halcyon spindle
From your notes.
#linear-algebra
austere willow
the course I'm watching wasnt that helpful and they just blasted me with these questions 😭
explained like 2 rules
vague crane
you should read a textbook on it!
gilbert strangs Linear Algebra and its Applications is online for free and from what i've heard it's very good
austere willow
alr ill check that out
austere willow
gilbert strangs Linear Algebra and its Applications is online for free and from ...
you know any free calc/trig books?
wintry steppe
Well yes u is projected on v
vague crane
you know any free calc/trig books?
uhhhh one sec
wintry steppe
And there's a nice formula for that too, to get both vector and scalar projection since the vector is projected on a vector (think of it as a line), a 1d space
robust flicker
And there's a nice formula for that too, to get both vector and scalar projectio...
do you have any recommendations to understand better the geometry of linear algebra?? i struggle a lot with visualizing it and ive heard its crucial to understand
wintry steppe
Have you watch 3 blue 1 brown vids on essence of Linear algebra
vague crane
<@&268886789983436800>
wintry steppe
On essence of Linear alg
robust flicker
Have you watch 3 blue 1 brown vids on essence of Linear algebra
i havent
wintry steppe
Go watch them. They're extremely helpful
robust flicker
is it on youtube??
wintry steppe
Yea
robust flicker
alright tysm!!
loud halo
Not gonna mute/ban for this ofc but please keep in mind that recommending websites for piracy is against Discord policies.
loud halo
We just wanna be careful with that
frosty vapor
no wories
loud halo
Ty
robust flicker
ofc ofc
wintry steppe
The projection is kv here btw, not u-kv
robust flicker
ohh do u happen to know why the u is added??
i genuinely dont understand why the solution is the way it is
wintry steppe
Added?
vague crane
you know any free calc/trig books?
all legal free math resources
also there's mit ocw
check that out too
wintry steppe
Because if you see, both u and the vector perpendicular to v add up together to form projection kv
It's just addition of two vectors
wintry steppe
Welcome
robust flicker
i struggled sm understading the solution lmaoo
wintry steppe
Yesterday, I was practicing projections from 3 dimensions to 2 dimensions. I've always noticed this that, whenever we project the orthogonal basis vectors (of the space on which the projection will happen) on the vector, we can write where they end after projection in a column matrix respectively. And the cool fact is that, for any other vector, we can multiply it by the transpose of that matrix to actually get the projected vector in 2 dimensions. The same happened in the projection from 2d to 1d too
Is this possible for any projection from n to n-1,n-2,... 2, and 1 dimensions? It seems like the transpose of the projection matrix is a m×n matrix where m is the dimension we are projecting on
icy totem
??
lavish jewel
you could use a cofactor expansion to find the determinant, too
whichever you prefer
icy totem
ı did ref but two rows are same . So det must be 0 and it is not invertible .
lavish jewel
if your arithmetic is right, it could very well be the case that no value of k makes it invertible
icy totem
ı think right but if you want check it
odd sparrow
How many people here know the expansion by minors? Not the regular one by elements of a row or column, I mean by minors of order 2 and larger. I think Friedberg's didn't have it but I read about it in Shilov's
dusky epoch
,w det {{-1,k,3,-2},{2,1,1,k},{1,k+1,4,k-2},{2,1,1,k+1}}
stoic pythonBOT
dusky epoch
@icy totem it looks like you are correct
wintry steppe
$M_{12}$ must be
$\vline{\begin{matrix}a_{21}&a_{23}\a_{31}&a_{33}\end{matrix}{\vline}}$ right
stoic pythonBOT
Harry127
lavish jewel
sounds right
wintry steppe
Thank you
dusky epoch
vmatrix btw
limber umbra
how do you solve this equation to get the nullspace?
lavish jewel
observe that you can free pick 2 parameters
spare widget
x1 - x2 - x3 = 0 -> x1 = x2 + x3 -> (x2+x3, x2, x3)
lavish jewel
as criver has written it, x2 and x3 are free parameters. you can use these to find x1 in terms of x2 and x3, and then see if you can rewrite this in a nice way that makes the dimension of the null space clear
wintry steppe
by 16 years, 99% of patients have at least a 2-year period of remission
anyone know how i can work this out step by step
cos clearly im fucking up somwhers
99% have(2/16) = 99% x(1/8)=99%x12.5%
dusk olive
If V is a finite dimensional vector space over C, can I view it as a subspace of gl_n(C) for some n ?
native rampart
The problem is scalar multiplication
dusk olive
Could you elaborate?
native rampart
I tried doing a c_1e_1+c_2e_2...
to e^{c_1}E_11+e^{c_2}E_22... Where E_ij is matrix with only ij entry as non zero and 1
Idea being if a_1 x maps to e^a_1 E_11 and a_2 x maps to e^a2 E_11 you can map (a_1+a_2) x to e^(a_1+a_2) E_11
Something like that
Actually,This won't work at all ig
wintry steppe
If you treat GL_n(C) as a vector space under addition and scalar multiplication only, then it is isomorphic to C^(n^2). Then V is isomorphic to C^m which is then isomorphic to some subspace of this
Otherwise, GL_n(C) generally has a ring structure, whereas V has a structure of a vector space, so V won't be isomorphic to some subring of GL_n(C)
dusky epoch
If you treat GL\_n(C) as a vector space under addition and scalar multiplication...
