Hello everybody. How's quarantine going for y'all? 😆
#linear-algebra
hardy meteor
hoary agate
brittle juniper
Ker(C) is contained in Ker(BC), so you take a supplementary subspace S of Ker(C) in Ker(BC)
the rank theorem tells you that C(Ker(BC)) and S are isomorphic and then
dim(Ker(BC) = dim(S) + dim(Ker(C)) = dim(C(Ker(BC)) + dim(Ker(B))
and dim(C(Ker(BC)) <= dim(Ker(B)) because C(Ker(BC)) is contained in Ker(B)
@vague fulcrum
hardy meteor
So, I got a question. I have a 3*3 matrix, how do I know it's eigenvalue's multiplicity?
gray dust
algebraic or geometric?
hardy meteor
Algebraic
limber sierra
construct and factor the characteristic polynomial
do you know what i mean by that?
then, just find the exponent (multiplicity) of the factor corresponding to that eigenvalue
for example, if a matrix has characteristic polynomial
[(x-5)^2(x+3)]
then the algebraic multiplicity of the eigenvalue $5$ is $2$, and the alg. multiplicity of the eigenvalue $-3$ is $1$
stoic pythonBOT
Namington:
hardy meteor
So, I got eigen values 10, and 2, what would the multiplicity be? Is it connected with the eigen value?
gray dust
what's the charpoly?
quartz compass
read #❓how-to-get-help
main root
hardy meteor
Don't know what charpoly means
limber sierra
characteristic polynomial
do you know what that means? (if not, how was algebraic multiplicity defined?)
limber sierra
so your characteristic polynomial is $(2-\lambda)(\lambda-1)^2$?
stoic pythonBOT
Namington:
hardy meteor
Yep
limber sierra
the algebraic multiplicity of an eigenvalue is the power of its factor
so for the eigenvalue 1
the power is 2
since there's a $(\lambda-1)^2$ factor
stoic pythonBOT
Namington:
limber sierra
so its algebraic multiplicity is 2
for the eigenvalue 2, the algebraic multiplicity would be 1
hardy meteor
Ahhh, I get it. Thanks a lot
hardy meteor
Another question that I have is how do you figure out the determinant if a matrix is in upper or lower triangular form
Upper being when only the 0s are below the diagonals, lower being when only the 0s are above the diagonals
subtle walrus
just multiply all the elements on the diagonal
hardy meteor
In both cases?
subtle walrus
yes
hardy meteor
Ok, thanks
calm plank
My textbook says d) is singular. This seems wrong.
The matrix respresents a map from a 3d domain to co-domain, and the matrix rank is 3, so why is it singular?
hardy meteor
How do I multiply random 3×3 matrix with a t exponent ?
A = PD (inverse of p)
How do I make it A^k = PD^K (inverse of p)
gray dust
try k=1,2,3, realize how things simplify, then generalize
torn hornet
what did you try?
fallow jolt
Well I was able to prove part a, but I cannot really figure out how multiplying a matrix by another matrix would result in the same norm
fallow jolt
orthonormal to A or just to each other?
fallow jolt
Well nothing is said beyond the problem i sent
torn hornet
can you use this now to solve it?
fallow jolt
Sorry
F norm of UA = F norm of VA
I can see how to get there
But what about the final thing, that both equal F norm of A?
torn hornet
put in UA in the form of part (a)
torn hornet
no
i told you what orthognal means
a matrix being orthogonal with respect to another doesnt mean anything
please look at what i sent before
torn hornet
good, can you show me what u did just to be sure
fallow jolt
Sure, let me write up a solution
Hm actually
This algebra gets really awful
I wrote U and A in general terms, then multiplied them
torn hornet
hmm thats not what you do
fallow jolt
But now to find (UA)^T(UA) is gonna suck
torn hornet
the actual solution is a one liner
fallow jolt
😮
fallow jolt
well the diagonal doesnt change from UA
torn hornet
in general whats the transpose of 2 matrices multiplied together
fallow jolt
Is there a special rule? I know what a transpose is
torn hornet
$(AB)^T=B^T A^T$
stoic pythonBOT
JohnDoeSmith:
fallow jolt
ohh so
torn hornet
theres definately some knowledge missing here btw,you should know what orthognal matrices are and that this is an identity. You prolly want to review some stuff
fallow jolt
(UA)^T*(UA) = A^T*U^T*U*A = A^T*A?
torn hornet
mhm
fallow jolt
@torn hornet Why are we allowed to multiply U^T*U in the middle?
Don't we have to go left to right?
fallow jolt
Ah okay
But so for AV
(AV)^T*(AV) = V^T*A^T*A*V
How do we simplify this any further?
torn hornet
hmm right this part is a bit indirect
theres a few arguments you can make here
im gonna let you prove the following:
covert skiff
hey everyone. Can someone tell me how to solve this question? More importantly the last part
For the following matrix, determine the eigenvalues and their multiplicities. Find the eigenspace and its dimension for the eigenvalue whose multiplicity is highest.
Haven't I already found the eigenvalue in the first part? Is it something different in the second part?
ok, I am dumb. Its eigenspace
though, how do I find that?
fallow jolt
@torn hornet Proved it! Thank you 😄
covert skiff
can someone tell me how I can find eigenspaces?
