linear-algebra

[jagged gulch]

here the a and b is swapped, but if you call the angle in the lower left theta, then h is the length of a times sin(theta)

I'm not explicitly using dot product here, just a bit of trig

[bold ivy]

Yeye

You will get the height of that sure

It is the same thing as in 3d?

[jagged gulch]

but yeah so that's basically the reason why c isn't the height of that parallelepiped

[bold ivy]

I understand

[jagged gulch]

c slants, and you need a straight line for the height (basically)

[bold ivy]

But

[jagged gulch]

same reason why we did |a|sin(theta) to find the height for that parallelogram I posted

[bold ivy]

If you do determinant of a 2d vector you will get area of the parallypoggrrjru.....

[jagged gulch]

the magnitude of the cross product does happen to be the same as the area of the parallelogram spanned by those two vectors

I don't know if that's what you were referring to

[bold ivy]

Ye

But when you do cross product in the volume of paraellpepied. Where on earth is ijk vectors?

[jagged gulch]

ah I got you

[bold ivy]

They just dissapperade

[jagged gulch]

So do you understand why, in order to get the height of the parallelepiped, you need to get the projection of c onto aXb?

[bold ivy]

Yes

[jagged gulch]

perfect

remember that the projection of u onto v is $\frac{u.v}{|v|}$

[stoic python]

mathew_soto78:

[bold ivy]

Yes scalar projection

[jagged gulch]

so the scalar projection of c onto aXb is c.(aXb)/|aXb|

let me write that in a prettier way

$\frac{c\cdot (a\times b)}{|a\times b|}$

[stoic python]

mathew_soto78:

[jagged gulch]

that is the height of our parallelepiped

The volume is the height times the base

what's the area of the base?

[bold ivy]

So what does absolute of cross product represent here?

[jagged gulch]

the magnitude of the cross product vector

this is just me taking the formula for scalar projection and plugging in c and aXb into it

(c for u and aXb for v)

[bold ivy]

And c * (a X b) is the height?

[jagged gulch]

no, what I wrote above is the height

$\frac{c\cdot (a\times b)}{|a\times b|}$

[stoic python]

mathew_soto78:

[bold ivy]

But what does c * (aXb) represent?

[jagged gulch]

that's the dot product of c and (aXb)

[bold ivy]

But why do you divide projection with magnitude of cross product?

How come you will get the height?

[jagged gulch]

no, that is the scalar projection

the scalar projection of u onto v is $\frac{u\cdot v}{|v|}$

[stoic python]

mathew_soto78:

[bold ivy]

Oj

Oh yeah

My bad

I see

[jagged gulch]

and what we're looking for in the height is the scalar projection of c onto aXb, so we just plug in those two vectors into that formula (into u and v respectively)

[bold ivy]

Now you have height now you gotta multiply it with cross product.

[jagged gulch]

yep, because the magnitude of the cross product of a and b is the area of the parallelogram

[bold ivy]

Scalar pjection * aXb.

[jagged gulch]

so the height of the parallelepiped is $\frac{c\cdot (a\times b)}{|a\times b|}$, and the area of the base is $|a\times b|$, so when you multiply the two together to get the volume you just end up with $c\cdot (a\times b)$

[stoic python]

mathew_soto78:

[jagged gulch]

really pretty stuff

[bold ivy]

Ah

I see

Bur

[jagged gulch]

math is fun

[bold ivy]

Where is kji

Ye

[jagged gulch]

I sent kji home packing

we don't need them

[bold ivy]

Wdym?

[jagged gulch]

I don't really know what you mean by where is kji, but I was just saying that it's not necessarily to use a formula where you see that

[jagged gulch]

you could say $a=a_1 i + a_2 j + a_3 k, b = b_1 i + b_2 j + b_3 k, c=c_1 i + c_2 j + c_3 k$ and then expand that formula that we found

[stoic python]

mathew_soto78:

[jagged gulch]

but that's objectively nasty and we don't do that

[bold ivy]

So you just remove ijk

But then it becomes 2X3

[jagged gulch]

No, I didn't remove i j k

the formula we found just doesn't explicitly use i j k

but it's there

i j k just stand for the unit vectors in the x y and z directions, respectively

so every vector in R^3 is a linear combination of those vectors (i,j,k)

that's where the components come from. If you look at the vector v = ai+bj+ck, the projection of v onto the x axis is a; the projection of v onto the y axis is b; the projection of v onto the z axis is c

[bold ivy]

Its same thing w ( 3,5) it is actually 3i,5k but we dont use it

[jagged gulch]

so in any vector we see, there's an implicit i j k there, but we don't necessarily need to write it that way

I don't know if that clears up what you were asking

[bold ivy]

Yeye

It does thanks

One question how come we divide it with 3 we get volume of pyramid?

[jagged gulch]

I think that's a very different question that you wouldn't necessarily answer with vector methods

do you mean a square or triangular pyramid, or does it not matter

[bold ivy]

Triungal pyramig

[wintry steppe]

visually, whats the difference between a homogenous and non-homogenous system of linear equations?

I assumed, the homogenous one, all the planes or lines intersect the origin

[dusky epoch]

the right hand side being zero or not

that's literally it

[wintry steppe]

in non homogenous systems, a line may not cross the origin, but in a homogenous one is has to doesn't it?

[dusky epoch]

?? you're either overthinking it or phrasing it poorly

but it is true that the solution set of a system contains the origin if and only if the system is homogeneous

[wintry steppe]

I was under the impression that, for a homogenous system, all equations in the system crossed the origin.

but in non homogenous systems, not all equations in the system cross the origin.

[dusky epoch]

i think you're kinda overcomplicating it ngl

[wintry steppe]

hmm yea i agree lol

it kinda worked nicely visually in my head for 2d, but completely broke down in 3d

[dusky epoch]

i mean what you're saying is correct if you identify each single equation with the hyperplane consisting of the solutions of that equation alone

[wintry steppe]

yea, i was trying to see what it would look like, and it seems like its true.
but it gets extra confusing when translating the fact:
"a homogenous system with more unknowns than equations has infinite solutions

it makes sense symbolically but visually it would be

"a homogenous system with more dimensions than lines/planes/etc has infinite intersections"

which idk is confusing

[dusky epoch]

visually it'd be like,, imagine 2 equations in 3 unknowns

[wintry steppe]

and probably wrong

[dusky epoch]

two planes intersecting in 3D space

gives you a line

an entire line of solutions

[wintry steppe]

yep so infinite solutions

but if there are more planes, its somehow possible there is only the trivial solution?

and then my mind short circuited

[dusky epoch]

if there's 3 planes they most likely intersect in a single point yes

unless they are specially aligned to all go through the same line

[wintry steppe]

i can't imagine 3 planes, intersecting only at 1 point

like there would have to be at least a few extra intersecting lines?

considering they all intersect at the origin too.

[frosty vapor]

try to visualise at first two planes intersecting

you can see a line forms

[wintry steppe]

yea i can see that

[frosty vapor]

then take the third plane and see how it intersects with the line that you formed

[wintry steppe]

but somehow the lines disappear when i add a third plane?

OOH

[frosty vapor]

bingo

[wintry steppe]

lol whoops, yea i kept seeing the lines

but those lines are irrelevant since not all 3 planes cross

at those lines

wowzers thats pre cool

thanks fellas i think i can appreciate homogenous equations more now

[viscid kernel]

Does an affine subspace has to shift “ paralel “ ?

