wind hound]

No lol

[vast torrent]

Then what

[wind hound]

as far as I know Doesn't matter

[sonic osprey]

but you know the dimension of the matrix right

[wind hound]

Yep

take MxN

[sonic osprey]

Okay it's pretty straightforward then

Just make all the rows 0 except one row

[wind hound]

That's why I said it only depends n certain things

[sonic osprey]

Which you can do by multiplying on the right by an NxN matrix with only one 1 and the rest 0's

Then multiplying by the vector (1,0,0..,0) on the right will give you the first element in that row

[wind hound]

But that will get you a whoe column

whole*

[sonic osprey]

okay yea

so you want to multiply it on the left by an MxM matrix

You can do this in either order, it's just transposed

[wind hound]

then

You still have to get rid of the matrix

[sonic osprey]

Uh I mean

after you do those two steps, you have a Mx1 or maybe its Nx1 vector with all zeros except one place

Where it's the value you want

so you should be able to multiply it by a 1xM or 1xN vector with all zeroes and one 1

[wind hound]

Remember, you have to get the answer in the real number

no matrix allowed

[sonic osprey]

I mean sure

You're left with a 1x1 matrix with the entry you want

Take the determinant I guess

[wind hound]

Yeah

That's the answer

[wintry steppe]

Could someone explain why it is the case that:
If Ax=b is a system of linear equations, where A is an mxn matrix whose rows add up to 0. Then it is necessary that the sum of all the components of b to add up to 0 for there to be a solution

[wind hound]

What do you mean by rows adding up to 0

pls explain

maybe with example

[wintry steppe]

adding all the m 1xn rows

[wind hound]

then what about columns?

[wintry steppe]

what do u mean?

[wind hound]

1xm rows?

there is no row in 1xm

[wintry steppe]

no

[wind hound]

rowXcolumn

[wintry steppe]

oh sorry

I meant the m 1xn rows

do u understand the question now?

[wind hound]

No, its still not clear

ok, so you want to add up the elements with all the elements downward

then you'll get 0?

[wintry steppe]

yes

[wind hound]

Oh

[wintry steppe]

(0 0 ... 0) n times

[wind hound]

then It's really simple

so

what you wanna do is take any number x from real number

then imagine then multiplied with the matrix

it gets distributed all over the matrix

now that matrix is equal to the other matrix 'B'

Right?

[wintry steppe]

not sure I follow

b is a column vector

A is mxn, x is nx1, b is mx1

[wind hound]

how do you get a column vector if you multiply x(real number) with a mxn matrix?

ohhh

okk

I thought x is a real number

[wintry steppe]

no

[wind hound]

Okk

Yeah, I got the answer now

So, try to imagine the matrix multiplication that's going on

I'll go step by step

let's try this with 2x2 matrix and 2x1 matrix and B is 2x1

got it?

[wintry steppe]

k

[wind hound]

So,

first element of the x is x1

and when you multiply, multiply the row of the first matrix with the column of the next

so, first element of A which is at row=1 column=1

gets multiplied by x's first element

right?

[wintry steppe]

ok, standard matrix multiplication

[wind hound]

Yep

now

think about the second multiplication that you'll do, 2nd row with the x

in that too you'll multiply the first element with a1

right?

[wintry steppe]

k

[wind hound]

so, when you get the final matrix/vector

and you add downwards

then you'll add the elements multiplied by a1

and then you take a1 common then you'll get 0

right?

[wintry steppe]

hang on

if I see the matrix multiplication as b = x1*A1+x2*A2+...+xn*An

where Aj , j=1,...,n are the columns of the matrix

[wind hound]

think of it this way, first element of x is always multiplied with the elements of A in the first column

Yep

[wintry steppe]

and I add the columns together

[wind hound]

Then when you add the rows then you can take the x elements as common

[wintry steppe]

it's like adding first all the elements multiplied by x1

like u said

[wind hound]

Yep

[wintry steppe]

right

[wind hound]

then you take it common

understand?

[wintry steppe]

ok, I think I get it

[wind hound]

do you get it now?

now try to do it on a paper

you'll see

[frank blade]

So, morale of the story: there is a relationship between matrix systems and simultaneous (linear) equations.