GL_n(C) is not a vector space at all 
wintry steppe
Oh right. I forgot that GL_n isn't the set of all matrices. It's the group of invertible matrices
dusk olive
Well I wrote gl_n and not GL_n, and gl_n is in fact a vector space
wintry steppe
can someone explain why rotation by an angle theta is one to one
If you see, considering a transformation transforms the vectors in the space by rotating them from the Origin by the same angle theta, there is always only one resulting vector after the transformation for each vector. It cannot at all happen that two vectors end up in the same place after rotating by the same angle theta. Each vector in the given vector space is unique, and transforms (rotates) resulting another unique vector. Hence it's one to one or injective
molten pilot
can someone explain why rotation by an angle theta is one to one
coz it's invertible
namely, rotating by -theta
dusk olive
Another way to think about it, there is no non-zero vector that you can rotate such that it becomes the zero vector,
wintry steppe
Yes to prove an injectivity there has to be a trivial kernel only. In this case, it holds
is there a python script or something for changing basis of a linear operator?
quartz compass
yeah it's called matrix multiplication 😎
stoic pythonBOT
bim
native gust
Does anyone know how to algebraically calculate a quadratic regression for a set of 5 points?
quartz compass
V is finite dimensional right?
depends on what V is
or are you asking for clarification on the question before you 😵💫 lol
native gust
Does anyone know how to algebraically calculate a quadratic regression for a set...
Like what is the formula?
grave kettle
why are they right?
wintry steppe
(A) seems right to me because the dimensions of the basis is always equal to the dimensions of the vector space
grim leaf
is there a name for this notation?
grave kettle
~~(A) seems right to me because the dimensions of the basis is always equal to t...
um
consider the case of 2 linearly independent vectors in r^3
does that span r^2?
to begin with, the basis vectors of r^2 have to be in r^2, which are 2-tuples
wintry steppe
I think no because 3 tuples cannot span r^2 
wintry steppe
Ah I see now
grave kettle
here, v_1 v_2 v_3 are 5-tuples
wintry steppe
So 5 tuples cannot span r^3
grave kettle
So 5 tuples cannot span r^3
r^3 and r^2 are different worlds
wintry steppe
Ty for giving out this vital info 👍
wintry steppe
r^3 and r^2 are different worlds
Yes
celest ore
ah I see
For this problem, since b is not in C(A) or the linear combinations of A, could I then translate the problem into- find A and b such that Ax=b has no solution? In that case, setting equivalence between a row of zeros in A and some number in b would work I think
teal grotto
it’s easier than this @celest ore
take either one or two vectors in R^3. for ease, make sure they are lin. ind. put them as the first two columns of a matrix. they will span at most a 2 dimensional subspace of R^3 whose span is all linear combinations of the one or two vectors you chose.
then just find something not in that span
celest ore
take either one or two vectors in R^3. for ease, make sure they are lin. ind. pu...
I'm slightly confused. 2 vectors in R^3 is able to span a 2 dimensional subspace? Is this because a basis with 2 vectors have a dimension of 2?
grim leaf
“Let $K$ be a commutative ring w/ identity and let $n \in \mathbb Z^+$. Then there exists at least one determinant function on $K^{n \times n}$.”
“There is exactly one determinant function on $n \times n$ matrices over $K$”
stoic pythonBOT
totally not anamono
grim leaf
are these two statements not the same?
only difference is the first one says “at least one” but the second one says “exactly one”, but since the second one says “exactly one” then the first one could also only have exactly one, no?
celest ore
then just find something not in that span
So intuitively I think what you're saying is have 2 columns of A construct a plane in R^3 and set b to be any vector outside of the plane?
teal grotto
So intuitively I think what you're saying is have 2 columns of A construct a pla...
yes, exactly
celest ore
I know that A would be linearly dependent for a nonzero vector b to be in null(A). How would I go from there?
carmine galleon
What if you take A to be the 3x3 matrix all of whose entries are 1?
Can you think up a vector b that will satisfy Ab=0?
icy totem
?
quartz compass
?
since X is arbitrary, X-A is arbitrary, so you can focus on showing the right hand side is not the square of a matrix
icy totem
thanks
wintry steppe
i am trying to graph this problem in R
but im a bit confused on x_k and x0 and how they relate to the graph
grave kettle
how do i show that two row-equivalent matrices in row echelon have the same pivot columns?
wintry steppe
are these two statements not the same?
The forward statement says that the determinant, if it exists, is unique. Thus, you just need to find a determinant to show that you've found the determinant
coarse ember
Hello, can someone explain why the geometric product is distributive?
fervent kernel
only difference is the first one says “at least one” but the second one says “ex...
Right, the first statement just claims existence. The second statement claims existence and uniqueness.
wintry steppe
need help
hoary void
fringe fjord
Apparently you're supposed to already know that the determinant is a linear function of each row, when the other rows are kept constant.
grim leaf
oh right yeah forgot about that
but i still dont get why alpha_j occurs twice then
is it a typo? something like alpha_1, ..., alpha_i, ... alpha_j, ... alpha_n
fringe fjord
No, (a1,...,ai,...,aj,...,an) is just A.
But you have B = (..., ai+c·aj, ..., aj, ...).
Then by linearity, det B = det(..., ai, ..., aj, ...) + det(..., c·aj, ..., aj, ...) = det A + c·det(..., aj, ..., aj, ...)
coarse ember
Hello everyone, is anyone here familiar with Clifford Algebra? I am struggling to find out why you can distribute a geometric product if there are vectors of different grades inside the parenthesis. For vectors of the same grade within the parenthesis, geometrically, it makes sense. When they are of different grades, it is hard to understand why you can distribute. If I understand this then I can prove why the geometric product is associative (and I would like to do that)
Is it that because they are of different grades, and geometric algebra tries to describe the real world, so in the end, when you equate vectors of the same grade, it is like the the other vector of different grade never existed? For example, (a+bc)d is equivalent to (bc)d when equating trivectors? Does this make it okay to distribute? To me, my justification seems weak.
grim leaf
is there a name for this notation?
gray dust
is there a name for this notation?
the resulting matrix or the notation itself?
grim leaf
the resulting matrix or the notation itself?
the notation itself
but knowing the name of the resulting matrix would be nice too i guess
grim leaf
cool thanks
gray dust
np
vague dock
Hey guys, anyone want to take a stab at this?
wintry steppe
Hey guys, anyone want to take a stab at this?