[dusky epoch]

what?

[viscid kernel]

If the rank before is bigger than the rank after. If you consider the nullspace being 1 dimensional you have a line full of solutions. But it is also possible that the nullspace doesnt include the origin, it could also be moved by away by another vector ( not the zero vector ofcourse ). And is that movement paralel ?? I hope Im being clear

[dusky epoch]

no

you're not being clear in the slightest

[viscid kernel]

Wait Ima send a picture, its dutch but I think you will understand it a bit. Also, ik my slights/book they use kernel for nullspace.

If you want I can translate it but, I think you will understand

[dusky epoch]

hhh

[viscid kernel]

You want me to translate it ?

[dusky epoch]

idk i don't feel like dealing w this rn

[viscid kernel]

Oh np, I can ask somewhere/someone else

[tepid juniper]

i believe this belongs in linear algebra?

So, T(x1,x2) is a transformation in r2 to r2 and have to figure out a sketch for what this transformation does to S1

So I plugged in e1 and e2 into T(x1,x2) and resulted in
T(e1) = (3,2)
T(e2) = (-2,1)

Now, im not exactly sure how to graph this which is why i kinda wanted to reach out for help

[native rampart]

What does π mean here?

[tepid juniper]

parallelogram

so .5e1 and .5e2 would be a square who side lengths are both .5 at max

so at most a .5x.5 square whose bottom left corner is at vector (2,1)

I think im supposed to plug in the e1 and e2 values from part a into part b?

and then move the entire parallelogram to the vector (2,1)?

bleh ive been working on this for way too long but still am not understanding how to go about this stuff

i'll ask my prof bout it but if possible, can someone help me out with this instead?

it wants us to find the vector p that makes this statement correct

i have no i dea how its not p = x(perp) when literally the format is T(x) = p . x

[magic light]

I'm a bit confused about vector spaces and sub spaces...
one of the question was whether Im(T) = R^2 col for a given Transformation T: R^3col -> R^2 col
We found that Im(T) has a dimension of 2, and Im(T) is a subspace of R^2 by definition(?)

does that mean Im(T) = R^2col?

basically if a vector space and sub space have the same rank they're equal?

[native rampart]

Yes

[modern palm]

this isn't obvious because we have no idea how addition and scalar multiplication are defined right?

but to prove this, don't we have to know exactly how addition and scalar multiplication are defined?

[subtle walrus]

no, use that scalar multiplication distributes over vector addition

or actually scalar multiplication distributes with respect to field addition i guess

[hollow oar]

for an arbitrary vector space we don't know how they're exactly defined, but we know they satisfy some axioms. it's "not obvious" because you're adding two vectors with vector addition and saying that it equals some scalar times a vector

but yeah to prove it, you can use the distributive law

[modern palm]

how would i start that?

[old flame]

May I ask how is 5.22 constructed ? Is it because each eigenvector is an element of the null space of $T-\lambda I$ ?

[stoic python]

Otoro:

[subtle walrus]

@modern palm how can you write 2w differently?

[modern palm]

w+w...

[subtle walrus]

no

2 is a scalar from a field, w a vector

your only chance is to write 2 differently pretty much

[modern palm]

(1+1)w

[hollow oar]

@old flame yeah. since they form a basis, you can write every vector as a sum of eigenvectors, which is the same as (5.22)

[subtle walrus]

exactly, now use distribution and the fact that 1 is the identity for scalar-vector multiplication

[modern palm]

thank you

[old flame]

Thanks @hollow oar

[jagged gulch]

@viscid kernel I think your question is: "does the kernel of a linear transformation have to include the zero vector"

[wintry steppe]

#precalculus @zealous vine

[jagged gulch]

If that's the question, the answer is yes, and it follows from the definition of a linear transformation

[frosty valley]

How would we do this in linear algebra? And where could I find more problem like this ?

[limber sierra]

Set up a system of equations corresponding to this situation

you might be able to find more problems by looking up the phrase "systems of linear equations word problems" or something

[frosty valley]

So you mean like 100x + 20y = 100 ?

[limber sierra]

uh

what are x and y and why do they equal 100

[frosty valley]

It should be the other way around e.g Taking the programmer and administrator in x and y and = 120

120*

[limber sierra]

well, you should have two equations

one for the 100 hours of programming

one of the 20 hours of administration

[frosty valley]

I see

Thank you

[wintry steppe]

yw

[scenic vapor]

So if rref(A - I) of a stochastic matrix gives two free variables, does it have a unique fixed probability vector?

how do i find that out

[spiral star]

depends on your matrix i guess. ker(A - I) is the eigenspace for eigenvalue 1, if you have two free variables, then that eigenspace has two dimensions. once you have a basis for the eigenspace you can usually tell how many solutions are possible because of the constraint that the components of the vector must sum to 1 and be non-negative. (there is also a way to embed that constraint into the matrix but its a little bit more involved.)

[scenic vapor]

so if the eigenspace is:

[stoic python]

Prostidude:

[scenic vapor]

there is no unique probability vector here right?

because you could get multiple vectors in which the sum is 1

or am i missing something?

[spiral star]

yea its not unique

you already have two given by the basis if you scale them accordingly

[stark acorn]

Homir how do i do this:

[native rampart]

Find Det(T),I guess

[stark acorn]

How?

[native rampart]

Find the matrix of T in standard basis(Or some basis)

And find the determinant of that matrix

[bold ivy]

can someone explain to me what plane of equation is all about?

[stark acorn]

Im still confused

Whats the first tep

[bold ivy]

i thought linear combination was the so called equation of the plane or describes the plane?

[native rampart]

Area scale factor(in this case) of a Linear transformation is the determinant of the linear transformation (which is determinant of the matrix,which represents the linear transformation in some basis)

[bold ivy]

why would you want an equation of the plane? what is it supposed to mean?

[subtle edge]

why would you want an equation of a line?

a plane is just another object similar to a line but has a higher dimensionality

ie a plane in R3 is just a 2d hyperplane

[bold ivy]

I see but why would you get the planes equation through norma vector?

why would you want ortogonal vector?

[subtle edge]

well suppose you have the point on a plane in R3

and we also have the normal vector

the angle between the point and the normal vector must be pi/2

[bold ivy]

yeah

[subtle edge]

in other words, the dot product between the normal and a point on the plane is equal to 0

[dusky epoch]

you need the normal for calculations of stuff like reflections

[subtle edge]

its just a way to express the equation of the plane

[dusky epoch]

which is needed for 3d graphics for example

[bold ivy]

uh

uhm

[dusky epoch]

lighting

this stuff has applications ok

[bold ivy]

can you eloborate

[subtle edge]

yeah there are a lot of good applications

[bold ivy]

i dont understand why you need normal vector for planes equation

[subtle edge]

ok so if we have an orthogonal vector to a plane we can say that the dot product between any point on the plane and the orthogonal vector is equal to 0 right?

[dusky epoch]

as far as i understood, you were asking why writing plane equations in terms of their normals was even a thing

[bold ivy]

@subtle edge I understand but how would that be plane's equation

[subtle edge]

oh a cool application is the use of it in multivariable linear regression using OLS

well, we can define such vector (x-a,y-b,z-c) on the plane

and n be the normal vector

lets suppose (a,b,c) lie on the plane as well as arbitrary x,y,z

such that their x-a, etc.. is also on the plane

[bold ivy]

a,b,c a point on the plane?

