linear-algebra

[elder robin]

{4, 8, 12}

[gray dust]

what's row 2 - that?

[elder robin]

Ah, i mistyped

row1*4 - row2

Is that allowed? or is it always supposed to be constant * some_row + some_otherrow

[gray dust]

that's ok but you still messed up

[elder robin]

oh

row1*4 - row2 = {4-4, 8-3, 12-6}

[gray dust]

actually hang on

if you did 4*row1-row2 then...

ok then step 1 is fine mb

you can really just think of that step as scaling row2 by -1 then adding 4row1 to row2

[elder robin]

np, this is my first time doing a problem like this so it's likely that i'm thinking about the problem wrong

[gray dust]

a composition of two legit row operations

[elder robin]

Oh yeah

What I found is reduced echelon form right?

because the columns with leading 1s are filled with zeros

except the 1

[gray dust]

yep it's in ref, in fact it's also in rref

[elder robin]

ok perfect

so if I want to see if two matrices are row equivalent, I compare their reduced row echelon forms?

because rref is kind of like the most basic form of the matrix

[gray dust]

depending on what those matrices are, rref'ing both may be overkill

"basic" isn't how i'd describe it

there are exact definitions on what ref & rref mean on google

to show row equivalence, you just need to perform a series of row ops on one matrix to reach the other matrix

[elder robin]

Oh, so I could either find rref or just make one equal to the other

[gray dust]

depending on what the matrices are, either rref'ing both or turning one into the other may take more work

[elder robin]

makes sense, thank you

Now I have to see if these are equal. I already found rref of the first one, and the second one looks like it is in rref, but I remember reading something like if it is a 1, then a zero on the far right it is a special case

it was related to 1=0

[dusky epoch]

these matrices are not equal

$2 \neq 0$

[stoic python]

Ann:

[gray dust]

you probably meant row equivalent not equal

[elder robin]

Oh yeah row equivalent

but either way I compare the rref right?

[dusky epoch]

now we're talking

[gray dust]

sure you can

[dusky epoch]

yes

the right matrix is in rref already

[elder robin]

Perfect, because I already find rref for the first one

also, if the second one was referring (don't think that is the right word) to a system, it would have no solutions right?

That's what I was thinking of earlier when I was talking about the special case

[gray dust]

if the right matrix is the augmented matrix representing a linear system of eqns then it has a soln

[elder robin]

Ohh I remember now. It's if there are all 0s with a number on the far right

of the augmented matrix

[gray dust]

it represents 1x+0y=1, 0x+1y=0, giving x=1, y=0

[elder robin]

right, because you can have y=0

but if you switched the 1 and the zero on the bottom

it would be 1x+0y=1, 0=1

[gray dust]

yep, no soln, we call the system inconsistent

[elder robin]

awesome

thanks for the help

[gray dust]

no prob

[elder robin]

one more thing actually

is there a symbol for not row equivalent

I've been writing \neq

if I'm comparing rref though \neq should work though i think

because then I'm checking for equivalence not row equ

[gray dust]

\neq is kinda reserved for not equal @elder robin

do smth like $\not\sim$ or just write out "not row equivalent"

[stoic python]

RokettoJanpu:

[elder robin]

ok thank you

[sharp merlin]

actually nvm i dont know

How do u do this

[river jasper]

can someone tell me how I can find the left null space of a given matrix?

[quasi vale]

@sharp merlin You could try converting those equations into Ax=b, turn it into echolon form

[sharp merlin]

Nvm got it

ty

[river jasper]

so something like this

[limber sierra]

you do the same sort of thing you'd do to find the regular null space

just with the matrix and vector swapped

[river jasper]

oh alright thank you

so would the vector becomes a 1x3 instead of a 4x1?

[bitter glade]

I believe the easiest way to find left null space is to find the null(A^T)

A^T being A transposed, idk how to use the bot yet

[gray dust]

$\mathcal N(R^T)$

[stoic python]

RokettoJanpu:

[stuck kayak]

how does r^21 differ from r^2 21

[dusky epoch]

what

what do either of these things refer to

@stuck kayak

[stuck kayak]

whoops

coulumbs law

ik r^21 is the unit vector from point 1 to point 2

so what does r^2 21 represent?

[dusky epoch]

okay, so PRESUMABLY, $r_{21}$ (non-boldface) is a scalar, and more specifically the length of the vector $\bd{r}_{21}$ (boldface)

[stoic python]

Ann:

[dusky epoch]

and $r_{21}^2$ is the square of $r_{21}$

[stoic python]

Ann:

[dusky epoch]

r_21^2 as it were

[stuck kayak]

oh I see

I thought it was notating a different value or smthn

do yk why it is squared?

[dusky epoch]

dunno, s'what the formula says

looks like coulomb's law if you ask me

[stuck kayak]

yeah it is

in vector form

just confused why r^21 is there

tbh

[dusky epoch]

$\hat{\bd{r}}{21}$ looks like the unit vector in the direction of $\bd{r}{21}$

[stoic python]

Ann:

[stuck kayak]

I see so it just gives direction

thanks 😳

[gray dust]

unexpected physics

[rough flicker]

just a quick question, if 2 vectors are linearly independent, does that imply that they are not on the same line?

[dusky epoch]

if you mean not on the same line in the sense of not being scalar multiples of one another then yes

[river jasper]

is the free vector the column where a free variable is?

so would this be correct?