The image on the projective plane is obtained by taking the line from the point to the eye and seeing where it intersects on the plane
As for why it has those properties, I can't really tell
vague dock
Yhea, it's confusing. I'm still wrapping my head where that cusp is generate from.
wintry steppe
Suppose φ and ψ are linear endomorphisms of the vector space V . If v ∈ V is an eigenvector of φ and ψ then v is an eigenvector of ψ ◦ φ. Can someone help me to identify if that true or false?
lavish jewel
apply the definition of composition and eigenvector
wintry steppe
apply the definition of composition and eigenvector
If φ(v)=λv, and ψ(v)=λv,then ψ(φ)=λv
lavish jewel
no, you'd have to apply ψ ◦ φ to v
ψ ◦ φ is another function, not a vector
also v, not V
wintry steppe
ψ ◦ φ is another function, not a vector
ψ ◦ φ (v) = ψ(φ(v)) = λ*(φ(v))
lavish jewel
yeah, now apply the definition again
wintry steppe
ψ ◦ φ (v) = ψ(φ(v)) = λ*(φ(v)) = λ * (λ*v) = λ^2 * v. So indeed v is an eigenvector of our composition
lavish jewel
close
except you don't know that the two endomorphisms have the same eigenvalue
only that it's the same eigenvector
wintry steppe
sorry I got confused I mean I care to show that the eigenvectors are the same not also the eigenvalues, or you mean bcs the eigenvalues are diff the eigenvectors will be diff?
lavish jewel
i mean that having the same eigenvector does not mean they have the same eigenvalue for that eigenvector
wintry steppe
yes u mean I should not use lambda for eigenvalues in all my statements
lavish jewel
the letter doesn't matter
just don't use the same one for both
look at what rokabe suggested
gray dust
sorry I got confused I mean I care to show that the eigenvectors are the same no...
all we know is v is an eigenvector of both maps. v doesnt necessarily have the same eigenvalue for each
wintry steppe
okii last thing the composition will have eigenvalues as a and b for example if I use Rokabe equations right?
gray dust
eg take I and 2I (I=identity), take any v
v is an eigenvector of both but corresponding eigenvalues are 1 & 2
wintry steppe
yes yes I got it
Statement: If φ(v)=av, and ψ(v)=bv, then ψ(φ)=cv for some a,b,c
ψ ◦ φ (v) = ψ(φ(v)) = b*(φ(v)) = b * (a*v) = a * b * v for some a,b
i used a in my composition since we have φ(v)
thank @lavish jewel
thank @gray dust
gray dust
Statement: If φ(v)=av, and ψ(v)=bv, then ψ(φ)=cv for some a,b,c
ψ ◦ φ (v) = ψ(φ...
i dont understand this, also i dont see the point of using d
φv=av, ψv=bv
just directly use these when composing, the eigenvalue for ψ ◦ φ comes out in terms of a,b
wintry steppe
> φv=av, ψv=bv
just directly use these when composing, the eigenvalue for ψ ◦ φ ...
okii fixed it above. But how do I justify its the same eigenvector? I mean its straightforward but like is it just enough?
gray dust
okii fixed it above. But how do I justify its the same eigenvector? I mean its s...
wdym same eigenvector?
lavish jewel
you just did
gray dust
all we're showing is v is an eigenvector of ψ ◦ φ
lavish jewel
they TELL you that the functions have the same eigenvector
lavish jewel
all you did is apply the definition twice, which nicely enough resulted in the same eigenvec (by using the same definition a third time)
gray dust
(ψ ◦ φ)v=abv
wintry steppe
okay thanks a lot again I just got confused with the eigenvals
versed topaz
For fixed n, m, is there an integer polynomial p in variables x_{i, j} for 1 <= i <= n and 1 <= j <= m and y_j for 1 <= j <= m such that for every field k, matrix A : k^n -> k^m, and vector w in k^m we have w in im T iff p = 0 when we set x_{i, j} = A_{i, j} and y_j = jth coeff of w
I want to say something like look at whether the rank of A drops if we add w as a column
But the polynomial can only depend on the numbers n, m and the rank of A can vary
teal grotto
polynomial over what structure?
versed topaz
Integers
(like how the determinant is an integer polynomial in the x_{i, j})
Like basically I want to measure this in terms of a determinant somehow
Also maybe I want to look at more than one polynomial? Idk
versed topaz
So like, suppose we're given A, w and let B be the matrix where we add w as an extra column to A. Then w is in the image = column space of A iff B has the same rank as A iff for all r <= min(n, m) we have rank(A) <= r implies rank(B) <= r
ie for all r we have rank(A) > r or rank(B) <= r. But rank(A) > r is a nonvanishing condition in the entries of A, not a vanishing one
Edit: Ignore this, I needed more commutative algebra than I realized
robust flicker
how do i find the values of a vector given the volume of a parallelepiped and the value of two other vectors
this is what i have so far
however im stuck after doing cross product
dusky epoch
,rccw
stoic pythonBOT
dusky epoch
@robust flicker can you show the original problem
robust flicker
sure gimme a sec
its the one highlighted lemme translate it
Let 𝒖=(1,2,1) and 𝒗=(0,1,0). Find all 𝒘=(𝑎,𝑏,𝑐) in R3 such that the volume of the parallelepiped generated by 𝒖, 𝒗 and 𝒘 is equal to 2. Describe in words and make a drawing of the locus formed by such points 𝒘.
dusky epoch
okay, so i see you have distilled the condition on w down to a + c = 2 (after the redundant parentheses are removed)
and indeed this condition is equivalent to the volume of the parallelepiped being equal to 2
robust flicker
so i could say vector w=<1,0,1> ??
dusky epoch
no
...well ok like
that's one possible value of w
but not the only one
the equation a + c = 2 in fact describes an entire plane
dusky epoch
i mean yeah of course we're doing this all in R^3
but think about it geometrically
moving the tip of w parallel to u or to v only shifts the top base of the parallelepiped parallel to the bottom base
but the volume is left unchanged
robust flicker
ohhhh i think i understand
so when it comes to drawing the parallelepiped is the only relevant value in vector w "b"
grave kettle
i struggle to understand the relation of pivot columns and freedom of variables
lavish jewel
if you have already seen rank and null space, you can relate the number of pivot columns to the size of the dimension of the null space
grave kettle
hm so the idea of pivot is related to linear transformations too
lavish jewel
matrices do represent linear transformations in a particular basis
grave kettle
strange question
all linear transforms are matrix transforms?