[subtle edge]

yeah

but n be a normal vector

[bold ivy]

x-a ?

[subtle edge]

right

(x-a,y-b,z-c) is the vector from (a,b,c) to (x,y,z) on the plane

which is also on the plane

[bold ivy]

ye

[subtle edge]

the normal is orthogonal to every point on the plane

[bold ivy]

ok?

[subtle edge]

so the dot product between n and the (x-a,y-b,z-c) vector is equal to 0

[bold ivy]

yes I see

[subtle edge]

which is why we need a normal to express the plane

[stuck stratus]

Hi I have a short question. Does the second line here mean that the vectors that make up the matrix span Rn?

[bold ivy]

we need normal because we need to use dot product

?

[subtle edge]

we need the normal to define the equation which uses the dot product

[bold ivy]

why do you need dot product to define the equation?

[subtle edge]

@stuck stratus if A is invertible it also means the A matrix spans

[stuck stratus]

You forgot R I think xd

[subtle edge]

yeah sorry haha

@bold ivy

[stuck stratus]

Alright good to know

[subtle edge]

because we need it to equal 0

orsomething

something we can use

to define a plane

it just so happens equal to 0 is the easiest

[stuck stratus]

so if a question is asked. Do these vectors span R? The question is actually can you reduce these vectors to I

[subtle edge]

yeah the RREF would be I

[stuck stratus]

RREF?

[bold ivy]

@subtle edge what if it is not 0?

[subtle edge]

reduced row echelon form

@bold ivy it has to be 0 because its the normal vector

[stuck stratus]

Ahh alright

Thanks @subtle edge :))

[subtle edge]

go ahead and take another vector if you want you just need to make sure you know the angle between the vector and the plane

and also do some more calculations

the normal just makes it so much easier for you

[bold ivy]

but you said we need normal vector because we need to do dotproduct

[subtle edge]

yeah

[bold ivy]

why do you even need vectors that goes up to describe the plane?

[subtle edge]

as in its applications?

i think ann told you those already

[bold ivy]

not its application, im talking about math theory, like why do you need vectors that is not part of the plane to make a equation of the plane?

[subtle edge]

all you need is some sort of relation to define the plane

[stuck stratus]

Also another thing. Why is this 3rd line even there? Doesn't every single Ax = b have exactly 1 solution?

Because there's no free variables

[subtle edge]

and the simplest one is to use the normal

the normal is there to make sure that the relationship persists

[bold ivy]

it is like having x-axis and y-axis that describes the line?

[subtle edge]

@stuck stratus it depends but since it is invertible then it has to have a pivot in every situation

yeah

[stuck stratus]

alright ty

[bold ivy]

yeah to whom?

[subtle edge]

im confused epic. do you mean to say that the equation of the plane doesnt make sense because you arent using a point from the plane itself?

the idea is that you have a key relationship between orthogonal vectors that you exploit to get the plane equation itself

the normal vector just helps it isnt part of the plane like when you find the equation of the line

[bold ivy]

can you make a pararell between a lines equation and planes equation.

[subtle edge]

in fact, you can do the same thing wiht a line

just do it in R2 and you will see the normal vector thing works too

[bold ivy]

where is normal vectors in R2+

[subtle edge]

well if its given you would know right

(x,y) could be a normal to (a,b) if ax+by=0

etc

[bold ivy]

how would it look visually?

[subtle edge]

i guess literally a line and then a line 90 degrees from it

similarly in 3d a plane with a vector 90 degress to it

think of like a table and then put your pencil on it

the table is a plane pencil normal vector

[subtle edge]

yeah in 2d

[bold ivy]

but how can normal vector in a line describe the equation of a line

the equation of a line is,

y=kx +a

[subtle edge]

the relationship exists specifically because there is a normal to the line

you already have something to help describe the lineitself

[bold ivy]

is there any mathematically proof that this normal vector has a relation to the lines equation

?

[subtle edge]

well yeah but it is intuitive to think a normal to the line has the relationship that it is perpendicular to the line

which is equivalent to the dot product equal 0

[bold ivy]

well yes it is perpendeicular

to the line

[subtle edge]

yeah that is specifically the relationship

[bold ivy]

but it is not intuitive it has any connection to the equation y = kx+a

[subtle edge]

the relationship governs the equation of the line

[bold ivy]

how

[subtle edge]

so suppose i have a point (a,b) for some line and i have (x,y) also

(x-a,y-b) is also on the line correct

[bold ivy]

yes

[subtle edge]

suppose i have some perpendicular line with vector (n1,n2)

[bold ivy]

ok

[subtle edge]

then (x-a,y-b) . (n1,n2) = 0 right

[bold ivy]

I mean sure

[subtle edge]

you are using points on the line to find the line itself right

[bold ivy]

but it doesn't describe the equatoin

[subtle edge]

well

n1(x-a)+n2(y-b)=0

in other words, n1x-n1a+n2y-n2b=0

y=-n1x/n2 + (n1a+n2b)/n2

[bold ivy]

sure

but

the equation of the line is

y = kx +a ?

[subtle edge]

k=-n1/n2

a=(n1a+n2b)/n2

[bold ivy]

so people found the equation of the line with normal vector?

[subtle edge]

well there is a relationship between a slope and its normal

[bold ivy]

yeah sure

[subtle edge]

suppose m1 is a slope and m2 is its normal slope then m1m2=-1

[bold ivy]

that I can agree

[subtle edge]

they just extend this idea to generalise to planes

[bold ivy]

they just use that as proof that you can do it on planes aswell?

[subtle edge]

https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/session-12-equations-of-planes-ii/MIT18_02SC_notes_5.pdf

[bold ivy]

so without ORTHOGONAL

You wouldn't have a plane

[lament lily]

if I have some vector w = 10x + 5y where x and y are vectors in in R^n, is span{w} = span{x,y}. I think yes because w is all linear combinations of w but part of me thinks no

[subtle walrus]

no

just take any example where x, y are linearly independent

[lament lily]

hm Im not sure I see how this would cause problems

[subtle walrus]

span{w} is a line, span{x, y} a plane in R^3

[lament lily]

but if x and y are linearly independent, then would w not be a plane aswell since a linear combination of x and y is also a plane?

[subtle walrus]

a single vector can never span a plane

the span of w is the line along the direction of w

[lament lily]

alright so span of any single vector is always a line, got it, not sure how I didnt see that

[subtle walrus]

of a nonzero vector, yes

[lament lily]

lets say then w = -4y, then span{w} = span{y} because both are just lines

[subtle walrus]

they span the same space but not because they are both lines

w = -4*y is in the span of y

and y = -1/4 * w is in the span of w

[lament lily]

alright I think I understand span a bit more than I thought I did, thanks

[bold ivy]

@subtle edge could you say planes equation is just describing the span?

because a plane is consists of span

[subtle edge]

it definitely describes the plane and therefore the span of the plane

[bold ivy]

yeyeye

but isn't

the span enough to describe the plane?