[karmic stream]

A free vector only has a magnitude and direction so that wouldn't be possible. Eg. The entries in the vector are unknown

[limber sierra]

what

oh god, youre using the physics definition of vector, arent you

i dont know what is meant by "free vector" in this context, admittedly

[karmic stream]

I've only vaguely heard of it from secondary school physics

[limber sierra]

even from that perspective, the "entries" in a free vector are certainly known, it's just not necessarily centred on the origin

in any case, this is unrelated to columns of a matrix

[karmic stream]

I thought the definition of such included "the position of the vector is not specified"

[limber sierra]

correct.

why would that make the entries undetermined?

the entries of a vector just state where the vector's endpoint would lie if it was centred on the origin

the vector $\begin{pmatrix}3\0\end{pmatrix}$ represents a 3-unit rightward horizontal line

[stoic python]

Namington:

[limber sierra]

no matter where it begins

[karmic stream]

Yeah i must be thinking of something else lol

[limber sierra]

aaanyway back to the question

conventionally, a "free variable" is a variable corresponding to a column without a pivot, yes

im not sure what "free vector" means in such a context

naively i'd assume they're referring to the vector formed by the free variable column

but without more context, i cant be sure

[elder robin]

Kind of lost on this question, here i'll post my work below

Is the goal to get something like x_1, x_2, x_3 on one side and a,b,c on the other?

a consistent system means it has at least one solution, and an equation is one side = another side

so I'm thinking I just rewrite the system I found (in the bottom left of my handwritten work) so that it is an equation

[quasi vale]

In order to get a consistent system, at least one solution should exist.

That occurs when the rank of the matrix of coefficients(assuming you're given a bunch of linear equations with variables) and the rank of the augmented matrix are equal.

i didn't see your working but try to make this happen

[elder robin]

So I want the rref of the matrix of coefficients to be equal to the rref of the augmented matrix?

I just reread some parts of my textbook and found an example problem so I got it now

[quasi vale]

Cool sorry I didn't respond

not sure about RREF, you can tell the rank from just echolon form

but i think RREF will help more

[elder robin]

The way I ended up solving it, is I found rref, which ended with the bottom row being {0,0,0 | 2a+b+c}, so the only way for the system to be consistent is if 2a+b+c = 0, so that is the equation of a,b,c which makes it consistent.

[quasi vale]

Ah yes nice

[fossil wagon]

I have a question, when we are using gaussian elimination to solve a linear system of equations how come all the equivalent systems maintain the same solutions?

[subtle walrus]

all you do is swap rows, multiply them by scalars and add scalar multiples of one row to another

this does not change the set of solutions of the original equation

[river jasper]

The vectors here are referring to the rows right?

[fossil wagon]

yeah but my question is like, in order to get the vector that's on the right hand side of the equal sign for both systems is the same

but my question is why

like why does this happen?

[subtle walrus]

so, swapping rows only changes the way we write things down, so it's fine

multiplying by scalars also only changes the way it is written down

you could just divide by it again

and you can add equations to one another because they must both be true

[fossil wagon]

hmmm...

[subtle walrus]

so you kinda just add information of one row to another row

to make it simpler

[fossil wagon]

and you can add equations to one another because they must both be true
this part is still confusing for me

[subtle walrus]

if you know that a = b and c = d, then you know that a + c = b + d

[fossil wagon]

yea I know about that property, I just don't know very well what it exactly means

[subtle walrus]

it just means that you can add the same thing on both sides of an equation

and the equation will remain true

[fossil wagon]

hmmmm...

[subtle walrus]

i mean you probably do this all the time

it's just in this case you add more elaborate stuff on one side

but the idea is the same

[fossil wagon]

okay, let's say I have this

2X + Y = 0
X + Y = 1

3X + 2Y = 1

what does "it will remain true" mean? what will remain true? in the third equation

[subtle walrus]

ok, if we do this real slow, you start with 2X + Y = 0

you add 1 on both sides

and get

2X + Y + 1 = 0 + 1 = 1

but the second equation tells you that 1 = X + Y

so you can exchange the 1 on the left hand side with X + Y

to obtain

2X + Y + (X + Y) = 1

and this simplifies to 3X + 2Y = 1

so yeah if the first two equations hold, the third one holds as well

[fossil wagon]

so... the third equation maintains the solution for all the equations involved?

[subtle walrus]

well, no

not in the sense that you can reconstruct the first 2 equations from it

it's more like you add the information of the second equation to the first

or vice versa

but given any two of those equations, you could "reconstruct" the other one

and in the gauss algorithm you keep one of the original equations, so this is fine

[fossil wagon]

hmmm...

not in the sense that you can reconstruct the first 2 equations from it
this confused me a little bit, I mean I can start with the third and generate one of the two previous equations, right?

[subtle walrus]

if all the information i give you is " 3X + 2Y = 1 "

[fossil wagon]

ohhh

[subtle walrus]

then you can't generate either of the first 2 equations

but if i give you either of the other ones as a bonus

[fossil wagon]

because I don't know them?

[subtle walrus]

you can give me all 3 back

[fossil wagon]

I'll be right back with more questions because this sounds simple but it's pretty difficult for me to understand what's going on haha

[subtle walrus]

this is just, to explain why you don't "lose" information in gaussian elimination

maybe another way to think about it is

you can always do gaussian eliminations backwards

to reconstruct the original matrix

so the information encoded is the same

[fossil wagon]

Maybe this is difficult for me because I'm trying to understand this concept by using lines

and see what every line looks like after adding to lines

😦

[subtle walrus]

this intuition will break once you have more than 3 variables

[fossil wagon]

i know

I just don't understand this very well

[subtle walrus]

also i don't think it's a good way to understand gaussian elimination

it's best to translate it into a system of linear equations and make sure that at every step you only make "legal" moves

(but maybe i suck at explaining this)

[fossil wagon]

hmmmm...

[zinc lava]

Because ... counting and object permanence. Both of these tell us why addition and subsequently subtraction, multiplication and division should work. Furthermore adding equivalent things tells us why linear equations should work.

[fossil wagon]

this just doesn't click in my head

[zinc lava]

okay. Sorry. Do you believe two different equations can have the same solution(s)?

Like 3x + 5 = 11 and 6x + 10 = 22

[fossil wagon]

ehmm yes

[zinc lava]

so do you believe one way to change one equation to another equivalent equation is by multiplying or dividing both sides by a non-zero number?

[fossil wagon]

yes because they are both the same(?

[zinc lava]

well let's think about that.

3x + 7y = 13 Does it have the same solutions as 6x + 14y = 26?

[fossil wagon]

hmmmm yes?

[zinc lava]

So okay here's the long way to explain this (Because honestly I can't think of an easier way).