I've got an exam soon, and I really don't know how to solve those
steady fiber
do you know how to find eigenvalues
covert skiff
yes
steady fiber
do you know how to find eigenvectors?
covert skiff
yep
steady fiber
the eigenspace is just the space spanned by the eigenvectors
covert skiff
huh? So, eigen value is 2, eigen vector is [1 0 0 0] (vertically) (1 column)
What would the eigenspace be?
steady fiber
there's only 1 eigenvector?
covert skiff
yeah
steady fiber
then the space is just $\operatorname{span}{[1\text{ }0\text{ }0\text{ }0]^T}$
just the span of the eigenvectors
covert skiff
huh, okay. But what happens if there are 2 eigenvectors?
stoic pythonBOT
PorosInMyAshe:
covert skiff
assume any value that is simple for you to explain 🙂
steady fiber
if you have $n$ eigenvectors $\mathbf{v}_1, \mathbf{v}_2, ... \mathbf{v}_n$, the eigenspace $V$ is:
$$V = \operatorname{span}{\mathbf{v}_1, \mathbf{v}_2, ... \mathbf{v}_n}$$
stoic pythonBOT
PorosInMyAshe:
steady fiber
the eigenvectors form a basis for the eigenspace
covert skiff
ahh, i get it. Thanks a lot
long blade
hello
long blade
long blade
this is my attempt of this question
i just want to know if what i did is correct:
to be vector space: non empty, closed under addition, closed under addition
(1) 0 is an element of H and K therefore it is an element of H intersect K
(2) lets say vector A is an element of H intersect K and vector B is an element of H intersect K, Therefore, A + B has to be an element of H intersect K
(3) we know that H and K are subspaces of V, then the scalar lets say C , then C * A is an element of H and also an element of K
therefore its an element of H intersect K
is my justification correct
thanks in advance
covert skiff
Next in line to lakinvu 🙂 I have no clue how to solve this, can someone explain the process to me?
flint flicker
@long blade yes, i think you have the right idea but for (2) you need to show more steps and (3) you might want to go the other direction and declare an element of K intersect H first
long blade
K and H are subspaces therefore cA and cB should be elements
dry river
this is linear algebra, though I dont even see an x
long blade
of K and H
flint flicker
That was a fun little proof problem
dry river
all I see is random english letter and people talking about them
this makes me feel stupid lol
long blade
@flint flicker i looked at ur explanation thank you
i realised what i needed to added ty
flint flicker
np
covert skiff
if anyone's got some time, can they explain how I do this?
long blade
@flint flicker can u help me with this question
this time i have no clue how to do it
covert skiff
me?
long blade
or me?
long blade
mine?
long blade
this is first year uni
dry river
oh f
covert skiff
mine is second semester college
dry river
so how is it linear?
dry river
i thought linear was like y = x + 2
long blade
lol
lol
yeah first chapter is that
like excersise one of first chapter
can someone pls help me with this quesiton
covert skiff
i could, but I can't
Can you help me with mine?
Cuz I don't know how to even start your question lol
@long blade
long blade
ty
covert skiff
It went way above
covert skiff
yikes
covert skiff
np
flint flicker
@covert skiff i'm pretty sure you have to use these two properties of linear transformations to solve the problem
and then you can figure out what T(0, 1) and T(1, 0) are
and then use that to solve
covert skiff
but I've heard into means something important. I haven't paid attention in class for linear transformations.
What can i do for it? Like find a scalar and multiply that with the values below or something?
flint flicker
T(3, 1) = 3T(1, 0) + T(0, 1)
expand it like that
do that with the other given transformation
and then you have a system of equations and can solve for T(0, 1)
i don't completely remember if that's the process for these questions but i think that leads in the right direction
@long blade i think for that problem you have to go and verify all 10 axioms of vector spaces
flint flicker
it's gonna be nasty and long and you have to work with complex numbers too
flint flicker
okay cool
covert skiff
Can you elaborate on it? Like what would I have to do next on it?
After I do 3T(1,0) + T(0,1)
@flint flicker
flint flicker
you'll get the result of what T(1, 0) and T(0, 1) are
and then you can use those to find the ones the question asks you for
grand imp
Let ${\mathbf{v}}= \mathbf{v}_1, ...,\mathbf{v}m$ be vectors in a vector space $V$, and let $\Phi{{\mathbf{v}}} :\mathbb{R}^m \to V$ be the associated concrete-to-abstract transformation. Then: $\$
- The set ${\mathbf{v}}$ is linearly independent if and only if $\Phi_{{\mathbf{v}}}$ is one to one. $\$
- The set ${\mathbf{v}}$ spans $V$ if and only if $\Phi_{{\mathbf{v}}}$ is onto. $\$
- The set ${\mathbf{v}}$ is a basis of V if and only if $\Phi_{\mathbf{v}}$ is one to one and onto (i.e., invertible)$\$
stoic pythonBOT
ApproximatelyASphere:
grand imp
this seems to suggest that an abstract vector space can only have a basis if it's isomorphic to R^n
But I could've sworn every vector space has a basis
is there a detail I'm missing?
sonic osprey
Infinite dimensional vector spaces get messy
To prove that every infinite dimensional vector space has a basis, you need to assume the axiom of choice for one
grand imp
oh hm, so this implicitly assumes that it has to be a finite vector space
sonic osprey
finite dimensional
civic tiger
viscid kernel
Does anyone know the difference between an euclidian space and a prehilbert space ?