[subtle edge]

well the span could describe any plane

suppose i have a 3 dimensional hyperplane in R4 i could have (1,0,0,0) (0,1,0,0) and (0,0,1,0) and that would span the 3 dimensional hyperplane

in fact it would span any

but it is not necessarily unique to one plane

[lament lily]

if I have like w = x + y, and z = 3x + y, then span{w} != span{z} right because the coefficients are different?

[bold ivy]

@subtle edge every plane look different

due to linear transformation

is this a point?

[wintry steppe]

Yes

[bold ivy]

So This point could be whatever regarding plane of equation

[lament lily]

@subtle walrus sorry to ping you but I just dont see how span{w} can not be a plane if w = ax + by and x,y are linearly independent

w would then be any point on the plane that x,y form

[subtle walrus]

no it would not

just take your example

10x + 5y

the span of w is all vectors of the form k(10x+5y)

so all vectors of the from k10x+k5y

in particular the coefficient of x is always a multiple of 10 and the coefficient of y is always a multiple of 5

4y is not in the span of w

it is in the span of {x, y} however

[lament lily]

alright I think that explanation is a bit better, so it just stretches

that k was what I was missing

[subtle walrus]

since 0x + 4y is

what span is basically

you take the vectors in the span, only regard there directions

and walk arbitrarily long in the direction of the first vector, then arbitrarily long in the direction of the second, ...

w is just a single vector

so you can only walk along a single direction

its a line

the span of 2 vectors (that point in different directions, and are therefore linearly independent) is a plane

you can walk any amount of units in the first direction and after that any amount of units in the second direction

just draw 2 different vectors on a piece of paper and convince yourself

[lament lily]

alright thanks for the explanation!!

[near yoke]

hi, i really do not know how to do this

this is what i have so far, i dont think it's right tho

[subtle edge]

multiplication matters

xA=/=Ax

uhh an easy way to show b as a lin com is to row reduce

[near yoke]

oh i need to reduce A before lin com?

[subtle edge]

no just to see how much of each column u need

let me try it out

[near yoke]

kk thx

[near yoke]

progress?

[bold ivy]

I understand the equation of plane

but I dont understand

xyz=point + t(vector)

what is this supposed to mean

I understand plane of equation, it is just like equation of a line, it describes the plane, but what does it had to do with xyz = point + t(vector)

[wintry steppe]

Can someone help me with this question:

7,c

What does it mean find the remaining root

acc nevermind, I don't need help anymore

I solved it

[wintry steppe]

why does it say "marks" instead of "points"

is that british

or indian?

because only my parents use the term "marks" these days

[frosty vapor]

!american

[wintry steppe]

[jolly rivet]

an infinite vector ?

or an infinite entry ?

[half forge]

cna someone help me with this one

[slate mauve]

You can write that two vectors are perpendicular by saying $\vec u \cdot \vec v = 0$ But how can you write that two vectors are parallel (that they have the same angle)?

[stoic python]

matt_:

[native rampart]

Show u=kv for some k

[magic light]

B = ( [1 0 1] , [0 2 2], [0 0 2] ) is a base
v = [5 -2 3] in the standard base
according to the book, to express v by B's vectors it'll be:
v = 5 * [1 0 1] - 1 * [0 2 2] + 0 * [ 0 0 2 ] and then

[v]_B = [5 -1 0]

how did they get these numbers? 5, -1, 0?

[dusky epoch]

solve a system of linear equations

[marble lance]

I will change the notation, so it's easier to write, but let
$(5,-2,3) = a(1,0,1)+b(0,2,2)+(0,0,2)$. Then simplify the RHS and equate the coefficients which gives you the system of linear equations that Ann mentioned

[stoic python]

Lunasong:

[magic light]

So basically I'm looking to solve
a + 0 + 0 = 5
0 + 2b + 0 = -2
a + 2b + 2c = 3

?

[dusky epoch]

yes

[magic light]

thanks

[vocal isle]

Can SVD (singular value decomposition) also be used to extract an approximate analytical equation from a bunch of x,y,z data? For example x = age, y = amount of hours walked per day, z = weight of person. Say the equation we would want to extract from the data would be z = x^2 - 4*x/y?

[dusky epoch]

try the set of all sequences of real numbers for your space

and for the operator, take the left truncation operator defined by L(x1, x2, x3, ...) = (x2, x3, ...)

[dim venture]

(thank you I'm trying to interpret what you just said, will need a sec)

could you elaborate on the left truncation operator? is it exactly as described? just removing the first element

[dusky epoch]

yes it is exactly as described

ker(L^n) will be the set of all sequences that are zero from the (n+1)th term onward

[dim venture]

I just started my lin alg 2 course so I'm not very familiar with sequences in the context of kernel

ker(L^n) will be the set of all sequences that are zero from the (n+1)th term onward
could you elaborate

[dusky epoch]

L^n is the operator that cuts off the first n terms of its input sequence

does that make sense to you or does it also require elaboration

[dim venture]

no thats good

[dusky epoch]

the zero vector of our space is the sequence of all zeros

[dim venture]

so does everything starting from the (n+1)th term map to 0?

[dusky epoch]

no

that doesn't make any sense

[dim venture]

I am still unsure what belongs in the kernel set with this operator

or is its content irrelevant

[dusky epoch]

ker(L^n) is the set of all sequences which get mapped to 0 by L^n

that's just me restating a definition

[dim venture]

okay I think I understand

though I don't need to prove it, how do I know that ker(L^n) =/= ker(L^m)

oh wait its immediately trivial because there must exist a sequence that does map to 0 that is in between the (n+1)th term and (m+1)th term

it does say for every m >= n so it could be (m+k)th term for all non-negative integers k

[dusky epoch]

...horrible wording

[dim venture]

I'll take it?

trivial may not be true

[dusky epoch]

you seem under the impression that the terms themselves are elements of your vector space

[native rampart]

What do you think is the basis of C?

It's over C

i+(-i)1=0

and for any z ,z=z.(1)

So {1} Spans all of C(over C)

Yes

If it were over R,then you would have been correct

Other way round

{1} over C and {1,i} over R

[a -b]
[b a]

I derived this one by simply putting in the basis vectors in {1, i} in T(z) = c * z (c being complex too).
@wintry steppe correct for R^2

For C it will be just [a+ib]

[hidden ember]

can two matrices have the same row echelon form?

[native rampart]

Sure,Take any 2 invertible matrices

[half storm]

That's a good answer actually.

[ember wedge]

hey

can someone help me please

[spiral star]

i think you can even say all matrices with the same row space have the same row echelon form

because you can just choose the same basis for the space

[native rampart]

Yes

[hidden ember]

oh i see

thanks!

[native rampart]

I think that's because of Gaussian elimination

[half storm]

That makes sense.

[ember wedge]

can someone help me please?

[tawny tulip]

It is given that every eigenvector of T is also an eigenvector of T* (T* is the conjugate of T).
Prove: T is a normal operator.```
now, I've been thinking of proving it using the fact that simultaneously diagonalizable matrices commute with each other. Any help please?