3a + 7b = 13 and 6c + 14d = 26. I want to show if (a, b) is a solution to the first then it is also a solution to the latter.

We know now 3a + 7b = 13. This means that 6a + 14b = 26 by just multiplying both sides by 2.

yeah that's it. It wasn't as bad as I thought.

Technically we should show that also if 6a + 14b = 26 we can also get to 3a + 7b = 13.

in any case matrix operations are just multiplying an equation by a non-zero real number and adding two equations together and collecting like terms.

[fossil wagon]

but my question really is why does gaussian multiplication work, like why does every equivalent system of equation preserve the same solutions

[zinc lava]

hmm cool. Is this for a college class? Can I use the word "uniqueness"?

btw, um every equivalent system by definition preserves the solution but you're asking basically if two systems differ by a matrix operation why does it preserve the solution.

[fossil wagon]

hmm cool. Is this for a college class? Can I use the word "uniqueness"?

I learn math as a hobby

[zinc lava]

ah okay. just wondering.

but does the paragraph above clarify what you're wondering about?

"If two systems differ by a matrix operation why does it preserve the solution?"

[fossil wagon]

btw, um every equivalent system by definition preserves the solution but you're asking basically if two systems differ by a matrix operation why does it preserve the solution.
yea this is what I'm asking

[zinc lava]

So that again boils down to ... we only use row operations that preserve the solution. Logically that's saying we only allow "useful" tools on matrices. And as we can "handwavy" tell, multiplying a row / equation by a non zero constant or adding two equations together is fine.

We can't just square all the numbers in a row for the reason that it would likely not preserve the solution set.

[fossil wagon]

hmmm...

[zinc lava]

lol

Could you frame that hmm in the form of a question, pls?

[fossil wagon]

it's just that I don't understand why these row operations work

[zinc lava]

Was the explanation about multiplying/dividing both sides okay?

[fossil wagon]

yes

[zinc lava]

Like would you agree that multiplying and dividing both sides of an equation by a nonzero constant would not change the solution set.

ok cool

[fossil wagon]

I understood that one, but what about adding?

[zinc lava]

so the other one is adding two equations ... and that's a short thing too.

[restive hound]

Are the eigenvalues of an orthogonal matrix always 1?

[zinc lava]

3x + 7y + z = 13
x + 2y + 3z = 5

3x + 7y + z + 5 = 13 + 5
(adding 5)

3x + 7y + z + x + 2y + 3z = 13 + 5
(substitution of the left 5 with x + 2y + 3z)

and collect like terms.

it's a bit more involved than that but that's ... sorta what goes on.

[fossil wagon]

hmmmm so the third one relates the solutions for both equations?

[zinc lava]

The three equations together has the same solutions as the original two.

[fossil wagon]

this is what I don't understand

[zinc lava]

I'm just simplifying:

3x + 7y + z + x + 2y + 3z = 13 + 5
4x + 9y + 4z = 18

So the two equations above have exactly the same solution set right?

just collecting like terms.

[fossil wagon]

yes

[zinc lava]

(also not clear is a perfectly good answer)

whew ... good. I don't have to talk about chocolate!

[fossil wagon]

yea it's not clear to me

I'm trying to figure out what exactly I don't understand haha

[zinc lava]

okay. so here's the (sorry for being kiddy) chocolatey math thought thing.

[fossil wagon]

it's alright!

[zinc lava]

you have x, y, and z

they count say for our discrete purposes chocolates but if you like a continuous example pounds or kg of chocolates.

X is a triangular box, y a square box, and z a pentagonal box.

it's not difficult to see:

3x + 7y + z + x + 2y + 3z

is equivalent in this metaphor to 3 triangular, 7 square, one pentangonal, 1 more triangular, 2 more square, and finally 3 more pentagonal boxes.

or ... 4 triangular, 9 square and 4 pentagonal or 4x + 9y + 4z

This is always true. It doesn't matter what the other side of the equation is.

Sorry had to throw that in there.

But we have:

3x + 7y + z = 13
x + 2y + 3z = 5

and
3x + 7y + z + x + 2y + 3z = 13 + 5

All 3 together must have the same solution set as just the two above. Proof by contradiction.

If

3x + 7y + z = 13
x + 2y + 3z = 5

and

3x + 7y + z + x + 2y + 3z = 13 + 5 + k

for some nonzero k, we could subtract both equations and get:

0 = k. Since k is defined as nonzero we have a contradiction.

[fossil wagon]

okay I'm gonna read this a few times :)

[zinc lava]

So adding equations is okay and when you add equations you cannot get any other non-equivalent equation. You won't get either more nor less solutions.

[fossil wagon]

@zinc lava okay I get it know!!!! thank you, now I have one more question, why do add two equations together, then substitute one of the two equations with added with the result gotten?

(Sorry for pinging you)

wait is it because the third one has the same solutions as the one I substituted?

but the point is that the third one is simpler?

[fossil wagon]

wait wait wait wait wait

[fossil wagon]

so... when we add two equations at the same time, we are basically getting a third equation that's true for some values of our variables where both equations are true, right? And one of the reasons why gaussian elimination works is that when we change one equation for another that maintains the same solutions we are not really modifying the overall answer, we are finding simpler equations whose solution is the same

[wintry steppe]

would c be p x q as well ?

[bitter glade]

Yes

[feral mountain]

no

assuming I is identity matrix

[bitter glade]

I'd assumed the identity matrices were of the same size as A to make it symmetric, am i missing something? @feral mountain

[feral mountain]

One identity matrix has size p x p and the other has size q x q

@bitter glade @craggy lark

[bitter glade]

that makes sense, i completely missed that

[viscid vale]

sorry my work has a lot of arithmetic, but im trying to use the gram schmidt process to find an orthogonal basis for the column space of this matrix

and im literally just pluggin numbers into the formula my textbook says

and i feel like ive quadruple checked my arithmetic, but i cant seem to get my v3 to be orthogonal to v1 and v2 (v1, and v2 are both orthogonal)

[zinc lava]

In R^3, v1 x v2 (cross product) always gives THE vector (up to opposite direction) orthogonal to both (if possible ... they might be parallel and then there's a lot of answers)

(ps the zero vector also causes trouble.)