[wintry steppe]

yea, I'll give it a shot later

ty

[wind hound]

np

[wintry steppe]

what do u mean Boni?

[frank blade]

I only summarised what ColorCookie had explained to you.

Your learnt at some point how to solve simultaneous equations:

2x + 3y = 3
5x + 2y = 10

Every such system of simultaneous equations can be written in terms of the Ax=b.

As they said, try writing them out, and you should see the relationship.

[wind hound]

@frank blade Yep

[wintry steppe]

I'm aware of this, I was just looking in the wrong spot for making sense of the statement

[wind hound]

It happens

[short gorge]

How does one express this as even or odd

is it to do with the number of disjoint cycles

or the size of them

[dusky epoch]

are you asked to find whether this permutation is even or odd

ie determine its parity

you don't have to do any cycle stuff to do that

@short gorge

[short gorge]

The 2 disjoint cycles I've found for it are (1,5,6)(2,8)

on this basis is it odd, because the product of the 2 lengths is even, therefore odd

[dusky epoch]

yes it is odd but not for that reason

[short gorge]

Right. Is it to do with it being an even number of disjoint cycles?

[south sedge]

Hi, I'd like to write the quadratic equation in canonical form. what is the symmetric matrix for this expression? How do I write sqrt(3) as a sum of 2 non-zero numbers?

[gray dust]

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

[stoic python]

RokettoJanpu:

[dusky epoch]

what?

this is not a symmetric matrix. what are you talking about

[flint wharf]

hello, I have a task in matrix to prove or disprove that adj(A+B) = adj(A)+adj(B)
I am not sure how to start with this, am I suppose to write random numbers or use parameters?

[native lodge]

maybe intro to linear algebra by strang?

or just search up practice problems for each lesson you do

intro to linear algebra is probably what is best for you then

[royal vigil]

so I haven't had a class on functional analysis yet, so be kind

but is the dimension of an infinite dimensional vector space the cardinality of its smallest basis?

like for example functions on the reals can be represented by a countably infinite basis (infinite polynomials) or as an uncountably infinite basis (Fourier transforms)

so could we say the dimension of the set of functions on the reals is aleph naught?

[steady fiber]

functions on the reals cannot be represented by Fourier transforms, only functions that are square integrable can be represented by fourier

also you cannot represent all functions over the reals with power series

[royal vigil]

ok, pretend I said continuous and infinitely differentiable functions on the reals

[steady fiber]

still no

e^-(1/x^2) is the stereotypical function

with a point inserted at x=0

infinitely differentiable, continuous, not analytic

[royal vigil]

ok, for the sake of answering my original question (which was not about the semantics of representing functions with series/fourier transforms) what about the set of functions that can be represented by both fourier transforms and power series

assuming it's closed on multiplication and addition and whatnot

[steady fiber]

sure, you can say the dimension of all function that can be represented by a power series is something

apparently the dimension of all functions from R->R is 2^2^aleph naught

kinda big

for polynomials it's 2^aleph naught

[royal vigil]

why would it be bigger than aleph naught?

couldn't all polynomials be represented with a power series?

[vast torrent]

@royal vigil yes they can, and?

[royal vigil]

If it can be represented by power series, then it can be represented as a linear combination of other polynomials

[vast torrent]

Okay, what's the dimension of R[X]?

I don't know

[royal vigil]

Isn't {1, x, x^2, x^3,...} a basis for the set of polynomials?

A set whose cardinality is aleph naught?