[dusky epoch]

hm

it's obvious that if $Tx = \lambda x$ then $T^*x = \overline{\lambda}x$ from the given condition

[stoic python]

Ann:

[dusky epoch]

so if x is an eigenvector of either then it's also an eigenvector of both TT* and T*T with eigenvalue |lambda|^2

hm

[wintry steppe]

Does anyone know the best way to solve this question

I plugged in x+yj in for z

and I expanded it

j

engineering 😎

I got (-4.5y - 1.5x^2 + 1.5y^2) + (4.5jx - 3xyj) [seperated real and imaginary]

Idk where to go from there

[wintry steppe]

<@&286206848099549185>

can someone help me with solving 2step Inequalities

[tall thunder]

Can anyone explain this formula for me in like stupid person terms lol, I have a test in 10 minutes and I don’t understand

[viscid kernel]

If you have a 3 x 4 matrix. You wanna determine how many column vectors are linearly dependent. After doing Ax = 0 and gauss elimination, if you get one free variable does it mean that that there is 1 columnvector in the span of the other 3. And if you have 2 free variables does it mean there are 2 columnvectors in the span of the other 2 ?

[thorn kayak]

Idk where to go from there
@wintry steppe I will assume that you want to compute the norm of w. Remember that $|w|^{2} = \Re (w)^{2} + \Im (w)^{2}$

[stoic python]

ImHackingXD:

[wintry steppe]

Oh okay that's helpful

[stuck stratus]

Hi everyone. I have a few questions about bases and subspaces. I was wondering if I can DM someone about it

[thorn kayak]

Hi everyone. I have a few questions about bases and subspaces. I was wondering if I can DM someone about it
@stuck stratus I think it is best if you post them here, where there is a bigger chance you'll be answered.

[stuck stratus]

Alright it's multiple questions that's why I wanted to DM 😅

I was wondering what exactly is being asked here

Because I don't really get the answers

23 is linearly independent, but they say it's not a basis

[thorn kayak]

Well, because in 23 the set of vectors doesn't span R^3

[stuck stratus]

27 is linearly dependent, but they say it is a basis

Is that the questions though?

It says 'is a basis for the subspace of Rn that the vectors span'

[thorn kayak]

It says 'is a basis for the subspace of Rn that the vectors span'
@stuck stratus Indeed, in the case of 23 it asks if that set of vectors is a basis for R^3

Which they are not

[stuck stratus]

Alright I get that one then

but 29 for example. I row reduced it and there's no free variables

is it not a basis because there's 3 vectors in R4?

[thorn kayak]

but 29 for example. I row reduced it and there's no free variables
@stuck stratus When you row reduce it you get rank 3?

[stuck stratus]

what's rank 3? 😶

[thorn kayak]

3 non zero lines

[stuck stratus]

I do the course in Dutch 😅

yeah

there's 1 zero line

[thorn kayak]

is it not a basis because there's 3 vectors in R4?
@stuck stratus Then your logic is correct!

To be a basis you'd need exactly 4 vectors

[stuck stratus]

but it doesn't say 'in R4' like in the previous questions

[thorn kayak]

but it doesn't say 'in R4' like in the previous questions
@stuck stratus I agree, the question is a bit confusing

[stuck stratus]

does it have anything to do with the fact they're row vectors and not column vectors

I just saw that

🤔

[thorn kayak]

I also noticed now that you are not looking at R^4 in that example, but at R^3

The row vectors only have 3 coordinates

Still, they are linearly dependent like you have shown, so you get to the same answer

It is about the space "that the vectors span", the number of coordinates tells you indirectly what vector space you're analysing

[stuck stratus]

So it's 4 vectors in R3, so it has to be linearly dependent

because they're row vectors

[thorn kayak]

So it's 4 vectors in R3, so it has to be linearly dependent
@stuck stratus Yes, sorry for my mistake earlier

[stuck stratus]

ahhh

and

27

has 2 linearly independent row vectors in R3

and it's a basis

so it doesn't have to span R3

[thorn kayak]

Now I'm confused

[stuck stratus]

😂

[thorn kayak]

I don't understand what they ask then

Sorry

[stuck stratus]

I think

they're kind of just asking

if they're independent or not

and not if they span the whole of Rn

[lusty flame]

https://gyazo.com/4ba223ed9219379241ebb5b5f89076e8 anyone can help?

[stuck stratus]

Thanks anyways @thorn kayak

:)

[thorn kayak]

Thanks anyways @thorn kayak
@stuck stratus Anytime

https://gyazo.com/4ba223ed9219379241ebb5b5f89076e8 anyone can help?
@lusty flame For w to be a linear combination of u, it must be a multiple of u. That is not the case.

[lusty flame]

@thorn kayak how do i show or prove that?

oh wait i think i get it lol

[thorn kayak]

Suppose that it is a linear combination. Then there is an $\alpha$ such that $\alpha u = w$. What happens when you try to figure out what $\alpha$ is?

[stoic python]

ImHackingXD:

[lusty flame]

i dont understand what you're asking

[thorn kayak]

oh wait i think i get it lol
@lusty flame What was your thought here?

Maybe I'm making it overkill

[lusty flame]

@thorn kayak thought I figured it out but i'm just dumb lol

I know what a linear combination is

But now im starting to doubt myself

[thorn kayak]

Don't, you'll understand it

If they were multiples of each other, you could multiply u by a constant and get w

[lusty flame]

if u and w were multiples of each other?

[thorn kayak]

if u and w were multiples of each other?
@lusty flame Yes

[lusty flame]

so u is 1, -1

and w is 4, 1

[thorn kayak]

Then, since you could multiply u by a constant to get w

You would have that $\alpha = 4$ and $-\alpha = 1$

[stoic python]

ImHackingXD:

[thorn kayak]

But that's a contradiction

[lusty flame]

I think

I understand

What you're trying to say

So let me tell you what I'm thinking

One second let me gather my thought

Lol

I understand that it's a contradiction because alpha can only be one number

Can only be a single value not multiple

[thorn kayak]

Yes

[lusty flame]

For it to be a linear combination

We have to find a scalar

w is a linear combination of u if w = alpha vector u

so basically alpha (1,-1) needs to = 4,1 for it to be a linear combination right?

[thorn kayak]

so basically alpha (1,-1) needs to = 4,1 for it to be a linear combination right?
@lusty flame Exactly

[lusty flame]

that isn't possible though because no number * (1,-1) can give 4,1

my problem is i don't know how I can show that or prove that

[thorn kayak]

I told you how

You try to solve and get that alpha is equal to two different numbers

[lusty flame]

oh

So would I do something like a(u1)=w1 and a(u2) = w2

[thorn kayak]

Yes, that's how matrices work, they are equal if and only if their entries are equal

[lusty flame]

I get it

Thanks

[whole solstice]

https://gyazo.com/4ba223ed9219379241ebb5b5f89076e8 anyone can help?
@lusty flame yeah i can help

oh someone showed you already

[lusty flame]

yeah thanks tho

[zenith bear]

Can anyone help with this problem? I've algebraically shown that the first part is true but I can't come up with a property that covers all commuting matrices and proves the second part.

[marble kayak]

can anyone explain how a span of vectors is a subspace?
im confused

[gray dust]

unpack defns & try showing it yourself

[zenith bear]

huh

[blissful pewter]

can anyone help me with this? i got this so far
(a) det(AB) = 4 * 5 = 20
(b) det(3B) = 3 * 5 = 15
(c) det(2AB) = 2 * 4 * 5 = 40
but im not good at math, and this seems too easy so far, am i doing it right so far?

[restive hearth]

det(3B) should = 3 * 5 no?