[viscid vale]

wait why is the cross product relevant?

arent we concerned about the dot product for orthogonality?

[zinc lava]

a dot b = 0 means they are orthogonal. But a x b gives you an orthogonal vector.

[viscid vale]

ah i see what u mean

[zinc lava]

but wait are you in R3?

[bitter glade]

yeah ^^

[viscid vale]

im in R4

i quadruple checked my arithmetic tho

[bitter glade]

gram-schmidt uses dot products

[zinc lava]

I can be ignored then. Because you want the orthogonal space?

[viscid vale]

ya, im trying to find an orthogonal basis that spans the same space that the column space of that matrix does

i cant just multiply that vector by a scalar? (10)

well im just scaling it to get a whole number

i dont think that would give me whole numbers (?)

wait sorry im really confused rn

if we call the line with v2 = blah blah line 1, which line are u talking about?

ah, for that part its cuz of that scalar from doing x3 dot v1 / (v1 dot v1) all times the vector

maybe its easier if i take a pic of this formula im using

[zinc lava]

Hmm. I think I know the problem.

So we do v1 and v2 to find a vector orthogonal to both?

Rinse and repeat for v1 and v3?

[viscid vale]

heres the process im using

[bitter glade]

that all makes sense

[viscid vale]

OH

LOL

[bitter glade]

but when you were finding v3 you scaled the vectors by different scalars

[viscid vale]

u right jinkies LOL

i dunno why i thought v1 dot v1 was 10

thanks a ton!

i was so focused on that last term with all the fractions i forgot to check the easy stuff

my textbook said thats its usually better to normalize at the end after u found the basis

wouldnt that make the arithmetic even more annoying?

ah i see

thats true, numpy everything lol

[stoic python]

Anesthetic:

Jinkies:

[dusky epoch]

uh

\| produces double bars

you don't need to do \|\| v \|\|

[gray dust]

try \norm

$\norm{\beta}$

[stoic python]

RokettoJanpu:

[spiral sonnet]

How would I go about doing this problem?

[gray dust]

hint: let w=(2,2,2)^T. A(v+w)=?

[spiral sonnet]

OH

I didn't see it that way

I was just doing some linear combination of v

[gray dust]

yeah never hurts to recall that matrices represent linear maps

[hollow heath]

Does this statement mean $A = \begin{bmatrix} 8\8\8 \end{bmatrix}$ ? I'm trying to do Gaussian elimination and trying to figure out how to tack on (or augment) the right hand side (b) while I'm doing the elimination. I've only seen examples where right hand side b is a vector

[stoic python]

hydr063n:

hydr063n:

[broken hawk]

matrix with vertical bars usually means determinant of that matrix, in my experience

[hollow heath]

Oh crap. That actually makes much better sense. Thank you

[restive hound]

Are you not allowed to put a matrix into rref to find its eigenvalues?

If so why?

[dusky epoch]

well the rref of any nonsingular matrix is the identity

and the only eigenvalue the identity has is 1

so yknow

the eigenvalue information gets lost

eigenvalues are not invariant under row ops

[restive hound]

I c

Ty

[spare crystal]

my professor said that if rank(A) = n (for a system with n variables), then there could be either exactly 1 solution or no solutions

is this correct? how can there be no solutions if rank(A) = n

wouldnt its rref always just be an identity matrix

[dusky epoch]

is n the column count

bc the row count could potentially be greater than n

[spare crystal]

hmm i guess

ok that makes sense lol i was just being dumb

thank you!

[jagged spindle]

Not gonna lie, I've been kind of ignoring maths lately on my uni and I need to fix it now. i just need to know what to do, do I just multiply these matrices? What is this A^T? Why is A repeated twice. I hope I translated it well to english so you can at least understand what I have to do because I don't.

[pallid rampart]

A^t is the transpose

Where you flip the matrix along it’s diagonal

So $\begin{bmatrix}1&2&3\4&5&6\7&8&9\end{bmatrix}^T=\begin{bmatrix}1&4&7\2&5&8\3&6&9\end{bmatrix}$

[stoic python]

Whoever:

[jagged spindle]

So I multiply the transposed A matrix by M matrix and then once again by A?

[pallid rampart]

Yes

[jagged spindle]

Thanks

[limber sierra]

uh

hold up

matrix multiplication is preformed right-to-left

i mean obviously it's associative but

make sure you're doing either $A^T(MA)$ or $(A^TM)A$

[stoic python]

Namington:

[pallid rampart]

Well he did say that he’ll calculate A^TM then multiply that to A

[limber sierra]

yes, but i've seen students do $A^TM$ and then literally multiply by $A$; as in then multiply by $A$ on the left

[stoic python]

Namington:

[limber sierra]

effectively calculating $A(A^TM)$

[stoic python]

Namington:

[limber sierra]

which is obviously wrong

[pallid rampart]

Damn ok

[limber sierra]

just worth clarifying

[jagged spindle]

I will have that in mind

[limber sierra]

order matters in matrix multiplication, and its best to preserve it

[jagged spindle]

thanks bud

[limber sierra]

[indeed, it's generally more natural to preform matrix multiplication right-to-left in general, since matrices are just linear functions and function composition is right-to-left; but it doesn't really matter as long as you keep the ordering the same]

[gloomy arrow]

Wouldnt this be false?

The pivot columns would create a basis for ColA wouldnt they?

[dusky epoch]

well they're cols

so yea

[half ice]

@gloomy arrow
Yes you're correct. Nul(A) is instead the vectors that get mapped to 0.

[icy osprey]

anyone help me out?

[rough flicker]

Does determinant of coefficient matrix being different from zero implies that system is linearly independent and is a basis for Rn?

[normal canyon]

no

@rough flicker

detemrinant of 0

[rough flicker]

sry not zero

[normal canyon]

means linearly dependent set of vecotrs

[rough flicker]

i corrected..

so if its not zero, is it linearly independent?