dusky epoch
a pre-hilbert space isn't required to be findim
viscid kernel
??
sonic osprey
finite dimensional
wintry steppe
@covert skiff The vectors u and v form a basis of R². If au+bv (a and b scalars) is an arbitrary vector in R² then T(au+bv) = aT(u) + bT(v). All you need to do is write (6,-8), (-6, -2) and (0,20) as linear combinations of u and v and plug the scalars into aT(u) + bT(v)
pale shell
Anyone here will teach for free over discord
main drum
my people
i have missed you
can somebody ask my prof what curve he is putting for the class
subtle walrus
he said no curve
main drum
i have an 89.72 if i go apeshit and get a 100 on the final
i need
just 1 point
and every semester he usually does 2
but idk if he gonna change things bc coronaannanana
@pale shell i will teach you
not for free
i want exposure
pale shell
dont have one
main drum
aren't you famous
pale shell
nO
main drum
i only accept payment in exposure
viscid vale
Can someone help explain this step to me
How did they know that A transpose A was equal to lambda?
thorn robin
A transpose A is a matrix
lambda is a scalar
So what you said does not make sense
Do you know what an eigenvector is?
viscid vale
@thorn robin a vector v such that Av=lambda v i think
Can u explain whats going on here
thorn robin
right
In the paragraph above the calculation, they say that v_j are eigenvectors of A^T A
what does that mean?
viscid vale
So (A^TA)v_i = (lambda_i)v_i
flint sequoia
Hey! I don’t have a professor right now because he’s too ancient to figure out how to use his email let alone run online classes. So I’m trying to teach myself
I know how to find a matrix representing the transformation but I am very rocky on my theory
Would someone mind explaining to me how to do problem 30? I already found the matrix A=[2, -1, 4]
I’m pretty much just confused about the definition of the range
odd kite
by range of T they mean the set of all possible output values
Disc,there isn't just one reason
example 1: suppose I want to find the dot product of $A\vec{u}$ and $\vec{v}$ where A is a matrix. We could write this as $(A\vec{u})^T \vec{v} = \vec{u} ^T A^T \vec{v}$
quasi vale
btw you can write \cdot to show the dot product
stoic pythonBOT
Timon:
sonic osprey
There are hundreds of reasons we care about transposes
odd kite
Another example: any matrix can be written as the sum of a symmetric and skew matrix $S = \frac 1 2 ( M + M^T) $ $A = \frac 1 2 ( M - M^T) $ . This is useful for a lot of reasons
stoic pythonBOT
Timon:
odd kite
$M = S + A$
stoic pythonBOT
Timon:
odd kite
So any matrix M can be split into S and A by using the transpose
flint sequoia
I’m pretty sure my nullspace and nullity are correct, but I don’t know how to express the range
odd kite
If it helps, the image is the vector space spanned by the columns of the transformation's matrix
range = image
flint sequoia
My understanding is that Rank(A) + Nullity(A) = the number of linearly independent columns in A which is trivially 3 since it’s a single row. And that rank is the number of column vectors forming the basis of the image
But I find the dimension of the image would be 3, so there’s definitely something I’m missing
odd kite
yes you are very confused
odd kite
no
look at the rank nullity theorem again
it doesn't say what you wrote
what you wrote should also send up some red flags
If T: V->W then the range is a subset of W. It can have at most the dimensions of W but not more
So T: R^3 -> R can't have a 3D image, does that make sense, @flint sequoia
ocean sequoia
Maybe another way to think about that is if you go from T:R^2 to R^2 you wont ever pick up a 3D image
flint sequoia
It does make sense in isolation, I think I have a spotty understanding of more basic concepts. The book talks a lot about the range of just a matrix before even talking about linear transformations
Going back to an old section to see if I can fix my understanding
pale shell
uhh hi
can someone explain what linear algebra is because im so confused
Like
I though it was just y=mx+b
But someone was like
NO
and all these things
odd kite
Have you heard of a vector yet, @pale shell
odd kite
how about coordinates in ordered pairs like (x, y)
pale shell
yeah
odd kite
okay, so in linear algebra we deal with linear functions but the output can be something like that, (x,y), not just one number as an output
pale shell
ok
odd kite
mx+b is not actually a linear function because of the +b part
linear function means f(qx) = qf(x) and f(x+a) = f(x) + f(a)
pale shell
ok
odd kite
that's not true for f(x) = mx + b unless b is 0
pale shell
- makes some sence i understand so far
daring solstice
ok urm i don't understand why the answer is a 2x2 matrix
pale shell
lemme take a look
daring solstice
I expected something like this because col(A)j x row(B)j where from j = 1 to j = 2
pale shell
yeah i get where you are coming from but im not very good t this
daring solstice
so I thought it'd be (col one of (A) x row 1 of B) + (col two of A x row two of B)
odd kite
it prob. means the outer product
so $ (a; b) (x ;y) = \begin{bmatrix}ax; ay \ bx; by\end{bmatrix}$
stoic pythonBOT
Timon:
daring solstice
I thought the outer product of a (1x2) x (2x1) would be a 1x1
wait no that's matrix mutiplication
odd kite
yeah that's inner product, it's different
the inner product of two N dimensional vectors is a scalar, the outer product is an NxN matrix
flint sequoia
@odd kite thanks for pointing me in the right direction, I understand all of this much better now
odd kite
yw
wintry steppe
I understand that vector spaces on a field are a set with 2 operations +, •, satisfying a set of axioms.