[blissful pewter]

idk im not sure if the det() makes it change

oh yeah i wrote that wrong oops

[wintry steppe]

careful

det(cA) = c det(A) is not always true

you actually have det(cA) = c^n det(A), where n is the number of columns of A

(you can prove this fact using multilinearity of det, it isn't hard)

so, for example, det(3B) = 3^3 det(B) = ...

@blissful pewter

[lavish drift]

I've been bashing my head against this for so long. I know the proof is so easy it just not clicking

thats as far as i got

i know i have to get u and v together somehow cause they're orthogonal

[gray dust]

why do you have ... as assumptions @lavish drift

[lavish drift]

this is literally my first time touching a proof

like ever

[gray dust]

they're not assumptions, you're meant to show these are true

[lavish drift]

ok, what should i call them

well i suppose it doesnt matter does it

[gray dust]

facts to show

[lavish drift]

goals 😤 💯

ok

where do i take it from the last line then

[gray dust]

1st write down the actual given info

[lavish drift]

ok, well uv are orthogonal and p = au + bv, and they're all real numbers

thats what i know

[gray dust]

there's more

[lavish drift]

uhm

they're vectors in r^n?

oh they're unit vectors

but i dont think im allowed to substitute the vectors in

[gray dust]

write it all. THESE are the assumptions

it's also nice to fully write out what these mean, ie u,v are orthogonal & unit

[lavish drift]

ok, its all laid out

i suppose i never used the unit vector info

im not sure if its relevant

fair enough

well if i substitute u and v for their vectors i get p=2p

i guess that means i should doubt my work instead of the info hehe

[gray dust]

idk where p=2p came from

[lavish drift]

cause p=puu+pvv

and if v and u are unit vectors

p = p+p

[gray dust]

where did these come from

[lavish drift]

yeah

oh hm

so i cant really substitute it in

ok

aw man but the prof substituted when i asked on the forums

^thats his message

i feel like a toddler playing with crayons im so lost

[gray dust]

the main mistake there is conflating scalar multiplication with dot product

take a look at the givens again, show the equations a=p.u & b=p.v are true, which you can do by starting at one end of an equation and eventually "reaching" the other

[wintry steppe]

Can someone please explain to me the difference between the terms Cryogonal and Orthogonal? I'm having a lot of trouble understanding what either of them mean. ||/s||

[half storm]

Cryogonal is a pokemon

[gray dust]

r/whooosh

[half storm]

I'm guessing he was memeing lol

but I still wanted to be that guy

lol

[gray dust]

see the /s

[half storm]

Don't know what /s is lol.

Is that signal that someone is memeing?

[lavish drift]

sarcasm

signals sarcasm

[half storm]

oh.

[gray dust]

it's like /j

[lavish drift]

ye

[gray dust]

also picea get back to the algebra

[lavish drift]

aaaAAAAA

i moved to another question

ill get back to it on a cool head

[pliant imp]

Hi everyone. Struggling with finding the basis for the eigenspace for A = [3 0; 8 -1] corresponding to the eigenvalue of lambda = -1. I know the answer is [0;1] from reading through a similar problem my textbook but I’m lost on the steps to get there.

[native rampart]

Solve a set of linear equations,you get required basis

Write Ax=(-1)x and solve for x

[pliant imp]

I get [-4 0; -8 0][x1; x2] = 0 from the characteristic equation.

This yields
-4x1 + 0x2 = 0
-8x1 + 0x2 = 0

My question is essentially how do I go from this to the solution of [0;1] instead of [0; 0] or something else.

[native rampart]

x1 will be 0 ,x2 can be anything

So,Any vector in in the required subspace will be of the form [0,a], which is generated by [0,1]

@pliant imp

[pliant imp]

The penny finally dropped. 😂 Thank you. @native rampart

[surreal thistle]

if i see S_20 as a dodecahedron

it has 190 permutations right?

because 19 unique rotations and 10 axes of symmetry?

or is this the wrong way to think about it? I cant really imagine a 20 dimensional shape..

[native rampart]

Don't

*most

You can choose (1,2,3) and (2,4,6) and span is same

[pallid rampart]

It is always false

Two subspaces always intersect

Namely 0 is always in the intersection

[native rampart]

Sorry,forgot trivial intersection

[thorn lichen]

okay bet

thats what i put

:)))

https://tenor.com/view/well-idk-meh-excited-gif-5242667

[winged belfry]

what does $A^T$ mean? I understand what's happening in this situation but not sure what it means exactly. Like I see that column 1 turns into row 1 in the new matrix (c2 to r2, c3 to r3, and so on)

Just don't know the logic behind it

[stoic python]

Apollo:

[gray dust]

@winged belfry ^T means transpose, turns rows into cols & cols into rows

[winged belfry]

oh ok

thank you ❤️

[gray dust]

you're welcome

[weary patio]

Can anyone explain what linear independence or dependence actually means? I understand how to find out, Im just not grasping conceptually

[subtle walrus]

so you want intuition?

in the span of linearly independent set each vector adds an additional degree of freedom (1 more dimension)

like the span of a single nonzero vector is a line, the span of 2 linearly independent vectors is a plant, with another linearly independent vector its 3d space, and so on

linear dependence is just the opposite of that

there is at least one vector in the set that does not add a degree of freedom, i.e. you could already "reach" that vector with linear combinations of the others

[limber sierra]

you could say that adding a linearly dependent vector doesnt give you any more "expressive power"

like say i give you blue and yellow paint

this would be analogous to a linearly independent set

you can make blues, yellows, or mix them to add greens, but there's no way to make blue from just yellow, or to make yellow from just blue

now say i gave you blue, yellow, and green paint

the thing is: you could already make green paint from blue and yellow

so this doesnt really give you the ability to make more new colours of paint

hence its linearly dependent

meanwhile, if i gave you blue, yellow, and red paint instead

you'd get a whole host of new colour options: reds, purples, oranges, browns...

it was not possible to make any of these colours before; theres no way to mix blue + yellow to get red

so adding red paint preserves linearly independence

in other words: {blue, yellow} had the same expressive power as {blue, yellow, green}

so the second set must be linearly dependent

whereas {blue, yellow, red} had significantly more expressive power

[half moss]

Hey guys, is there a relationship between the norms of two vectors, and the induced matrix norm of the matrix produced by the outer product of those same two vectors ?

I know there is some sort of relationship, but I'm having trouble finding what it is

[half moss]

The Matrix p-norms induced by the vector norm

[wintry steppe]

you tell us

[gray dust]

recall that norm is related to dot product by ||x||^2=x.x, apply properties of dot product and you get an answer

[oblique sonnet]

can someone help with this?

and this

would be helpful

[zenith bear]

Anyone got any idea on this one?

I've established that the result vector has to be to a point on the "left" side of the line parallel to the first vector, but I have no idea how to generalise that.

[flat sedge]

LOL DEAD. Another funny reference is if ‘bae’ used the replacement theorem after ‘his’ response

[dry pulsar]

Find the eigenvectors of T. Let $V=$ $M_{2 \times 2}(\mathbb{R})$ y $T\left(\left[\begin{array}{ll}a & b \ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \ c & a\end{array}\right]$

[stoic python]

alef0:

[dry pulsar]

how can i do it

[native rampart]

Use (T-aI)v=0

for eigenvectors v with eigenvalue a

[dry pulsar]

Thank you!