[normal canyon]

yes

[rough flicker]

ok thanks.. 🙂

[viscid kernel]

Can anyone explain this geometrically ?

[half ice]

Ugh

Normally I take that to be part of the definition

But you want to know how to get that, with the "determinant is the volume made by the three vectors as columns" definition?

[viscid kernel]

Nah just in 2d is fine

I just wanna keep it simple first

[half ice]

Oh thank God

[viscid kernel]

Lol

[bitter glade]

Have you watched the video?

[viscid kernel]

Yes I did, I still cant figure that out

[bitter glade]

okay

a 2x2 matrix transforms R2 right?

[viscid kernel]

Yup

Just in 2d, so I can understand it better

[bitter glade]

and in the video did you understand how col1 and col2 of A scale ihat and jhat

[viscid kernel]

Yeah

[bitter glade]

okay so imagine you scaled the 2x2 identity matrix of [1 0][01]

using matrix A [20][02]

the determinant would be 4 because everything is scaled by 4 correct?

[viscid kernel]

Aight Im out, I understand it lol. I was just overthinking

[bitter glade]

okay lol

[gray dust]

you messed up

sometimes it's nice to think of matrix multiplication as: the $(i,j)$ entry of the product of two matrices $A$ and $B$ is the dot product of the $i$th row of $A$ and the $j$th column of $B$

[stoic python]

RokettoJanpu:

[half ice]

@real plaza
I like to pick up the vector, rotate it counter-clockwise, then "run it down the matrix"

( 1 2) (x) = (1x + 2y)
(0 -1) (y) (0x - 1y)

[slow scroll]

wdym formal way?

[stoic python]

kxrider:

[half ice]

The multiplication itself is the formal definition. "Rotating the vector" is just an easy way to remember it

[slow scroll]

the reasoning you use disc is just "by the definition of matrix multiplication." I doubt even the most stickler of teachers care for you to clarify how you are defining matrix multiplication, as long as you are aware how to do it

[half ice]

While we're talking about easy ways to remember things,

That's an easy way to multiply any two matricies

Put the second one above like that. You get the size instantly

[restive hound]

Would like to confirm if my proof is right.
If x and y are eigenvectors of matrix A, then
x+y is an eigenvector of A.
My answer is false because

Ax = mx where m is the eigenvalue of x
Ay = nx where n is the eigenvalue of y
Thus
A(x+y) = mx + ny
For the vector x+y there is not a unique eigenvalue thus x+y is not an eigenvector of A

[half ice]

Your line of thinking is correct, it would be best to use a counter example to really show it

[restive hound]

I see

Tyty

[half ice]

Note that if x and y have the same eigenvalue, then x + y is an eigenvector! You can use the above to show it

[restive hound]

Ic

[restive hound]

Oh also, when I'm finding determinant, can I apply one row operation and stop there. Then if I see one row is identical to another then can I just conclude that the matrix has a determinant of 0?

[slow scroll]

ye

[restive hound]

Kk ty

[ocean sequoia]

what is everyones thoughts on linear algebra the right way

[slow scroll]

doing it the right way is certainly better than doing it the wrong way

[ocean sequoia]

hahaha

i mean the book

[slow scroll]

oh yea I would say axler is good

[ocean sequoia]

yea itll be my second go round

thanks

[coral ferry]

do the solutions have to be 0?

[pallid rampart]

No

[coral ferry]

ok so it doesnt have to be homogenous

confused me for a second

[pallid rampart]

Oh

[coral ferry]

oh

?

[pallid rampart]

No it has to be homogenous

[coral ferry]

why?

[pallid rampart]

Did you read the proof?

[coral ferry]

the row operations preserve the solution set

yea but i still don't understand why it needs to be homogenous tho

[pallid rampart]

Ok so row operation can be thought as multiplying an invertible matrix of a specific kind

[coral ferry]

ehhhh invertibility hasn't been introduced yet

[half ice]

I don't know if it particularly needs to be homogeneous, but the theorem is demanding it is

I think things get messy if that's not the case

[coral ferry]

idk i checked the proof and i don't see why any of the steps require it to be homogenous

like i can duplicate them without it

[pallid rampart]

Well invertibility is already introduced

See theorem 2

[coral ferry]

oh is that what invertibility means

[pallid rampart]

Yeah the inverse operation

[coral ferry]

ah okok

continue

[pallid rampart]

So basically $B=E_1E_2\cdots E_nA=EA$ for some matrices $E_1,\dots,E_n$

[stoic python]

Whoever:

[pallid rampart]

And if x is a vector such that Ax=0, then Bx=(EA)x=E(Ax)=E0=0

[coral ferry]

what's E?

a row operation?

[pallid rampart]

Oh row operation as matrices are not introduced yet

So each row operation can be represented ad multiplying by a matrix

Called elementary matrix

[coral ferry]

i think they are

i didn't really get the notation for it though

but I guess i understand what it's saying

[pallid rampart]

Um

Think of the elementary operation as a function that takes in a matrix and outputs another matrix

So e(A)_{ij} is the i,j th element of the output of A

[coral ferry]

a function that takes in a function hmm

ah so A_ij is the return of i,j as the input for A

[pallid rampart]

So A_ij is the number at row i, column j of matrix A

[coral ferry]

ye

[pallid rampart]

So the elementary operation of type 1 is to multiply a row of a matrix by a number c

Specifically, row r

The second type adds c times the s'th row to row r

the third type interchanges row r and s

[coral ferry]

yea but what does everything before the last bit of each sentence mean?

ie e(A_ij) = A_ij

and the i doesnt equal r part too

[pallid rampart]

That means leave everything else the same as before

[coral ferry]

ohhhh

ok so i understnad the elementary row operations thing

what about the homogenous thing

i sent way before?

why does it have to be homogenous?