But can anyone explain why and how we can interpret vector space as a set of functions from A to B? And in this case what would A and B be?
slow scroll
all you need is a way to add functions together and multiply by scalars. For example, the set of polynomials with degree less than n and coefficents in R form a vector space. The addition and multiplication are basically exactly what you would expect.
ocean sequoia
and someone can correct me here but if we add two vectors or multiply by a scalar we always stay in the same space right
if the addition vectors are in the same vector space sorry
odd kite
yes a vector space must be closed under both operations
wintry steppe
Ok, so because functions have the property f1(x)+f2(x) = (f1+f2)(x) and cf(x) = (cf)(x) we can interpret vector space as set of functions?
slow scroll
you can interpret certain sets of functions as vector spaces, not really the other way around I don't think
wintry steppe
So not all vector space can be represented as a set of functions is what you mean?
slow scroll
wat, that doesnt even make sense xd
ocean sequoia
sorry to join in on the questioning
slow scroll
every finite dimensional vector space is isomorphic ("same") to its so-called "dual space", which IS a space of functions, but thats not really how people normally think of vector spaces, no.
wintry steppe
I'll provide some context to what I'm confused about
wintry steppe
yea this is sort of proof based lin alg
So I don't really understand what this vector space is
slow scroll
yes, this is just an example of a set of functions which happens to be a vector space.
wintry steppe
Oh, ill think about it
cold topaz
these are the constants I found for this matrix, but it doesnt pass for the second row. But it does for the first and the third. How's that possible?
wintry steppe
the claim (f+g)(i) = f(i)+g(i) means that the statement must be true for all i in {0,1,2} and all functions f,g in V right?
odd kite
yeah
slow scroll
(f+g)(i) = f(i) + g(i)
is really a definition of f+g.
you can prove that its associative, commutative, and closed inside the set
wintry steppe
And is it because of these nice properties of functions that in general any set of function is a vector space?
including 0 function i guess, idk
odd kite
no of course not any set
if my set includes f(x) but not 2f(x) for example
perhaps you meant any set of functions can serve as a basis (still not right but closer)
half ice
Note that a vector space is a set, with a couple operations, that follows 8 axioms. Anything that follows these rules is a vector space
slow scroll
it all depends on how you define your operations. Certainly not every set of functions from some set to R form a vector space where f+g and cf are defined the way they are in that example
wintry steppe
I see
In the example (cf)(x) = c(f(x)) is that always true for functions?
or depends on how • is defined?
half ice
That's how we define (cf)(x), at least here
slow scroll
like, what if the domain of one of the functions in the set is different, then you are clearly going to run into problems even trying to define + and x.
cold topaz
can someone please help me. I've posted my question above.
half ice
Now, it's usually easier to think in terms of the subspace axioms
cold topaz
thanks!
odd kite
@real plaza no, linear maps don't have to be bijective
half ice
A subset of a vector space is a subspace if it's:
- Closed under addition
- Closed under scalar mult
- Contains the 0 vector
odd kite
@real plaza linear transformation is only bijective if the kernel is {0} (only the 0 vector), otherwise multiple vectors are mapped to 0
half ice
Wait, no, ignore me. I'm on a tangent I thought the question was asking for a proof it is a vector space
Instead it's looking for a basis
odd kite
and if dim(codomain) > dim(image) then it's not a bijection either
@real plaza pretty much
slow scroll
a linear map is literally just a matrix without the basis.
? its a linear function (mapping) between vector spaces, but you aren't writing down an array of numbers that tell what maps to what, because the exact vectors depend on the basis you are assuming for each vector space
to even do computations with a linear function, i do believe you have to assume some basis for each vector space (domain and codomain). However, there are many theorems and proofs in linear algebra that don't depend on a basis for a linear map. Thats the only real distinction i can think of
odd kite
my 2 cents, linear maps can have a basis
sonic osprey
maybe it's easier to say that
linear maps can be represented in different ways depending on which basis you choose
slow scroll
^ this.
@real plaza the kernel and image of any linear map is a subspace. thats an example.
the distinction between "linear map" and matrix is rarely that important. When you learn about change of basis though, you often talk about / introduce notation something to the effect of:
If f: V to W is a linear map. f_{AB} is the matrix for f which takes vectors from a basis "B" for V, and transforms them into vectors from a basis "A" for W.
trail swan
If A is a 10x2 matrix, is it expected that there is a unique solution x for Ax=b?
viscid vale
u could have a 10x2 matrix with the first column with a 1 on top and the second column filled with 0's, which would mean x2 would be free
so i dont think so
if anyone is familiar with singular value decomposition could i get some help in #help-5 please?
trail swan
But the question is asking if it is expected so for the majority of cases, rather than certain specific cases
sonic osprey
need to show
gaunt geyser
I am trying to understand this equation
to solve this
not sure how to get the D-CA^(-1)B term
trying to apply the leibniz formula but getting lost
found this explanation
is this the best way to think of it?
via the product rule
sonic osprey
The equation isn't going to apply to your question
in your question, a' and b are vectors, not square matrices
gaunt geyser
right
sonic osprey
Oh no that's a lie, it will work
gaunt geyser
sonic osprey
I think there are easier ways
junior pilot
Questions can lie?!