but some vector spaces have vectors that aren’t defined with components
how do we incorporate a matrix
lavish jewel
in finite dimensions, yes
you make a correspondence between the vectors and their coordinates when you pick a basis
grave kettle
you make a correspondence between the vectors and their coordinates when you pic...
but sometimes vectors don’t have coordinates?
idk, vector space of colours
dusky epoch
if you have already seen rank and null space, you can relate the number of pivot...
the size of the dimension? 
grave kettle
with some notion on addition/multiplication
dusky epoch
when you pick a basis
lavish jewel
that was a typo, sorry. i meant dimension but had typed size first
and didn'T delete it
when you pick a basis, you can then make a vector of coordinates
using simply the definition of linear combinations
grave kettle
so the entries will be however the colours are represented
lavish jewel
for example if you say that orange is red plus yellow, with whatever plus means here
you can relate this to some coordinate [1,1], for example
wintry steppe
So like mapping colors in the form of coordinate vectors representation?
spare widget
I think he means infinite dimensional vector spaces, e.g. the colours can just be represented through wavelengths
grave kettle
?_?
lavish jewel
chromium never mentioned dimensions
spare widget
are you asking about an analogue to matrices for infinite dimensional spaces?
lavish jewel
no
spare widget
but some vector spaces have vectors that aren’t defined with components
e.g. this
grave kettle
how do we incorporate a matrix
question
lavish jewel
the question was the relationship between linear transformations and matrices
and coordinate vectors when the original vector space does not have "vectors" in the R^n sense
spare widget
But it's still isomorphic to F^n if it's finite dimensional
lavish jewel
that was precisely the point, yeah
spare widget
So has components
grave kettle
ok
grave kettle
hm so the idea of pivot is related to linear transformations too
back here lol
spare widget
<:thonk:406575732563902485> ok
e.g. polynomials: a x^2 + b x + c, the components are (a,b,c)
lavish jewel
pivot columns are related to the rank and dim of the kernel of the linear transformation the matrix represents
spare widget
colours as RGB are R^3 vectors, while colours represented through wavelength are functions
lavish jewel
those are invariant regardless of the basis you pick
spare widget
What would be a proper way to prove that AV is of rank r, where A is of rank r, and the columns of V form a basis for R(A^T)? My idea was that since R(A^T) is orthogonal to N(A) then the kernel of the composition doesn't increase, but it seems a bit too informal.
lavish jewel
what is R
gray dust
range likely
spare widget
range yes
lavish jewel
all right, so they form a basis for the row space
this already tells you A and V have the same rank
you can pair that with your argument. you have r linearly independent columns in v which are also orthogonal to the kernel of A
spare widget
Also is this the only case where rank(AB) = rank(A) = rank(B)? As the general result is rank(AB) <= min(rank(A), rank(B)).
That is does it follow from rank(AB) = rank(A) = rank(B) that R(B) = R(A^T)
or are there other possibilities when this holds
lavish jewel
you can try to check
perhaps by observing that the spanning set for the image of AB is the same as for the image of A, and rewriting AB in some way as linear combinations of the columns of A
and checking what is needed for the rank of A to be the same as that of AB
spare widget
AB = [Ab1 | Ab2 | ... | Abn] so I do get a linear combination of the columns of A
celest ore
I don't really understand why (a) is false. Isn't the given basis containing 3 R^4 vectors meaning it's a 3 dimensional subspace that lies in R^4?
lavish jewel
that's the basis for one 3 dimensional subspace, but not necessarily S
spare widget
S can be any hyper-plane in R^4 not necessarily the one orthogonal to (0,0,0,1), in which case the above vectors would not be a basis for it.
gleaming knot
Hello everyone, is anyone here familiar with Clifford Algebra? I am struggling t...
The simplest way to prove this is to use the definition of the Clifford algebra as the quotient of the tensor algebra by the ideal generated by v \otimes v - |v|^2 (in your notation). Geometric product is just tensor product in the original tensor algebra, which distributes across any type of addition by construction
Also, tensor product is associative so that automatically proves geometric product is associative
coarse ember
Hmm, I am not familiar with tensor algebra, unfortunately. Is there any other way of proving this just by algebraic manipulation? If I can say what the product of a vector with a bivectors is and what the product of a vector with another vector is, can I prove what (a + bc)d is? I find myself questioning what distribution really is, especially in the case of the geometric product where it was introduced to me as a function of two 1-vectors, not a function taking the sum of multiple vectors.
wintry glen
Can I just check if this is correct?
Since Ax = b has at most one solution for all b, in particular, Ax = 0 has at most one solution. Since the trivial solution is one solution, it is the only solution. Hence, by the definition of the matrix-vector product Ax, the only linear combination of the columns that yields the zero vector is the trivial one. By definition, the columns of A are linearly independent.
spare widget
Suppose that the columns of A are linearly dependent
Then there exists y !=0 such that Ay = 0
But then Ax = A(x+y) = b
and since y!=0, the x+y != x
And both x and (x+y) are solutions
But this contradicts the assumption that x is the only solution
However in the above they say at most one - which means that there may be 0 solutions, so this is a second case you have to look into
I think that the 0 solution case is not true. Consider 2 3d vectors spanning a plane, and pick another vector on that plane. Then the 3 vectors are linearly dependent and you can use them as columns of A. Now pick b to a point outsude of that plane, then the system has no solutions but the columns are linearly dependent.