[pallid rampart]

sure

So let's see a specific example

$\begin{bmatrix}1&2&3\4&5&6\7&8&9\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=0$

[stoic python]

Whoever:

[pallid rampart]

So this is equivalent to saying

$$x+2y+3z=0$$
$$4x+5y+6z=0$$
$$7x+8y+9z=0$$

[stoic python]

Whoever:

[coral ferry]

yea

[pallid rampart]

Ok

Let's say we apply row operation 1

On row 1

so multiply everything by 2

So we have

$$2x+4y+6z=0$$
$$4x+5y+6z=0$$
$$7x+8y+9z=0$$

[stoic python]

Whoever:

[coral ferry]

mhm

[icy osprey]

I got a question but i will wait for snoopy

[pallid rampart]

now if the constant there wasn't 0, like 1, then the resulting value will be 2

[coral ferry]

yea

[pallid rampart]

so that is basically the reason

when you apply row operations, if all the constants are 0, then no matter what row operation you do, the constants remain 0

But if the constants are not 0, then they might change

In other words, if A and B are row equivalent, then Ax=b and Bx=b might have different solution sets

If b is not the zero vector

[coral ferry]

mm

i guess ill think about it for a second

thanks for evyerhting

:)

everything*

[pallid rampart]

lol sure i don't think i explained very well xD ask me if you have any question

[icy osprey]

hold on

[coral ferry]

yea sure

thanks

[icy osprey]

if anyone can help a brother out, that would be much appreciated

[coral ferry]

@pallid rampart rip i still don't get it

:(

like if you have a an equation from a system

and you show that its a linear combination of equations from another system

then the solution set for the equations must all be the same

isn't that what we're effectively doing with the row operations from rows in A to get a matrix B?

i don't understand why they need to be homogenous

[slow scroll]

remember that when you apply row operations to a system, you change the LHS and the RHS. if you start with Ax = b, for non zero b, and apply row operations to A to get to B, then you don't have Bx = b, but you have Bx = c for some c that you obtain by applying row operations to b.

in the case when b = 0, applying row operations to 0 does nothing, so A reduces to B while 0 reduces to, well, 0.

[long blade]

hello

can someone help me with this quesiton

the awnser i worked out was

r = (1,4,5) + s(4,0,-7) + t(1,1,-2)

im just wondering if this correct

and im also wondering

if theres a lot of awnsers for this

for example

it could be

.... + t(2,-2,-3)

or

....+t(0,4,-1)

if someone can clarify this

also

is there a proper method to work this out

i just worked this out in my head since it has to be perpendicular, i just experiment with numbers that add up to 0

i also need help with the cartesian equation

thanks

in advance

can someone help me pls

[quasi vale]

Can you show your working

[long blade]

um

i didnt really have working out

since the question

says

the point

(1,4,5)

and i know the general equation for this

would

be

r = x_0 + sv_1 +tv_2

x_0 = (1,4,5)

and the v_1 and v-2

would be vectors

that are perpendicular

to

(7,1,4)

im not really sure what the proper way to work this question is

[coral ferry]

@slow scroll but thats saying that A and B have the same y values

which is not always true for equivalent matrices

[slow scroll]

here um ill ping you about this in another channel

[quasi vale]

How'd you get the 2 vectors that are perpendicular to 7,1,4?

[coral ferry]

yea that sounds good

[long blade]

@quasi vale i just randomly made up the numbers

thats why im confused since there can be unlimited

thats why im wondering if theres a proper way to work it out

[quasi vale]

Okay

Listen

when it says the plane is perpendicular to the vector "..."

It's just giving you the normal vector of the plane

[long blade]

a bit confused but keep going

[quasi vale]

cartesian form of the plane is ax+by+cz=d, where the vector perpendicular to the plane(or the normal vector) is <a,b,c>

[long blade]

oh

so the cartesian would be

[quasi vale]

(x,y,z) is any point on the plane

[long blade]

7x +y +4z+ d?

[quasi vale]

So you plug in your point and the normal vector, solve for the constant d

[long blade]

what would d be?

oh okay

[quasi vale]

Find d by plugging the point

[long blade]

so if i use

0,4,-1

wait

d =0 ??

[quasi vale]

?

[long blade]

wait a minute

[quasi vale]

What are you doing

[long blade]

i subbed x= 0

my mistake

how would we find x y z

[shy atlas]

if (x, y ,z) is a point on the plane then the vector <x - 1, y - 4, z - 5> is parallel to the plane so dotting it with the normal vector would result in 0. so <x - 1, y - 4, z - 5> . <7, 1, 4> = 0 is the equation of the plane

[long blade]

oh

so

7(x-1) + (y-4) + 5(z-5)

which equals

7x - 7 + y - 4 + 5z -25

so

d = 31

[shy atlas]

4(z - 5)*

[long blade]

yeah

meant that

d = 331

31

then

oh okay

so i understand

the cartesian

part

but um

the vector parametric form

still confuses me

https://cdn.discordapp.com/attachments/540211747613704221/696940972218777620/IMG20200407143120.jpg

this was the question

r = (1,4,5) + s(4,0,-7) + t(1,1,-2)

was my awnser

so can v1 and v2 be any vector

that is perpendicular

to (7,1,4)

[shy atlas]

lemme draw a diagram

[long blade]

okay

[shy atlas]

so the red vector goes from the origin to a point on ur plane

and hte green vectors lie completely in the plane

if u span these green vectors u get your entire plane

and the red vector is effectively moving this plane around

[long blade]

hmm

so the vector

represents

in my question

(1,4,50

(1,4,5)

?

RED*

red* vector

[shy atlas]

yes thats a valid red vector

any point would do tho

[long blade]

hmm

still a bit confused

honestly

[shy atlas]

hmmm how should i explain this

[long blade]

wait

how can

the red vector be

(1,4,5)

because

in the question

it says

the plane through the point

if its "through"

[shy atlas]

yeah so that means the point (1, 4, 5) is in the plane ?

[long blade]

im confused about the word through

ohh

okay

its in the plane

okay

the green vectors

[shy atlas]

u see the normal vector alone cant define a plane

[long blade]

will

be the perpindicular

vector

[shy atlas]

you can get infinitely many parallel vectors which would have the same normal vector

[long blade]

s

aah

so my awnser

[shy atlas]

so u need a point which the plane goes thru to define the plane

[long blade]

is correct??

r = (1,4,5) + s(4,0,-7) + t(1,1,-2)

[shy atlas]

the green vectors would have to lie in the plane

i.e. normal to the normal vector

[long blade]

so normal to (7,1,4)

?