[vast torrent]

With 1 you mean

Hmm

[royal vigil]

Yeah

[vast torrent]

Okay yes, it seems the dimension of R[X] is aleph0

Is that bad

[royal vigil]

I mean @steady fiber said it wasn't so I was confused

[vast torrent]

Oh indeed i see he said that

[steady fiber]

https://math.stackexchange.com/questions/2283359/cardinality-of-the-set-of-all-polynomials-with-coefficients-in-mathbb-r

I just went on math stack exchange ngl

and it says cardinality of all polynomials with real coefficients is that

[vast torrent]

That's cardinality

Not dimension as a vector space

[steady fiber]

oh ya you're right

it is indeed aleph naught

for dimension

I'm kinda stupid at times

[vast torrent]

nah

[floral prairie]

line (x-1)/2=2y=z+1

how do i convert that to smt i can use

i want it like ax+by+cz+d=0 or smt

if i set t=(x-1)/2=2y=z+1 i believe this is called parameterize in english? i can then set up a system of quations x,y,z and express them in (t) and find its coefficients

ill try that

[gleaming topaz]

If I have a linear transformation S:R3->R3, would I just put all this into an augmented matrix with the untransformed vectors to the left and then calculate RREF for it to get the transformation matrix for S?

https://gyazo.com/d41800b38fbdc6dc5f0259421e4fa670

[wintry steppe]

How is this not the rotation matrix about y axis?

(0,0,1) lands behind the zy plane from the looks of it so the x coordinate must be negative?

oh wait anti clockwise

MY BAD BYEEE

[wintry steppe]

Hello we had a 2x2 matrix in the exam, and we had to diagonalize it, however the 2x2 matrix has only one lambda, not two

Can it be diagonalized?

[gleaming topaz]

Yes and no, depends on what you got as a basis for the null space, so for it to be diagonalized you need a 2-dimensional null-space

[wintry spruce]

f( [a,b], [c,d] ) = [ a*c , b*d]

what's the name for this operation?

[dusky epoch]

what are those square brackets supposed to mean

[wintry spruce]

@dusky epoch two vectors

[dusky epoch]

oh like column vectors

,,,,, idk i'd call this "coordinate-wise multiplication" if i had to

why do you need a name for it

[wintry spruce]

@dusky epoch i wanted to write it down ha - "element wise product" was about all i had

[terse mirage]

Interesting

apparently this is called the Hadamard Product for matricies

[dusky epoch]

no

hadamard

[terse mirage]

yes that

[vast torrent]

hey guys, looking for more help please with these practice exams

I dont understand the student posted solution for this one

Let A, B, C be nxn invertible real matrices

consider the equation

$\lambda^3 Ax + \lambda Bx = Cx$

[stoic python]

gfauxpas:

[vast torrent]

Show: if n is odd, there is at least one real number lambda and real non-zero vector x such that this equation holds

well before I go to the answer key

how would I approach such a thing

I notice that this equation is the same as

$\left({\lambda^3A + \lambda B - C}\right)x = 0$

[stoic python]

gfauxpas:

[steady fiber]

it does show that $\lambda^3A + \lambda B - C$ is non-invertible

[stoic python]

PorosInMyAshe:

[vast torrent]

oh yes it does

I'm not sure that's helpful, but it's true

well

it means that

[steady fiber]

if you can show that you can find a lambda st that becomes non-invertible

[vast torrent]

hmm

[steady fiber]

then you've solved the problem

[vast torrent]

$\ell(\lambda) = \det(\lambda^3A+\lambda B-C)$

[stoic python]

gfauxpas:

[vast torrent]

should I do this and find roots?

well I dont know how to find roots

I do know cubics have a real root

but this has 3 different matrices

so I can't use anjy theorems I know about char. polynomials

cayley-hamilton or whatever

[steady fiber]

$\det(\lambda^3A + \lambda B - C)=0$ is what we have to prove for us to solve the problem

[stoic python]

PorosInMyAshe:

[vast torrent]

that it has a solution in lambda

yes

[steady fiber]

and that determinant has a characteristic equation with odd degree

which means it has at least one real root

[vast torrent]

but I dont know how to have a characteristic equation of 3 matrices

[steady fiber]

you just sum the terms up

[vast torrent]

I only know for a polynomial in one matrix

[steady fiber]

and find the determinant of it

[vast torrent]

ooh

[steady fiber]

it's not really a characteristic equation

it's just a polynomial

that has odd degree

and you know that an odd degree polynomial

has at least one real root

[vast torrent]

and it's a polynomial that's degree odd degree... how do we know it's odd degree just because n is odd

[steady fiber]

yes

[vast torrent]

yes what

how does it follow?

[steady fiber]

yes we have to prove it has odd degree

[vast torrent]

okay

odd degree in lambda

yes?