sonic osprey
Just cofactor expand along the last column or the last row
gaunt geyser
I guess cofactor expansion with vectors is where I get stuck
would the cofactor of b in the matrix be equal to -1(det(a'))
then I run into the issue of the determinant of a vector
I thought you needed atleast a 2x2 to find determinant
sonic osprey
b is a vector, you need to expand element by element
gaunt geyser
how can I do that when I don't know what the matrices are
if A is MxN
then b has coordinates i, N+1
for i = 1...M
seems like thats impossible to calculate with cofactor expansion
as I am taking a "chunk" of A and adding a' to the bottom
sonic osprey
A is square
gaunt geyser
seems like this can't really be proven with laplace expansion
idk how I could
like if A is 2x2
I am taking the determinant of
this then
this then
this
I don't feel like I can generalize the determinants of these minors to be zero
and because I have no knowledge of the actual values im just left with a mess of determinants
I guess they would cancel out to some extent
well I can't even say that
im basically looking to prove that second equation
this makes sense to me
I think I just was not aware of the decompisition they use
and how this generalizes when B and C are vectors
idle echo
Regardless of how nullity is formally defined, is it always equal to the number of columns in a matrix that don't have a pivot when that matrix is in rref?
limber sierra
It is @idle echo
The number of free variables in a homogenous RREF system, basically
quartz compass
I say just make up some numbers for those lines and then write down the matrix explicitly
that way it's less abstract to you maybe
flint sequoia
Is the set of all defective matrices a subset of the set of all non-diagonalizable square matrices?
Or here’s a better question, do matrices exist that are both non-diagonalizable and non-defective?
dusky epoch
define defective?
main drum
hi
i have a question
how protective of curves are professors usually from year to year
Like for instance
if a professor usually does a 2 point curve if things in his class go as expected
How likely is he to do that same curve if what is expected takes place in a different year
subtle walrus
if you spent halt the time worrying about a curve actually studying, you wouldnt have to worry about a curve anymore
main drum
boy im slapping 100's out here to recover one bad grade
my question still stands
if i get perfect grades on everything from now on
i get 89.72
i need a smidge of a curve
for A
subtle walrus
what changes if there is no curve
subtle walrus
ask your professor
subtle walrus
did he tell you to ask randoms on discord
so yes curve = + motivation, no curve = -motivation
?
main drum
well I was meaning to ask people that actually have more than one neuron performing mitosis 😦
hey though that correlation is showing you have at least 2
subtle walrus
i see no reason not to assume yes curve
main drum
too many double negatives in that sentence give me a sec
ok
i see
i was wondering how this corona stuff would effect it though
because our exams are totally cheatable to the point averages are going from 85 to 90
and the final average is supposed to be 79 but I expect it to be approx 86-88
i have one more question
if i show up as the only kid to everyone of this prof's office hours
and he is a pretty chill dude not an asshat
do you think he would lower out of pity
for the people like me
subtle walrus
can you ask him for extra credit or something
subtle walrus
sorry, i assumed real unis have no curve
then i dunno
assume he is a nice guy, and doesnt want to fuck you over
main drum
yea
subtle walrus
he will do it if it wont lead to people passing that shouldnt pass
main drum
he said that today he doesnt want anyone to be screwed by the ronavirus
it is usually a 2 point curve
not the biggest difference in people passing
but more so in people getting A
how is corona affecting grades
arent profs handing out freebies left and right as a result
main drum
yea US
alright ima go to all the office
hours
coax him into giving that extra .38 curve
and then run for the hills with my A
subtle walrus
then you should do better stuff
subtle walrus
k is the field
no, the dimension is arbtirary
infinite dimensional vector spaces have dual spaces as well
this notation is not unique to napkin btw
(and napkin is a bad book)
depends what you use it for i guess
but it's not a good source to learn from
you are referring to evan chens napkin?
it only gives a small overview of each subject
and has the problem of presenting a lot of "cool" stuff
without the nitty-gritty work you need to do it rigorously
yeah
as long as you are aware of that, it's good
wintry steppe
No; a simple example is A = 2I
A has eigenvalues 2, whereas A^-1 has eigenvalues 1/2
yeah the eigenvalues of the inverse are reciprocal
you can see this by looking at Av = kv
v = A^-1 kv; v/k = A^-1 v
gray nebula
Makes sense, thanks
dry spear
in this video why did he put the free varaibles in the same rows as the leading 1's
doesn't a free variable mean a row of all zeros, or a row without a leading 1?
placid oracle
If K is a field, and we choose three distinct elements a,b,c in K, how can I show that this is true? I think this is the Lagrange Interpolation formula
limpid vine
just wait 15m before @ helpers
placid oracle
my b
pliant fox
for subspaces of vector spaces, why is being a vector space a prerequisite for it to considered one? isn't it possible to be a subspace without being a vector space?
vital swallow
that would just be a subset, no? TheJohn
pliant fox
i guess so...