Actually the say for all b, I missed that
It should be fine then I guess
celest ore
dumb question, but how does a single column vector be linearly independent?
lavish jewel
use the definition of linear independence
spare widget
a * v = 0 only if a=0 or v = 0
lavish jewel
for a set of vectors to be linearly dependent, you need coefficients not all 0
spare widget
v being the vector, a some constant
This actually means that the zero vector is linearly dependent by itself?
lavish jewel
if it's the 0 vector, it's always linearly dependent
yep
if it's not the 0 vector, then it's immediately lin indep cuz the coefficient must be 0
celest ore
Ah ok
spare widget
it is a sphere
one centered at (1,-2,-4) with radius sqrt(k)
the canonic form of a sphere is
$|\vec{p}-\vec{c}|^2 = R^2$
stoic pythonBOT
criver
spare widget
where R is the radius, c is the center, and p is a point on the (hyper)sphere
tame pond
That question doesn't seem doable to me with the given information, or the problem is at least poorly written? By Caley-Hamilton, any matrix A with dimension greater than 2 with that quadratic as a factor of its characteristic equation is going to satisfy the given equation so you won't have enough information to get all the eigenvalues.
zinc timber
That question doesn't seem doable to me with the given information, or the probl...
are you sure about it?
tame pond
I already gave my opinion and my reason why I hold that opinion. Answering if I am sure doesn't add anything
zinc timber
have you heard of minimal polynomial?
lavish jewel
it's pretty late for me so i might be mistaken, but couldn't one also observe that -(A + 2I) is the inverse of A, and what the eigenvalues of A^-1 are when compared to A, along with what the eigenvalues of A + 2I are
tame pond
Part of my point is that A could have eigenvalues -1,-1, 0 so A wouldn't have an inverse in that case.
tame pond
I don't know, it just seems like an oddly broad problem to allow A to be of arbitrary size
lavish jewel
it's enough for AB = I_n to say A is invertible if A and B are square
and since A^2 exists, A is square
tame pond
A being square was explicitly written
lavish jewel
so rearranging -A(A + 2I) = I makes -A - 2I the inverse of A
tame pond
Anyway I have a boring zoom meeting now
zinc timber
(A+I)²=0
only eigen value =-1
mini poly roots blah blah blah
something something annihilators
lavish jewel
i see
zinc timber
but yeah, -1
zinc timber
lavish jewel
wait
it was addition, not mult
i found it
yes, -1
ok, i can sleep well now
-(lambda_i + 2) = 1/lambda_i
zinc timber
but anyway
lavish jewel
then (A + 2I) is the neg inverse
zinc timber
ohh
lavish jewel
or does that not work
zinc timber
ye
lavish jewel
ye?
lavish jewel
i know nothing about polys
zinc timber
mult my A' and rearrange
lavish jewel
was just arguing that wasn't even needed here 😛
zinc timber
ig
molten pilot
how do we know that d linearly independent vectors of some vector space with dimension d, span that vector space?
teal grotto
if they don’t span, then there is a vector v not in their span
union this vector to the collection of d lin ind vectors
this will be a collection of d+1 lin ind vectors in a d dimensional space
show that this collection can’t be linearly independent because of the existence of a basis
viral olive
Jup that sums it up quite nicely
teal grotto
how do we know that d linearly independent vectors of some vector space with dim...
another way i think might work would be to define a linear transformation T on your vector space so that if u = c1v1 + … + cdvd, then T(u) = c1w1 + … + cdwd
where v1,…,vd is your basis and w1,…,wd is your set of d lin independent vectors.
show that this is an isomorphism of V. should be enough to conclude that w1,…,wd is spanning
lapis wedge
umm
i have a math question but im not sure if its about linear algebra
can i still ask it?
languid sphinx
why not ask it first
lapis wedge
just dont want to get in trouble with the mods
languid sphinx
lapis wedge
M=13
languid sphinx
Not really #linear-algebra
lapis wedge
yea ik
languid sphinx
But I don't really know where
languid sphinx
#discrete-math is a lot less populated sadly
molten pilot
another way i think might work would be to define a linear transformation T on y...
don't d linearly independent vectors in a vector space with dimension d form a basis anyways?
teal grotto
i thought that’s what u we’re trying to show
viral olive
don't d linearly independent vectors in a vector space with dimension d form a b...
Well whats the definition of a basis?
molten pilot
wait
I'm having such a hard time knowing which result came first
it's all jumbled up in my head
viral olive
okay which results are there in your head?
molten pilot
ok, so I know a basis is supposed to span it's vector space
viral olive
Thats correct
So how can you tell that a subset of a vector space is a basis of the same vectorspace?
molten pilot
and the number of vectors in a basis is defined to be the dimension of the vector space
viral olive
Thats also correct
molten pilot
ok so how do we know the dimension of a vector space is well defined?
viral olive
well the only way to disprove the dimension of a vector space would be that you have a vector space that has two different dimensions
so let V be a vector space
Dim(V)=a
Dim(V)=b
a does not equal b
gray dust
ok so how do we know the dimension of a vector space is well defined?
one proof of this uses the fact that independent sets arent longer than spanning sets
viral olive
how would you proof that
teal grotto
a set of vectors is a basis iff it’s minimally spanning iff it’s maximally lin. ind.
viral olive
Yeah correct thats one way.
gray dust
theres an iterative argument that shows a spanning set cant be exhausted faster than an independent set
gray dust
yes lin indep
viral olive
correct
teal grotto
a collection of vectors such that any non-empty finite subset of them is linearly independent
viral olive
just out of curiosity its not given that its finite in normal case right?
teal grotto
that what’s finite?
viral olive
an independent set.
teal grotto
no
viral olive
so no its not given or no it is given?
teal grotto
no it’s not given lol
stiff vigil
What is a Schur complement in inverting matrices?
viral olive
you mean whats the inverse of a Matrix that has its own schur complement or whats exactly the question
stiff vigil
What is a Schur complement, basically?
viral olive
So M is a quadratic matrix which can be split in 4 smaller Matrix
Let M be a (n+m)x(n+m) Matrix
and A be for example nxn , B nxm , C mxn, D mxm
The Schur Complement is that matrix that comes from the Formular D - Cx(A)^-1 xB
gray dust
ok so how do we know the dimension of a vector space is well defined?
independent sets arent longer than spanning sets
we use this to conclude dimension is well defined. say we have two bases of size n,m. using the above gives n=<m and m=<n, thus n=m
stiff vigil
The Schur Complement is that matrix that comes from the Formular D - Cx(A)^-1 x...