[lucid cedar]

I have been thinking about this problem for a while:

As i was typing out my dilemma, I realized the flaw in my logic

Would a valid transform be something like T(0,2) -> T(2,0) -> (0,0)

that would satisfy this condition that it must be even, but its hardly a proof

[native rampart]

You are mapping (2,0) to (0,2) and (0,2) to (0,0)?

[lucid cedar]

T(0,2) = (2,0)

T(2,0) = (0,0)

[native rampart]

Should work

For the proof, just use the rank nullity theorem

[lucid cedar]

never heard of him

[native rampart]

Rank(T)+null(T)=dim(V)

[lucid cedar]

oh axler did introduce that somewhere

hold on let me look that up

fundamental thm. he called it

[wintry steppe]

axler 🤢

[lucid cedar]

its the book my class is using

i dont hate it really, what books would u use over it?

[gray dust]

ladw

[lucid cedar]

So is the idea here that because dimNull(T) and dimRange(T) are both mapped to V that they should have a basis of equal legnth? n + n = 2n?

[native rampart]

Yes

[lucid cedar]

That was my intuition. That the basis have the same number of elements because were mapping V -> V. It makes sense when i think about it

im still pondering how to show that tho

[native rampart]

Rank(T)+null(T)=dim(V)
@native rampart You mean this?

[lucid cedar]

yea thats what im talking about right now

[native rampart]

Take a basis of null(T) and extend it to a basis of V

And manipulate

[lucid cedar]

Basis V: u1,...,um,v1,...,vn
Basis Null(T): u1,...,um
Basis Range(T): Tv1,...,Tvn
Range(T) = Null(T) so each Tvi maps to a uj term. Therefore, m=n.

im very unsure about the statement "each Tvi maps to a uj term". I think thats right but im shaky on that

im trying to read around the text and see if i can confirm that or not

oh yea no that must be true

if they are equal then they have equivalent basis

[versed hearth]

hey, axis aligned bounding boxes are parallelepipeds where each edge is parallel to one axis. They are easy to represent and in 3D can be represented by only 2 points. My book claims that there is another representation where you just need three axis aligned basis vectors and one corner point. The other points can be reconstructed as a linear combination from this. I don't really see what they mean

[spiral star]

@versed hearth im lazy so im gonna show you how it works in 2D, but you just need another vector for 3d ...

one corner is p, the other corners are p + u, p + v and p + u + v

[errant mist]

Hi, I was wondering if anyone could give me some intuition/ geometric interpretation from the following example? I certainly understand projections when it comes to having a vector i.e $v$ being projected on some other vector $w$ (proj(v,w), but was wondering how it works in this case with both functions and polynomials? Projecting the function $2t-1$ along $t^{2}$. I assume then the resulting projected vector will be a scalar multiple of g(t)?

[stoic python]

Fredrikpiano:

[native rampart]

Treat polynomials as vectors

[errant mist]

yes, I get the computation part of this and are we taking the integral I assume because we are dealing with functions, so the projection will another function in some n-dimensional space?

[native rampart]

These functions are just concrete vectors

Yes

[errant mist]

And why is it that we can relate the Fourier coefficients as c? Is it because one can view the whole $1, cos(t), cos(2t)..., sin(t), sin(2t),...$ as an orthogonal subset of a vector space, i.e say $V$ making projections possible?

[stoic python]

Fredrikpiano:

[native rampart]

Yes

[errant mist]

cool! Certainly easier to visualize in 2d though...

[native rampart]

They don't exactly use that inner product,though

[errant mist]

What do u mean?

[native rampart]

The inner product (according to which cos(t),sin(t)... Form an orthonormal basis) is
(f,g)=$(1/π) $$\int_{0}^{2π} f(x)g(x) dx$

[stoic python]

DrunkenDrake:

[errant mist]

Ahh yes the inner product needs to be defined that way, yes agreed )

[dusky epoch]

bad tex!!!

[native rampart]

What should I change?

[dusky epoch]

$(f,g) = \frac{1}{\pi} \int_0^{2\pi} f(x) g(x) \dd{x}$

[stoic python]

Ann:

[dusky epoch]

none of this "two dollars right in the middle of the formula" nonsense

[errant mist]

haha

[strange crystal]

$ T : \mathbb{R}^3 \rightarrow \mathbb{R}^2 $

[stoic python]

DreyNotDre:

[strange crystal]

Say this transformation had a standard 2x3 matrix with a pivot in both rows, would it be valid to say that T is onto because the columns of A span R^2?

[dusky epoch]

yes

[strange crystal]

Awesome, ty

[mighty ridge]

Are there any rules for what an induction proof's equation can look like?

[limber sierra]

equation?

as long as youre quantifying over the naturals, induction is "valid"

doesnt have to be an equation

[mighty ridge]

Doesn't it? Every time I've hunted down a video on the subject there's always an equation that gets solved for N+1

[mighty ridge]

Thank you for the clarification then. Induction is a strange thing for sure

[lime herald]

Hello. How can i rotate second order surface using angles? I know that rotation is in general eq but how i can transform it for using angles?

Oh. I realized that rotation matrix should be enough

[ocean sequoia]

write it as an augement matrix

[10, -11, 9|14] we then have two "free" variables

[wintry steppe]

i'm having trouble visualizing how to write out the matrix with this 18% mistake

For a general diet problem with f foods and n nutrients, how many vertices do you expect for the feasible region? Why?
This is an open ended question.

can someone help me with this?

[versed hearth]

@spiral star thanks

can you tell a bit more about this @wintry steppe ? Is this a question for a flow network optimisation problem?

[wintry steppe]

im not sure what that is @versed hearth

but its basically the diet problem

so if u had this problem

u get a graph like that

and there are 3 vertcies that satify this

which are these 3

[versed hearth]

I see

[wintry steppe]

but what if we have some number of nutrients and some number of foods

in this case we had 2 nutrients(iron,protein) and 2 foods(spoo and gaph)

[versed hearth]

are you sure that you have 3 vertices in figure 4?

[wintry steppe]

yeh because i used the test point 0,0

and found that its above both lines

thus the feasbile region is at and above A B and C

[versed hearth]

oh it's to the right? That is strange

I thought it's the area which contains the origin

ah right, you have a minimization problem so it makes sense

[wintry steppe]

ye if u plug in 0,0 for x1 and x2 u find that the its above the lines

[versed hearth]

I'm not sure how to solve this myself... If you stick with 2 nutrients, then you have f foods. Each food gives you a new linear constraint in this 2D space. This reduces the question to how many intersections can f + 2 non-parallel and distinct lines have (your x1 x2 >= 0 are two additional ones). That's your upper case of vertices

if you have parallel lines, then you can disregard them

but I don't know how this upper bound relates to what you can expect from this problem

[wintry steppe]

https://cdn.discordapp.com/attachments/540211747613704221/756625326460764311/unknown.png

i'm having trouble visualizing how to write out the matrix with this 18% mistake

[obsidian wren]

rq, can anyone gimme a hint on this?