the material my class is using implied that a space that is a subspace of a vector space, which is a subspace, is only a subspace if it is also a vector space
i wanted to verify whether the word 'vector' as in 'vector subspace' was implied and omitted
sonic osprey
Subspaces are always vector subspaces
placid oracle
<@&286206848099549185>
sonic osprey
What's x in your formula
torn tangle
wintry steppe
didn't your question already get answered in #prealg-and-algebra
gaunt geyser
if a and b are vectors
and A is a square matrix
and alpha is a single number
how is this true
$\det(\alpha - a'A^{-1}b) = (\alpha - a'A^{-1}b)$
stoic pythonBOT
gREEnOnions:
gaunt geyser
or is the answer incorrect
@torn tangle those two notations are interchangeable pretty sure
you could just write
pallid rampart
\times
gaunt geyser
ah
so that answer on stackexchange is wrong
damn
still trying to prove this
@torn tangle
$A_{m\times n}$
stoic pythonBOT
gREEnOnions:
gaunt geyser
man having to go to like actual academic papers for proofs is annoying lol
trying to get what i need out of this paper
so we have
stoic pythonBOT
gREEnOnions:
gaunt geyser
$\det(\begin{bmatrix}
A & b \
c' & \alpha
\end{bmatrix}) = \det(\begin{bmatrix}
\alpha A-bc & b \
0 & \alpha
\end{bmatrix}) / det(\begin{bmatrix}
\alpha I_{m-1} & 0 \
-c' & 1
\end{bmatrix})$
stoic pythonBOT
gREEnOnions:
gaunt geyser
$\det(\begin{bmatrix}
A & b \
c' & \alpha
\end{bmatrix}) = \frac{\det(\alpha A- bc)\times \det(\alpha)}{\alpha^{m-1}}$
pallid rampart
Determinant equals a matrix 
gaunt geyser
idk man I've been trying to figure this out for a while
no one seems to be able to prove it
if you got any insight would be much appreciated @pallid rampart
stoic pythonBOT
gREEnOnions:
gaunt geyser
this is where I get stuck
I cannot take the determinant of alpha
because its a constant
gaunt geyser
can you show me the like property that shows thats true
odd kite
basically it's just a shorthand for const * I
odd kite
there's really an identity matrix there
gaunt geyser
but for the original partioned matrix
alpha has to be 1x1
because b and c are column and row vectors respectively
gaunt geyser
so I can say that
stoic pythonBOT
Timon:
gaunt geyser
$\alpha = \alpha \times \begin{bmatrix}
1 & 0 \
0 & 1
\end{bmatrix})$
stoic pythonBOT
gREEnOnions:
odd kite
well if it's 2x2, yeah
stoic pythonBOT
Timon:
gaunt geyser
I just am confused because I thought you could only take the determinant of at the minimum a 2x2 matrix
and you are saying the determinant of a constant is the constant
odd kite
no, a 1x1 matrix just has det equal to the value of the 1 number in it
gaunt geyser
oh shit
that simplifies things
so $\det(\begin{bmatrix}
A & b \
c' & \alpha
\end{bmatrix}) = \frac{\det(\alpha A- bc)\times \det(\alpha)}{\alpha^{m-1}} = \frac{\det(\alpha A - bc)}{\alpha^{m-2}}$
stoic pythonBOT
gREEnOnions:
odd kite
looks okay to me but you left off the ' in c'
gaunt geyser
or I guess that simplifies to
$\frac{\det(A)\times (\alpha - bA^{-1}c)}{\alpha^{m-2}}$
stoic pythonBOT
gREEnOnions:
gaunt geyser
which is almost what I want
but not exactly
don't know where I diverged
but this is then valid
@odd kite
because
$\alpha - a'A^{-1}b$
stoic pythonBOT
gREEnOnions:
gaunt geyser
is a 1x1
daring solstice
Is subtracting a row by itself a valid elementary row operation?
daring solstice
ah ty for telling me
mystic night
can anyone give me a pointer how to start on b? i'm confused as to what the standard basis (to apply T on) would be?
dusky epoch
you should do the same thing you do when constructing the matrix of a transformation in a basis
mystic night
so i should construct the matrix by applying T on basis vectors of C?
dusky epoch
the basis vectors of B
mystic night
ah tyvm @dusky epoch, i'm just so clueless about complex numbers XD
dusky epoch
cos^2(t), sin^2(t) and cos(t)sin(t) are real-valued functions and so are their derivatives
the matrix will be entirely real
wintry steppe
in the cooling system of a car there is five liters of coolant. Coolant is water mixed with glycol. The glycol content is 15%. how much of the coolant liquid should be drained out and replaced with pure glycol for the liquid to withstand -30 degrees. round to one-tenth of a liter
in the coolant liquid you mix in glycol so that the pipes do not rust and that the liquid should not freeze in winter. if you have 50% glycol the liquid can withstand -30 degrees
wintry steppe
Oh sorry
gaunt geyser
I am trying to understand this proof
I do not understand how they are getting the original equality
stoic pythonBOT
gREEnOnions:
gaunt geyser
this is essentially saying the transpose is equal to the original matrix
dusky epoch
hmmm
it is indeed only an equality up to transpose in the general case
but that's okay since det(M^T) = det(M)
mighty saffron
@gaunt geyser just do the multiplication
The multiplicationw works out
unless im misunderstanding
gaunt geyser
when I multiply it out I get the matrices I posted in latex @mighty saffron, I think that this is what they start with, then they decompose it into the first equality in the proof, but I don't know where what I posted in latex is coming from
@dusky epoch not sure what you mean by "equality up to transpose in the general case"
mighty saffron
Surely you're computing (B I_n) . (I_m 0)
gaunt geyser
ah think I might of made an error in the multiplication
dusky epoch
assuming i didn't fuck up the matrix multiplication i did in my head
gaunt geyser
yeah
mighty saffron
im pretty sure it works out
gaunt geyser
yeah the LHS and RHS are equal, then they decompose it two different ways
thank you
vague fulcrum
can anyone help with 3d
i feel like i can prove it in a general form, idk why they have specified the inner product space
professor said that a complete proof requires us to use the defined inner product space
mighty saffron
Well what's your proof without it?