[shy atlas]

yeah

[long blade]

so basicaly

any vector

that is a normal

to the (7,1,4)

will be in the plane

[shy atlas]

will be parallel to the plane

[long blade]

and thus it can be the awnser

[shy atlas]

well yeah kinda

[long blade]

the "kinda"

makes me confused

hahahahaha

[shy atlas]

if both of the green vectors u choose lie in a line then u cant reach every point on the plane with them

[long blade]

??

this makes me more confused

[gentle yacht]

hey guys im new here, im trying to do this HW and im not sure if this is correct or not, anyone can help me pls ?

[long blade]

hmmmmm

[long blade]

wait

[shy atlas]

like this

[long blade]

how do i know

though

[gray dust]

@gentle yacht go to #calculus

[long blade]

if it lies or not

[shy atlas]

if its a scalar multiple of the other

[long blade]

ohh

so for example

if the 2 normal vectors i chose were

um

[shy atlas]

normal vector to the normal vector

[long blade]

(2,2,2) and (4,4,4)

then this would be wrong

since their a scalar

i mean

their a mulitple of the other

[shy atlas]

yeah you wouldnt get a plane if u just blindly do r(s, t) = <1, 4, 5> + s<2, 2, 2> + t<4, 4, 4>

[long blade]

okay

cool

this makes more sense

[shy atlas]

nice

[long blade]

ty

oh way

i have another question

is there a way

to get the vector

from a cartesian

equation

[shy atlas]

[gray dust]

yes

[shy atlas]

what vector tho

[long blade]

um sec

[gray dust]

cartesian to vector form

[quasi vale]

I think you can find a basis that span the entire plane from the cartesian equation

[shy atlas]

oh ye ye

oh god bases

[gray dust]

the basis
no just A basis

[long blade]

in part b) it gives me a cartesian equation

i want to make it into a vector

wait a minute

im an idiot

from this

wouldnt u get the vector

(1,1,-1)

[gray dust]

you gave no context for where that came from

[long blade]

oh

because of the coefficient in front

the plane

wait

let me just try to work it out

[gray dust]

focus on turning the plane you were given into vector form. i did some algebra to get z=x+y+1. substitute into r=(x,y,z)=?

[long blade]

ohhh

i got it

so

x = z - y + 1

let z = t

and y - s

so (z-y+1,s,t)

i mean

[gray dust]

algebra's off

[long blade]

(t -s +1, s, t)

so

= s(1,1,0) + t(1,0,1)

[gray dust]

stop typing at 100wpm, read chat

[shy atlas]

dem 100wpm

[long blade]

lol

[gray dust]

x = z - y + 1 is wrong

[long blade]

oh

ty

z = z -y -1

my bad

i mean

x = z- y-1

let z = t, y= s

(t-s-1,s,t) = s(-1,1,0) + t(1,0,1)

[gray dust]

the -1 after t-s is missing

[long blade]

hmmm

(-1,0.0) + s(-1,1,0) + t(1,0,1)

?

im not sure

[gray dust]

yes

[shy atlas]

do u guys plan to cross something together to get the normal vector

[gray dust]

no

[shy atlas]

wew thats nice

i wonder how ur gonna get the normal vector tho

from the vector form

[gray dust]

this plane (-1,0,0)+s(-1,1,0)+t(1,0,1) is parallel to the one we want

[long blade]

what am i meant to do next

ohh

ok

so

what do i need to do next

[gray dust]

think this one out

[long blade]

ok

[gray dust]

or at least follow along with me

[shy atlas]

damn rocket always teaches new and fancy ways of doing stuff

[long blade]

sorry about before

[shy atlas]

i bet it has something to do with changing <-1, 0, 0>

[long blade]

since thats basically

the transition

[gray dust]

for the plane given by (x,y,z)=(-1,0,0)+s(-1,1,0)+t(1,0,1), the vectors (-1,1,0) & (1,0,1) are what determine the orientation of the plane

if (-1,0,0) weren't there, you'd have a plane with the same orientation except it also contains the origin

[long blade]

oh

since we want the plane

to go through the point

(6,5-2)

if we replace

(-1,0,0)

with (6,5,-2)

would we get the vector that is parallel and goes through the (6,5,-2)?

i mean plane*

[gray dust]

then you'd know for sure the new plane contains (6,5,-2) AND has the same orientation of the original plane

[shy atlas]

[long blade]

thank you

[gray dust]

you're welcome

[long blade]

im actually begining to undertand how planes

work now

just out of curiousity

for c)

the awnser should be

(0,0,0) + s(1,1,1) + t(1,2,3)

right

[gray dust]

sure

[long blade]

wait

but do we need

to include

the (0,0,0)

[shy atlas]

nah

[long blade]

since without it

its same right

[shy atlas]

its redundant

[long blade]

ok cool

[gray dust]

the 0 vector is defined to be the additive identity

[long blade]

?

additive identity??

what does that mean

[gray dust]

when learning algebra in the real or complex numbers you learned of some number 0 such that for any number x, 0+x=x+0=x. we call 0 the additive identity

similarly for vector addition, the 0 vector is called the additive identity

[shy atlas]

ye additive identity in the vector spacce

i have dem k n o w l e d g e

[slim bridge]

Hey All

Can someone explain how I would do this questions?

Imagine the dumbest person u know and explain it in the most dumbed down language for him. thats meeee

[dusky epoch]

what properties of the determinant do you know

[slim bridge]

um

that for a 2x2 its 1/ad-bc

[dusky epoch]

no

the determinant of $\mat{a & b \ c & d}$ is $ad - bc$, not $\frac{1}{ad - bc}$ and certainly not $\frac{1}{ad} - bc$

[stoic python]

Ann:

[slim bridge]

ah so thats how u do matrix in latex

was just wondering htat

[dusky epoch]

no

it's not

[slim bridge]

oh

[dusky epoch]

i have a macro that lets me do that

[slim bridge]

ok yea

👀

what a genius

how did u set the macro up

[dusky epoch]

but the default way is

$ \begin{bmatrix} 
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{bmatrix} $

[stoic python]

Ann:

[dusky epoch]

it was just a \newcommand

[slim bridge]

christ

anyways

yea its ad-bc

[dusky epoch]

anyway yeah that's woefully insufficient

like,,,

uh

[slim bridge]

i do agree

[dusky epoch]

not even det(AB) = det(A)det(B)?