And that would be the Schur complement of M
viral olive
Correct
tribal willow
is there an equivalent of summation notation for direct sum?
plush canyon
Can someone pls show me how to do 2
I have no idea where to start or what to do
That's all I need and then I can hand in my hw
I've never seen anything like that. I normally calculate evalue and e vectors with a matrix
tribal willow
can someone explain what a module is and why it's significant to me as if i were 5
eli5 of exterior product would be dope too
tough veldt
vector spaces are over a field
modules are exactly the same axioms except wherever 'field' is mentioned, read it as 'ring' (with 1)
Hence all vector spaces are modules.
tribal willow
so then what's the significance of using "ring" as opposed to "field"?
tough veldt
u do or dont know what a ring is
theres not much point considering modules if you dont know what a ring is
tribal willow
i get that a ring is a generalization of a field
so then is a module a generalization of a vector space?
tough veldt
Hence all vector spaces are modules.
well yes.
tribal willow
interesting
tender ledge
can you do RREF on a non-augmented matrixc?
vocal bone
Hi ,shouldn't the equation of the plane for this question be 5x-y-z=3? Why do the solutions give 5x-y-z=7? Thanks in advance!
fringe fjord
can you do RREF on a non-augmented matrixc?
Gaussian elimination can be used to convert any matrix to RREF by row operations. Whether the matrix is "augmented" or not is a question of how you interpret the values in the matrix, not an inherent property of it.
tender ledge
ty
fringe fjord
Hi ,shouldn't the equation of the plane for this question be 5x-y-z=3? Why do th...
5·1-(-1)-(-1) = 7
vocal bone
5·1-(-1)-(-1) = 7
can u expand on this pls?
dusky epoch
what is there to expand on?
the value of 5x - y - z at (x,y,z) = (1,-1,-1) is 5 * 1 - (-1) - (-1), which equals 7.
vocal bone
ohh ok
I didn't know where Trophosphere had gotten the numbers from
so I was like completely lost
placid shadow
Guys
may I have help
plz
So if 'w' belongs in the matrix H
does that mean it's a linear combination of v1,v2,v3
so do I test for a determinnant of zero
dusky epoch
H is not a matrix
but yes, $\bd{w} \in H$ iff $\det[\bd{v}_1, \bd{v}_2, \bd{v}_3, \bd{w}]=0$
stoic pythonBOT
Ann
vernal yew
Big test tomorrow, anyone on here willing to chat in voice with me about some topics?
Its on Linear Independence, Dot Product, Range, Nullspace, Rank-Nullity theorem, Dimensions, Orthogonal/orthonormal, projections, and linear transformations
lavish jewel
i can discuss stuff here, but not on vc
vernal yew
okay, may I dm you @lavish jewel
lavish jewel
let's just do it here, this is the type of stuff the channel is for
vernal yew
i get dimension of range and nullspace but what about like other bases?
lavish jewel
they behave much in the same way
vernal yew
example?
lavish jewel
the key observation is that, when you speak of the range of a matrix, what you are doing is treating the columns of the matrix as a spanning set
so in general, the only difference is that the vectors don't need to come from the columns of a matrix
or tell me if i misunderstood your question
vernal yew
where do they come from then? theyre just given to you or?
lavish jewel
yeah
you could really be given a set of arbitrary vectors
and just ask you what would be the dimension of the vector space they span
vernal yew
oh so like, given this span in whatever dimension, is it a basis? and then the number of linearly independent vectors in the basis is the dimension of the span??
lavish jewel
your wording is a little shaky, but that is the main idea
a basis ONLY contains linearly independent vectors
a spanning set can contain linearly dependent vectors
vernal yew
as i thought. so remove linear dependent vectors and then the number of vectors remaining is the dimension?
lavish jewel
so here it is correct to say "the number of linearly independent vectors in the SPANNING SET is the dimension of the subspace"
yes
and once you do that, the spanning set is called a basis
vernal yew
why say spanning set if the spanning set is a basis once said vectors are removed?
lavish jewel
it's the classical quadrilateral and square situation
all squares are quadrilaterals
vernal yew
right right
lavish jewel
not all quadrilaterals are squares
so a basis is a spanning set
but in general, a spanning set is not a basis because it can be linearly dependent
lavish jewel
mhm
vernal yew
alright, next question. what in the heckin heck is a subspace? intuitively not proof because i know its closed under addition and scalar multiplication but that doesnt intuitively mean anything to me
lavish jewel
say you have a vector space
vernal yew
so a span of some arbitrary vectors?
vernal yew
so say span{u, v, w}
lavish jewel
for example, consider R^3 spanned by [1,0,0], [0,1,0], [0,0,1]
vernal yew
then span {v, w} is a subspace?
lavish jewel
sure, yes
vernal yew
if it meets closed under addition and scalar
lavish jewel
right
limber sierra
intuitively it's a 'part of' a vector space that acts like a vector space all on its own
like a plane in 3d space
lavish jewel
the idea is that "you cannot escape it" and it satisfies all the properties of a vector space
vernal yew
Namington, welcome, can you join voice?
limber sierra
as long as it has a 0 (ie goes through the origin), a plane in 3d space 'acts like' a vector space itself (it's just 2d space)
lavish jewel
for example, consider R^3 spanned by [1,0,0], [0,1,0], [0,0,1]
like here, [1,0,0] and [0,1,0] span the xy plane, and nothing you do with them will get you that z component to be nonzero
limber sierra
and a line (through the origin) in 3d or 2d space acts as a 1-dimensional subspace
limber sierra
Namington, welcome, can you join voice?
n
vernal yew
isnt the zero vector apart of all subspaces or its not a subspace though??
lavish jewel
it has to be, so you always have to check for that
limber sierra
it is, thats why i said 'as long as it has a 0 (ie goes through the origin)'
limber sierra
just verify it fits whatever definition
vernal yew
given some arbitrary "subspace" how do you see if zero vector is in it?