vague fulcrum
i use the bottom equality they give
<Hp,q> = <p,Hq>
since p and q are eigen vectors with distinct eigenvalues
x1<p,q> = x2 <p,q>
so
(x1 - x2) <p,q> = 0
x1 - x2 cant be 0
so <p,q> = 0
steady fiber
Ya that works
vague fulcrum
but why do they define the product space
mighty saffron
Are you allowed to conclude <p,kq> = k<p,q>?
vague fulcrum
i mean yea, that's a property of a product space
steady fiber
Guess you can use the definition of the inner product space to prove that if you really want
vague fulcrum
howd you use the definition
steady fiber
Just stick lambda1 p q in there and you can see that you can factor a constant out of an integral
It's not particularly interesting
mighty saffron
Yeah but you're told its an inner product space
I think the reason it's defined is because <Hp,q> = <p,Hq> doesn't necessarily hold on all inner product spaces
vague fulcrum
hmm yea
theres no logical flaw in what i said before right
i feel like what i wrote sufficiently answers the question
Also another small question
for part (c)
when they say find all eigenvectors of H
I find the eigenvectors with respect to the B matrix of H and then simply transform it back right
in other words the eigenvectors of H are polynomials
not 'vectors'
vague fulcrum
but like
-9 is an eigenvalue of it for example
so when i describe the eigenvectors related to it, i write them as polynomials right
cus the vectors in this example are in Pn, so theyre poly
mighty saffron
Yes
vague fulcrum
ok dope, ty
mighty saffron
wait how do you find an eigenvalue of -9?
never mind
im dumb
is presumed when they said n=3, it meant up to x^2
vague fulcrum
any clues? im stumped
idk how to use the spectral theorem in this case
how will using it show that an orthonormal basis exists
hollow finch
If a matrix $A$ has a set of eigenvalues $S={\lambda_1,\ldots,\lambda_n}$, will the matrix
$A'=ABCB^{-1}C^{-1}$ also have the same eigenvalues?
stoic pythonBOT
nix:
vital swallow
I don't believe so
sonic osprey
hm
vital swallow
unless B and C share eigenvectors
but I guess the way to answer that is to check the characteristic polynomial of A and A' and see if they are the same
sonic osprey
You should be able to come up with an example where the trace is different, and thus the eigenvalues can't be the same
main drum
hi I have a question
@sonic osprey
or anybody that can answer
its basic linear algebra
hello
anybody out there
@everyone
<@&286206848099549185>
pallid rampart
sacred quest
low rez 
sacred quest
Unfortunately we never really covered transpose matrices in my LA class, so I can't help you with this one
limpid spindle
You've gotta put in at least some effort to see where you stand in the question instead of begging for an answer @main drum
main drum
no i just need help understanding it
i put what i think the answer is based on what my textbook says
limpid spindle
I'm not a math major and am pretty trash at LA so I'm not the guy to answer this question
main drum
everybody keeps telling me "read the rules" and "do it yourself"
i came here though for an explanation
:9
😦
limpid spindle
Well why do you think the answer is true?
limpid spindle
What did your textbook say about it though
limpid spindle
Have you checked any other resources?
main drum
yea i search it up
nothing
i ask this channel
nothing
i ask other channel
nothing
quartz compass
you're not entitled to free help
limpid spindle
Email your TAs maybe?
That's always a sure-fire way to getting the answers you need. Assuming your in University it's kinda their job
main drum
this what i mean
merosity "fuck you"
zopherus" fuck you"
like what did i do to you people
im sorry for spamming
i already said i am new here
i dont know why people so mad
limpid spindle
He/she/they are right tho. You're not entitled to anything , they're not slaves and they shouldn't be treated as such
main drum
bro you blocked me when i asked for help
where was i entitled
in asking for help
do i need to say please
here is your please
lol
sonic osprey
Then ask your question, and wait patiently
instead of dm'ing ten different people asking for help
main drum
i was trying to finish my hw fast
limpid spindle
Do a little self reflection dude. This isn't the way
sonic osprey
Yeah it comes off as being selfish
limpid spindle
^^
limpid spindle
I saw u on the multi variable chat
main drum
and you help people usually
limpid spindle
It was very spammy
limpid spindle
Don't fucking spam everyone lool
People have shit to do bean. If you just wanna settle on an answer simply settle for it you have a 50/50 chance a true and false question. If you wanted to frame it as non-HW help you could've. You have TAs you can email but aren't doing it. I don't wanna come off as rude but you're being an entitled dick rn
limpid spindle
You could have said "hey how does this work" instead of "hey I have a HW question I want to solve and be done with so could someone please spoon feed me the answer despite me already having an answer" -- anyways I'm not math savvy and I'm wasting my time here. So have a good rest of your day genie man 👋
arctic hare
pale shell
I am decent i suppose
pale shell
Yes
main drum
not god stuff
pale shell
Well nxn matrix implies what, right so it is a square basically
main drum
yea
pale shell
So when we consider that it is symmetric
Right doesn’t that tell you that it is commutative?