[slim bridge]

im terrible at maths

[dusky epoch]

consider looking up "determinant properties" honestly bc that shit's important

[slim bridge]

well i guess now i know that exists

not even det(AB) = det(A)det(B)?
@dusky epoch

i still dont really get

what a determinant is used for

lik why is ad-bc so important..?

[dusky epoch]

uh

i mean that's just the 2 by 2 case

[slim bridge]

oh and also, is there any tricks to remember finding dets for larger matrices?

[dusky epoch]

anyway beyond this problem dets are most often only useful in terms of knowing whether or not the det of a matrix is 0

[slim bridge]

cuase i get confused over which to multiply and stuff like what

[dusky epoch]

i don't think i can tell you any tricks

[slim bridge]

anyway beyond this problem dets are most often only useful in terms of knowing whether or not the det of a matrix is 0
@dusky epoch which baiscally tells us if there in an inverse right

inverse matrix

[dusky epoch]

uh

yeah

[slim bridge]

ah ok

thats alright

thanks Ann

is there a particular way to reasona for matrix multiplication maybe?

[dusky epoch]

watch 3b1b's essence of linear algebra

[slim bridge]

is it just matrix one defines row, matrix 2 defines colum

[dusky epoch]

he explains the intuition behind dets way better than i could

[slim bridge]

oh ok

i watched some of the earlier ones

but they werent related to what we learning in uni class

so i just thought it was more advanced or something

[gray dust]

did you define \mat to do \begin{bmatrix}#1\end{bmatrix} ?

[dusky epoch]

yes

yes i did

[gray dust]

aight ty

[pallid swallow]

@slim bridge determinant is like signed volume

It is basically how much volume the basis vectors make

After being transformed

[slim bridge]

ITS AREA

i just watched 3b1b vid

wtf

its the constant that area is scaled by

woah

[slim bridge]

@dusky epoch thanks for the suggestion Ann

it really did help a lot with the intuition

[solid bough]

For part (b), i cant seem to decide which generator polynomial to use , since n = 6 and k = 2 (which i got from turning G into standard matrix to check for k)

since my g(x) = n-k = 4, by reducing my x^6 - 1 into irreducible polynomials, i can form 3 different generator polynomial with degree 4. Am i heading the right direction or is my understanding off?

[viscid kernel]

Any linear algebra experts or whatever who I can have a discussion with about about 3 + n dimensions on the voicechannel ?

[dusky epoch]

if you're willing to wait like 2h and if i'm no longer tired out of my mind by then, i could try clearing up some of your doubts

[viscid kernel]

Ok, Ill wait then @dusky epoch

[restive hound]

How can a real 2x2 matrix have a characteristics polynomial of 1-lamda^2
When it's det(A-lamda*Identity matrix)?
Lamda is always negative so after expansion lamda squared is always positive

[dusky epoch]

with a characteristic polynomial the only thing that really matters is its roots

and λ^2 - 1 and 1 - λ^2 have the same roots

[limber sierra]

in fact, this holds in general

some sources define the char. polynomial as $\det(A - \lambda I)$; others define it as $\det(\lambda I - A)$

[stoic python]

Namington:

[limber sierra]

it doesnt matter since we only really care about its roots

[restive hound]

So if I want to find a matrix that has the characteristics polynomial of 1-lamda^2
Then I can just find a matrix that had the characteristic polynomial of lamda^2-1 ?

[limber sierra]

and multiplying by -1

doesnt affect roots

i mean, sure?

im not sure thats an easier task

but yeah

[restive hound]

it's just a 2*2 matrix

[reef prism]

is Ker(A) = R^2 or λ(1,0)?

i think its R^2 because it doesnt matter

ah i think λ(1,0) is the right answer because its the correct spelling

[cursive narwhal]

[dusky epoch]

@viscid kernel yeah i got done with my shit, ya here?

i'm not going into voice until you give me the ok

[viscid kernel]

Ok

@dusky epoch

[potent totem]

what exactly does "Re(...)" mean and how does 6.19 follow from the line above it?

[prisma drift]

real part of the vector, assuming you have a complex valued vector

[dusky epoch]

<u,v> isn't a vector, it's a scalar

complex in this case

@potent totem have you worked with complex numbers before

Re() is the notation for real part

[potent totem]

ok ty

@potent totem have you worked with complex numbers before
@dusky epoch Yeah. I guess I just forget the notation (used to different notation)

[dusky epoch]

Re is pretty universal is it not

[restive hound]

If A is a 3x2 matrix and B is a 2x3 matrix, is AB invertible? Is this a valid explanation?
It is never invertible because
Determinant has to be non zero to be invertible
But
det(AB) = det(A)det(B) thus since the determinant of A and B cannot exist because they aren't square matrices

[dusky epoch]

no

you can't say det(AB) = det(A)det(B) bc A and B would have to be square for that identity to apply

[slate haven]

you can also find a 3x2 matrix and 2x3 matrix such that their product yields the identity matrix which is invertible

[dusky epoch]

no you can't, the product will have rank at most 2

[slate haven]

oh yeah, i think you're right. It is because B corresponds to a map f from K^3 to K^2 while A is a map g from K^2 to K^3, since the image of f can't have dimension >2 then the image of g also can't have dimension >2, right ?

and since g o f : K^3 ->K^2 -> K^3*

[viscid kernel]

When you have a non linear transformation of a certain vector lets say [2,3] given with his transformed matrix. Would you still be able to calculate what the output vector is ?

[dusky epoch]

what

what's the "transformed matrix" of a vector

[subtle walrus]

they don't know