lavish jewel
very common questions in this type of evaluation are where you are given a really weird definition of vector addition and scalar mult
vernal yew
isnt the zero vector the trivial solution to all linear combinations?
lavish jewel
so you have to check that the zero vector for those operations is there
vernal yew
hrmph, do you have an example?
lavish jewel
isnt the zero vector the trivial solution to all linear combinations?
the thing is that you need to first establish you have a vector space
sometimes you are given a set of vectors and weird operations
and you need to check that it even is a vector space to begin with
vernal yew
i dont get it
lavish jewel
say for example that we have vectors in R^2 of the form (x,y)
vernal yew
alright
lavish jewel
and i tell you the scalar multiplication is the usual one
but the vector addition is given by (x,y) + (u,v) = (x+y, u+v+1)
vernal yew
??
lavish jewel
hmm?
vernal yew
i have no idea what youre saying. how do i check if something given to me is a subspace
lavish jewel
did you cover the vector space axioms?
vernal yew
i dont know what that is so um.. no?
vernal yew
oh ive seen it
vernal yew
no
lavish jewel
ok, it may be that it is not so important in your course then. we can leave it aside for now
vernal yew
so a basis is in a span is in a vector space is in a R^n
lavish jewel
those are all true, but the relationship is not so useful, i think
vernal yew
hrmph
lavish jewel
the span of a set of vectors is a vector space
and depending on how you look at it, also a subspace
vernal yew
i have foundly renaimed linear algebra to synonym soup in all my notes
i just dont get how to check if a zero vector is in something (regardless of whatever something we are talking about)
vernal yew
if you divide by a fraction then you are multiplying by the reciprocal
lavish jewel
please try #prealg-and-algebra or a help channel
chilly lance
if you divide by a fraction then you are multiplying by the reciprocal
thank you so much
vernal yew
no problemo, use the right channel next time though homie
chilly lance
im new so idk which
limber sierra
I mean it just kind of depends on how the object is defined
Like let's say your set is the set of all real number triples (x, y, z) such that x + y = 1
limber sierra
Right
vernal yew
so ill just know?
lavish jewel
before we go on, is this HS linalg? engineering linalg? we don't wanna through you off by guiding you into stuff that, although very important, might not come in your exam
vernal yew
dont understand the question
vernal yew
i am an undergraduate mathematics education major finishing up my math minor. it is a MATH3400 course. feels mostly proof based
limber sierra
Well it's 0 + 0
Which is 0, not 1
So we know this vector can't be in the set
Therefore it's not a subspace since it doesn't contain the zero vector
There are more complicated ways to define subsets but the process of checking for the zero vector is always the same
lavish jewel
i am an undergraduate mathematics education major finishing up my math minor. it...
ok, then this stuff is relevant to you
vernal yew
its also not a vector space?
limber sierra
Just ask "is the zero vector in this?"
It's also not a vector space, yes
A subspace is just a subset that is a vector space
lavish jewel
you're not wrong. many of these terms are interchangeable in many cases
they can be helpful in others though.
vernal yew
do you have any example problems i could try?
lavish jewel
say you are given the set of vectors {[1,0],[0,1],[1,1]}. what is the dimension of the vector space they span? is this a basis?
vernal yew
let me think on it
lavish jewel
consider the subset {[1,0],[1,1]}. what is the dimension of the subspace they span? is this a basis?
vernal yew
say you are given the set of vectors {[1,0],[0,1],[1,1]}. what is the dimension ...
the matrix i need to look at here. is the vectors the columns or the rows?
columns right?
lavish jewel
it doesn't matter, but i suppose in your course you have done it with the vectors as columns, yes
vernal yew
okay so, first one has a free variable i think
so, not linearly independent
so at least one needs to be removed to get the span
lavish jewel
not really
vernal yew
hrmph
lavish jewel
you can get the span of the set of vectors perfectly fine as it is
but this DOES give you information about the 2 questions i put up there
vernal yew
what
is span just all the vectors given
and then basis is with the last one removed?
lavish jewel
i.e. what is the dimension of the vec space, and is this a basis
vernal yew
and dimension is 2
lavish jewel
dim is 2, yes
lavish jewel
and then basis is with the last one removed?
this is true, but not the only solution
the span is not correct though
lavish jewel
which vectors can you make that are of the form a[1,0] + b[0,1]?
lavish jewel
no, i mean in general
vernal yew
what
lavish jewel
which vectors can you make that are of the form a[1,0] + b[0,1]?
this is the same as "what is the span of {[1,0], [0,1]}"
lavish jewel
all of R^2
vernal yew
y
lavish jewel
all vectors in R^2 are of the form [a,b]
which is precisely what you get when you take a linear combination of [1,0] and [0,1]
vernal yew
right span definition is all linear combinations of the form whatever
lavish jewel
indeed
vernal yew
so span{v1,v2,v3} is all vectors whomst i can get from the form av1+bv2+cv3
where a b c are arbitrary scalars
lavish jewel
yes
vernal yew
nicce
i have learned a thing
or at least was reminded what it means
different example, same concept? to verify understanding?
lavish jewel
consider the set of polynomials { 1, x, 2 + 2x } with usual polynomial addition and scalar multiplication. what is the span of this set? what is the dimension of the vector space they span? is this a basis?
or did you mean again with vectors in R^n
(the procedure is exactly the same, don't get scared)
vernal yew
this is fine. is that curly thing youve given mean verticle?
lavish jewel
i use {} to denote set
vernal yew
with 1 on top, x in middle and 2 +2x on bottom
vernal yew
alirght
lavish jewel
{} is a set of vectors
vernal yew
rn is not given?
vernal yew
can you draw? do you have microsoft whiteboard
lavish jewel
it can't be drawn
lavish jewel
not without picking a basis first, which we haven't done
it will look exactly the same
i would just write
${1, x, 2x + 2}$
stoic pythonBOT
Edd
lavish jewel
same thing
vernal yew
alrihgt, let me think i guess
lavish jewel
it's ok, i might've taken it too abstract in one go
let's just do R^3 instead
let's instead consider the set {[1,3,0], [1.5, 4.5, 0], [-0.2, -3/5, 0]}
what is the dimension of the vector space these vectors span?