main drum
yea
pale shell
Ohhhh ok
Well when you take the dot product, let me ask you,
Is that commutative or what?
pale shell
I agree
pale shell
Is it correction or what.
main drum
it was correct
cold topaz
here, how is x2 = x2, when everywhere in the system it is equal to 0?
https://www.slader.com/textbook/9781118879160-elementary-linear-algebra-applications-version-11th-edition/247/exercises/9b/
half ice
If x2 wasn't equal to x2 I'd worry
steady fiber
lol
cold topaz
thanks a lot! that helped extensively.
half ice
The important question is "how do we know not to write x1 = x1?"
x1 has a pivot, so we express it in terms of the variables without a pivot. x2 and x3 are without pivots, so we call them "free". We write x2 = x2 and x3 = x3 to show that they do have a place in the null space @cold topaz
cold topaz
so in this case, x2 and x3 have the same values; even though, one has the coefficient of -1, and the other one has 0?
real wedge
if W is a subspace then W = W + W right?
vital swallow
yes, since w = w+0
real wedge
ok thanks!
real wedge
would something like 1+x^2, 2x^2+5x, 4x^2+18x span all of P2?
if not, if I had a subspace of P2 that was just span(1+x^2, 2x^2+5x) would a basis for that just be those two vectors?
steady fiber
yes those 3 would span P2
n linearly independent vectors form a basis for a vector space of dimension n
real wedge
oh right so 1+x^2, 2x^2+5x, 4x^2+18x are linearly independent
so they form a basis bc dim of P2 is 3
steady fiber
yes
real wedge
Right, thanks!
flint flicker
The question here was to find the kernel of T and its dimension
I think dim(Ker(T)) should be 2 but I’m not sure
How would I show that?
Also is my geometric interpretation correct?
real wedge
Given x1 x2 in V and y1 y2 in W, does there always exist a linear map from V to W such that T(x1)=y1 and T(x2)=y2?
My answer would be no, right? I think it would be true if x1 x2 was a basis for V
As stated its possible that x1=x2 in which case that is clearly impossible ...
half ice
@flint flicker
You can give a basis for Ker(T), of course! Let's say v2 and v3 are free. Then,
Ker(T) = v2(-b/a, 1, 0) + v3(-c/a, 0, 1)
Yes that geometric interpretation is correct. You can also think of Ker(T) as the space of all vectors orthogonal to u.
fathom flax
I don't understand the method I'm given for determinants. It's the Laplace expansion. Can someone explain?
gritty sorrel
where can I read about where the motivation behind the formula of the determinant comes from instead of just accepting it as a 'definition' (thats a weird definition to give)?
limber sierra
what type of motivation are you looking for? an intuitive one? a pure mathsy one?
a geometric one?
gritty sorrel
well i think understanding the geometric properties of determinants is a consequence of accepting the definition
so maybe i'd like a 'pure mathsy' one
limber sierra
the determinant is the unique alternating multilinear operator satisfying det(I) = 1 [where I is the identity matrix]
that's the pure math definition
let me explain what it means a bit
when we talk about the determinant, we have to ask: what information do we want this function to actually encode?
well, we want the determinant in some way to translate information about matrices into real number information
one convenient way to do that is the multiplication formula, det(AB) = det(A)det(B)
why is this convenient? well, it gives us a few interesting pieces of information
for example, how would we use determinants to check for invertibility?
well, the only real number without a multiplicative inverse is 0
so, by the multiplication formula, if a determinant is 0, the matrix isnt invertible
cool!
in order to do this, we need the determinant to evaluate to 0 if the rows of the matrix aren't linearly independent
[since then the matrix's RREF form isn't the identity matrix]
this is where the term "alternating" comes in
"alternating" in this context means exactly that: if one of the rows is a lin. comb. of the others
it evaluates to 0
the "satisfying det(I) = 1" part also comes for free from the multiplication formula
since obviously det(AI) = det(A)
but also det(AI) = det(A)det(I)
so we need det(I) = 1
finally, the thorniest word of this definition: "multilinear"
this just means that the determinant is a linear function in each of the rows of the matrix
i'm assuming you know what a linear function is
but this is important, since when we look at the multiplication rule
det(AB) = det(A)det(B)
gritty sorrel
linear function in each row vector?
limber sierra
yes, essentially
since matrix products AB are products over all the rows in A and the columns in B
we actually need a map that is linear over each row (and column)
we actually need a map that is linear over each row (and column)
i'm cutting out some of the formulaic justifications here
but this should make intuitive sense; if you multiply one row by a constant, it should multiply the determinant by that constant, since if you multiply one row by a constant and then multiply matrices
the formula for the resulting matrix has a bunch of constant multiples thrown in
again, cutting out some of the algebraic justifications, if you want to convince yourself you can do the expansion
but basically, this motivates the definition "alternating multilinear operator satisfying det(I) = 1"
it lets us encode information about the multiplicative structure of matrices as the multiplicative structure of real numbers
we translate properties from matrix multiplication
into properties of real number multiplication
that's handy
there's one problem: how do we know that the determinant is the best way to do this?
well, as it turns out, the determinant is the only way to do this
this is a theorem often called "the determinant is unique"
the proof isnt long but im not gonna state it here
but what this means is, any definition that defines an operator satsifying these properties