[steady fiber]

yes, I'm almost certain it has odd degree in lambda

just have to prove it

[vast torrent]

well it better lol

[steady fiber]

okay, so if we do the laplace expansion, along any row of our summed matrix, we will get a $A_{ij}\lambda^3+B_{ij}\lambda+C_{ij}$ multiplied by a cofactor

[stoic python]

PorosInMyAshe:

[steady fiber]

and that's a degree 3 polynomial

and if we keep doing this, the final polynomial will have degree 3n

which is odd when n is odd

[vast torrent]

wait, that's for each cofactor coefficient, that formula?

[steady fiber]

you can prove it by induction, for a 1x1 matrix, the polynomial would be degree 3, for an nxn matrix, the degree of the polynomial would be 3+degree of the (n-1)x(n-1) minor

ya, the laplace expansion formula

it's not very rigorous here, but do you get what I mean

[vast torrent]

well induction will make it rigorous

[steady fiber]

yes

[vast torrent]

thanks Poros

part 2 now?

[steady fiber]

what's part 2

[vast torrent]

for n =2 , find a counterexample

where there is no real lambda and x!=0 such that the equation holds

I don't want to jus tguess randomly

I want to have an approach rather than just hope I pick the right ones

I see that the answer key picks one of the matrices to be I

[steady fiber]

A = Identity, C = zero matrix

[vast torrent]

that seems like a goodidea

has to be invertible

[steady fiber]

oh true

then I'll make C something else

but ya, starting with identity is probably a good idea

[vast torrent]

$\ell(\lambda) = \det(\lambda^3 I + \lambda B - C)$

[stoic python]

gfauxpas:

[steady fiber]

technically you can just put in the 8 constants

find the polynomial that results

[vast torrent]

I only need one counterxample

[steady fiber]

and manipulate the coefficients however

[vast torrent]

but I don't want to just shoot in the dark randomly until I find one

[steady fiber]

to make a function that's always positive

or whatever

it's not a shoot in the dark

it's getting the polynomial and then using the properties of polynomials to make it always positive

or always negative

[vast torrent]

okay so

[steady fiber]

either works

you can even make C = I or something

[vast torrent]

uh

[steady fiber]

if you want to make it easier

[vast torrent]

let's try making them all diagonal

hopefully thats enough

[steady fiber]

that's a good guess

limits you to 6 degrees of freedom

[vast torrent]

just to make computation easier

[steady fiber]

your polynomial is just $\left(a_{1}x^{3}+b_{1}x+c_{1}\right)\left(a_{2}x^{3}+b_{2}x+c_{2}\right)$ then

[stoic python]

PorosInMyAshe:

[steady fiber]

x is lambda

[vast torrent]

$\begin{bmatrix} \lambda^3 + \lambda a - b & 0 \ 0 & \lambda^3 + \lambda c - d \end{bmatrix}$

[stoic python]

gfauxpas:

[steady fiber]

oh true, we made A = I

[vast torrent]

,w (lambda^2 + lambda a - b)(lambda^3 + lambda c -d) expand

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[steady fiber]

you put ^2

in the first one

[vast torrent]

,w distribute (lambda^3 + lambda a - b)(lambda^3 + lambda c -d)

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

gross

[steady fiber]

lol

[vast torrent]

so uh

if we set a = -c

[steady fiber]

,w distribute (lambda^3 - lambda c - b)(lambda^3 + lambda c -d)

[stoic python]

Display results online and refine query

Requested by PorosInMyAshe#2423

[steady fiber]

becomes that

[vast torrent]

set b = -d

[steady fiber]

make d less than 1

so that it can't be larger than the x^6 term

[vast torrent]

,w distribute (lambda^3 - lambda c +d)(lambda^3 + lambda c -d)

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

okay this is quadratic in lambda^2

[steady fiber]

you can probably put that in desmos

[vast torrent]

oh wait

no it isnt

[steady fiber]

and play around with sliders

lol

[vast torrent]

yeah but I wont have a compuiter on the exam

I will have enough time to multiply polynomials

so I'm cheating now

but I wont have desmos

[steady fiber]

just set both c and d to 1

see what happens

[vast torrent]