#2

[soft burrow]

use $x=M^{-1}y$ for some non-zero $y$ and the fact that the operator norm is submultiplicative, along with its definition as the supremum of a certain quotient

[stoic python]

bastian.uwu:

[wintry steppe]

Wtf is a kernel M isn't corn

[obsidian wren]

what is a supremum again?

fuc

[gray dust]

for a linear map T, ker(T) is the preimage of {0}

[soft burrow]

what is a supremum again?
@obsidian wren least upper bound, sorry

$\lVert A \rVert = \sup_{x\neq 0} \dfrac{\lVert Ax \rVert}{\lVert x \rVert}$

[stoic python]

bastian.uwu:

[obsidian wren]

ah, i see.

thank you

@soft burrow omg i got here 3 different ways

but i didnt put it together until i tried this

[soft burrow]

great

[dry pulsar]

3. Find Eigenvectors of T and Matrix associated to T Let be $V=$ $M_{2 \times 2}(\mathbb{R})$ y $T\left(\left[\begin{array}{ll}a & b \ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \ c & a\end{array}\right]$

[stoic python]

alef0:

[wintry steppe]

How on earth do I row reduce with decimals? I'm trying to follow to example of getting ' 1 ' , in the first row. But Im clueless on how to get it. I thought doing 0.9x = 1. then solving for x, but Im getting a repeating decimal.

[wintry steppe]

My operations just seem a bit unorthodox as I’ve never done that with any of the practice problems. I ended up dividing the first row by 1/0.9, and got half of the matrix right. But bottom half is still tricking me

I figured it out, just multiply R2 by -1/0.9 to get 1. ugh. dont know how i missed that. yikes

[limber sierra]

this isnt linear algebra

[slow scroll]

wrong channel. This goes in #prealg-and-algebra

[limber sierra]

set up an equation for "what they wanted to happen"

and another equation for "what actually happened"

note that the right hand side of the first equation will be 9*0.25, while the RHS of the second equation will be 9 * 0.22

oh, and of course you'll want an equation for the total volume (so the RHS of this one should just be 9)

[old flame]

Got a quick question, I'm not too sure what is the meaning of $P_{U,W}$, for abitrary $U,W$. From my understanding, take for example $P_{U,W}$ from the text. Is this transforming the vector v to coincide with its u component ? So like looking at the vector v, from the perspective of vector u. So let's say v is a combination of the basis vectors $(v_1,...,v_n)$. If we take $P_{V_1,(V_2,...,V_n)}$, (capital as in the subspace with respect to the basis vector) then it would yield $v_1$ ? With its prospective scalar of course.

[stoic python]

Otoro:

[pallid rampart]

Here’s my little diagram to help you

Notice how $P_{U,W}v+P_{W,U}v=v$ and $P_{U,W}v$ is on $U$ and $P_{W,U}v$ is on $W$

[stoic python]

Whoever:

[old flame]

Sorry to interrupt,but as I recall from memory, it seems to be that there's a formula involving $cos \theta$ for vector projections, is this similar ?

[stoic python]

Otoro:

[pallid rampart]

Well not necessarily

But the thing is that, the diagram I drew is a specific case when you are in the vector space R^2

The definition gave in the book works for any vector space

You can think of $P_{U,W}v$ as "giving you part of $v$ that is on the subspace $U$"

[stoic python]

Whoever:

[old flame]

So basically, it is just looking at the vector in different subspaces? Given it is a part of the vector

[pallid rampart]

Kind of

That is the intuition you should have about it

[old flame]

Ahhhhhhh thank you very much

[pallid rampart]

So you are projecting the whole space onto a subspace

Do you understand the rigorous definition though

[old flame]

Sorry, but which is the rigourous definition ?

[pallid rampart]

Definition of P_U,W

[old flame]

So $P_{U,W}v=v$ iff $v \in U$ ?

[pallid rampart]

Yes

[stoic python]

Otoro:

[old flame]

Lets see

So the projection of any vector is of itself, if only the vector exists in that particular subspace ?

[pallid rampart]

Yeah

[old flame]

So as in both the subspace and the vector space contains that vector, hence the projection is the same

Then may I ask what is the use of this operator ?

Just fueled from curiosity

[pallid rampart]

Well you will learn later in the book that the projection actually gives you the "closest" point in the subspace to v

and this can be used to find the least square regression line

[unreal sierra]

Find a system of equations in three variables x, y, and z whose solution is (2, 3/5, 1/2)
Can you please help me with this question pls

[limber sierra]

x = 2
y = 3/5
z = 1/2

i'd assume this isnt the sort of answer they want

but it technically meets the criteria

feel free to add equations to each other or multiply them by constants or whatever to make this more complicated

[old flame]

@pallid rampart oh that's very interesting, wasn't expecting an application in linear algebra, thank you so much

[unreal sierra]

They want 3 eqaution of x+y+z=0 so I get trouble in this equation hays

[dusky epoch]

@unreal sierra if it helps: there is no unique answer to this question

[unreal sierra]

Why?

[dusky epoch]

there are many possible systems with (2, 3/5, 1/2) as solution

such as the trivial one namington posted

but also you can just pick (almost) any coefficients you want, construct the left-hand side with those, and then fill in the right hand side with the values of the LHS for x=2, y=3/5 and z=1/2

[unreal sierra]

Yes I will pick one of those but I didn't know how to get any possible system regard that

[dusky epoch]

you don't need every possible system! just one is enough.

[unreal sierra]

2x - 5y + 2z = 2
x + y + z = 31/10
-3x + 5y + 6z = 0

[dusky epoch]

,w 2x - 5y + 2z = 2
x + y + z = 31/10
-3x + 5y + 6z = 0

[unreal sierra]

This is my first answer regards that

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[dusky epoch]

this works

you don't need to do anything else really

[unreal sierra]

Ah okay thank youu no formula regard how you get that?

[dusky epoch]

??

what do you mean formula

i just took your system and threw it at wolfram alpha and it solved it

and the solution matched what you gave

[unreal sierra]

Ah okay okay thank you for that

[stiff zenith]

Which of these courses would you recommend?

http://joshua.smcvt.edu/linearalgebra/

https://www.math.ucdavis.edu/~linear/

https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/

[flat sedge]

If ur going for intro level I’d say first or second link

Are those campuses open campus for non students

[wintry steppe]

Help me TT

[native rampart]

d

[wintry steppe]

can u explain TT thankyouu

[stoic python]

DrunkenDrake:

[wintry steppe]

thankyou so much!!

:3

[mossy trail]

Can someone help me or give me any tips on proving: If T is a linear transformation and u1, u2, u3 are linearly dependent vectors in the domain of T, then the
vectors T(u1), T(u2), T(u3) are also linearly dependent.

[magic light]

well u1, 2, u3 being dependent means a1u1 + a2u2... anun = 0

iirc

[limber sierra]

erm

[magic light]

and every an = 0

[limber sierra]

no

thats what it means to be linearly independent

[magic light]

o

misread

they're dependent

either way it's the same

[limber sierra]

it means there exist $a_1, a_2, a_3$ such that $a_1u_1 + a_2u_2 + a_3u_3 = 0$ and at least one of $a_1, a_2, a_3$ are not $0$

[stoic python]

Namington:

[magic light]

yes

plug that into the transformation

[limber sierra]

thats a little vague

but yes you can use the fact that $x = 0 \implies T(x) = 0$

[magic light]

I just suck with the latex so it's hard

[stoic python]

Namington:

[limber sierra]

for linear transformations T