I dont want to lose invertibility but I dont think I will

,w distribute (lambda^3 - lambda 1 - b)(lambda^3 + lambda c -1)

[steady fiber]

they're all diagonal

so you won't

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

ah of course, as long as I dont turn them into 0

set b = -a

b=-1 I mean

,w distribute (lambda^3 - lambda 1 +1)(lambda^3 + lambda c -1)

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

getting better

[steady fiber]

https://cdn.discordapp.com/attachments/540211747613704221/657388334544453652/wolf.png over here why not just test out a c and d

[vast torrent]

how'd you get that

[steady fiber]

you had that

it's from above in this channel

[vast torrent]

,w distribute (lambda^3 - lambda c +d)(lambda^3 + lambda c -d)

[stoic python]

Display results online and refine query

Requested by gfauxpas#9497

[vast torrent]

ah, so I did

but I dont want lambda=0 to ruin things

oh nm

-d^2 is not in lambda

hmm

[steady fiber]

ah ya, this simplification doesn't work

too many assumptions

[vast torrent]

the answer key does use 3 diagonal matrices

but doesn't tell you how to come up with them

oh, well

he does do

[steady fiber]

I personally would use another approach

[vast torrent]

one matrix is the transpose of the other

[steady fiber]

make A and B both I

[vast torrent]

to simplify things

[steady fiber]

and allow C to have any entry non-0

then you get $\left(x^{3}-x+a\right)\left(x^{3}-x+b\right)-cd$

[stoic python]

PorosInMyAshe:

[steady fiber]

and all you have to do is make cd a very large negative number

and you're done

if you shift the function up enough, it just cannot have a root

[vast torrent]

can you go back and slow down a little bit please

[steady fiber]

sure

[vast torrent]

my ritalin wore off by now

[steady fiber]

make A=B=I

so both are identity

[vast torrent]

sure, and let's make C diagonal too

[steady fiber]

no

don;t do that

make C = [a b; c d]

[vast torrent]

okay

so the method to this madness is

[steady fiber]

then the determinant is $\left(x^{3}-x+a\right)\left(x^{3}-x+b\right)-cd$

[stoic python]

PorosInMyAshe:

[steady fiber]

and if you set a=b=0

[vast torrent]

to assume 4 degres of freedom is enough to have a coutnerexample?

that's our guess?

[steady fiber]

and make cd extremely huge

it cannot possible have a root

[vast torrent]

trying to build intuition here for how to find counteramples onan exam

[steady fiber]

by setting a,b to 0, you get $\left(x^{3}-x\right)\left(x^{3}-x\right)-cd$

[stoic python]

PorosInMyAshe:

[vast torrent]

we have in theory 12 degrees of freedom to make a counterexmaple

[steady fiber]

which you can simplify to $x^{2}\left(x^{2}-1\right)^{2}-cd$

[stoic python]

PorosInMyAshe:

[vast torrent]

we're narrowing it down to 4?

[steady fiber]

the first term is always non-negative

since it's a product of 2 square terms

[vast torrent]

you cant set a,b, to 0

we need invertible matrices

[steady fiber]

oh true

make one of them really small

0.0001

[vast torrent]

sure

[steady fiber]

and then just make cd massive

like c=-1000, d = 100000

the function is shifted too far up to have a root

[vast torrent]

cool. I got it, I'm just trying to learn intuition here to create counterxxamples on my exam

[steady fiber]

and ya, there you go

[vast torrent]

if I have 12 degrees of freedom

and im trying to make a counterexample

I should make an educated guess that 4 degrees of freedom is enough?

[steady fiber]

make the complicated parts I

like in this case the lambda^3

so we made A = I

i went further and made B = I too

so coefficients of lambda are already set

and then keep a reasonable amount of degrees of freedom from the rest

so instead of 2 for C

[vast torrent]

it could be that any assumption we make is going to ruin our chance of finding a counterexample, but if the space of non-solutions is big enough our chances are pretty good, right?

[steady fiber]

keeping 4 for C is good

it could be

but it's a safe assumption it won't be