is this a basis?
vernal yew
okay, i will do that one, just a sec
the last row of the matrix they form is all zeros, indicating their will be a free variable in the last column. dim = 2
it is not a basis
because they are linearly dependent
the span should be all solutions to a[1,3,0] + b [1.5, 4.5, 0] + c[-0.2, -3/5, 0]
dont know how to simplify it though
@lavish jewel
lavish jewel
your first observation is correct
there is at least one free variable
however, notice that 1.5*[1,3,0] = [1.5,4.5]
and also -1/5*[1,3,0] = [-0.2, -3/5]
vernal yew
how to write the set of vectors as a system of linear equations?
lavish jewel
that alone doesn't make sense
vernal yew
hrmph
lavish jewel
what you CAN do is write a system of equations to use the definition of linear (in)dependence
if we call the vectors u, v, w and scalars a,b,c
we want to test whether there exist a,b,c, not all zero, such that au + bv + cw = 0
vernal yew
and also -1/5*[1,3,0] = [-0.2, -3/5]
these two statements make dimension one but i have no idea how to get to those
lavish jewel
now idea how to get what?
vernal yew
yeah, i got linear independence down pretty good
well, if i dont notice that that " 1.5*[1,3,0] = [1.5,4.5]
and also -1/5*[1,3,0] = [-0.2, -3/5] " just from looking at it then how do i algebraically come to those conclusions
can you react to each line of my answer with a thumbs up or thumbs down for the parts that are right or wrong respectively
lavish jewel
you could always do it the brute force way
put the vectors as columns of a matrix and RREF
admittedly this one might be nasty because of the fractions, but it can be done
lavish jewel
when given a set of vectors in R^n, rref always works
vernal yew
by "works" you mean?
lavish jewel
if you don't mess up the arithmetic, you can always do rref and it will give you the info you want
vernal yew
in what form?
lavish jewel
but part of the exercise was to notice it's not always necessary. be on the lookout for patterns
vernal yew
let me pull up a rref calculator
lavish jewel
well, you will RREF and find there is only 1 pivot column
vernal yew
ah, okay
vernal yew
right
lavish jewel
try it out and check
vernal yew
so i get why dimension is one. i also understand that the given is not a basis. but what is the span?
lavish jewel
use the definition of span 😛
lavish jewel
first, the rref has told you we need only 1 vector to find the span
vernal yew
i know it the set of all linear combination of something
lavish jewel
yes
that's precisely it
what do the linear combinations of one of the vectors in the set look like?
lavish jewel
no no
vernal yew
?
lavish jewel
the linear combinations of the vector [1,3,0] are of the form a[1,3,0]
and that's all
that's what the span of this set of vectors is
vernal yew
.... only as many terms as vectors?
vernal yew
span = {[1,3,0]}
lavish jewel
sure
vernal yew
= basis of given?
lavish jewel
hmm?
vernal yew
span of the given vectors = {[1,3,0]} = basis of given vectors?
and dim of the given vector space = 1 ?
lavish jewel
the wording is wrong
the dimension of the vector space is 1
dimension is a property of vector spaces
also there is no basis for a set
vernal yew
span of the given vectors = {[1,3,0]} = basis of given vectors?
and dim of the given vector space = 1 ?
WHAT
lavish jewel
there is a basis for a vector space though
lavish jewel
a basis is a set. a vector space is a set with extra structure
a basis spans a vector space
all vector spaces are sets
not all sets are vector spaces
vector spaces have bases
sets do not
vernal yew
what is the answer to the question "let's instead consider the set {[1,3,0], [1.5, 4.5, 0], [-0.2, -3/5, 0]}
what is the dimension of the vector space these vectors span?
is this a basis?"
in your words
lavish jewel
the dimension of the vector space spanned by the given set of vectors is 1. the set of vectors is not a basis, as it is linearly dependent.
vernal yew
okay but thats like what i said
lavish jewel
not at all
lavish jewel
do the vectors in R^2 form a basis?
vernal yew
all of them?
lavish jewel
all of them together
vernal yew
no
lavish jewel
why not?
vernal yew
they are linearly dependent
lavish jewel
perfect
lavish jewel
i see you have rank nullity, so let's spice it up
vernal yew
lets goo
lavish jewel
consider the matrix whose columns are the vectors in the set i gave you before. what's the dimension of this matrix's null space? can you find a basis for the null space?
vernal yew
confirm these vectors [1,3,0], [1.5, 4.5, 0], [-0.2, -3/5, 0] ?
lavish jewel
yep
vernal yew
okay one sec
must think
dim(N(A)) = Null(A) = 2 because two free variables, and
hold on, not done
lavish jewel
dim(N(A)) = Null(A) = 2 because two free variables, and
this is correct in this case, but we'll come back to it later 😛 it's not always so easy
vernal yew
basis for null space is = [0 , -3/2 , 1/5 ] from nontrivial solution
vernal yew
lavish jewel
so a basis for the null space must have 2 linearly independent vectors
vernal yew
right, dim = 2
then um i dunno
WAIT
its gonna be [0 , 0 ,0 ] , [0 , -3/2 , 0 ] and [0 ,0 , 1/5] but the zero vector is never in a basis so just [0 , -3/2 , 0 ] and [0 ,0 , 1/5] ?
lavish jewel
i don't think these are right still
vernal yew
hrmph
vernal yew
lavish jewel
right, you were missing considering the 1st column
vernal yew
this is what i should of gotten but i struggle at the very end to put the nontrivial solution back into vector form
i dont get that part
lavish jewel
that's the same as we were doing before with "showing the general form of a vector in the span of {...}"
you construct a generic linear combination
lavish jewel
did you get to that part you circled?
vernal yew
how get what is circled in red from what is above?
no, thats the part i cant get to
everything else ive got
lavish jewel
so first you write that as a vector
in terms only of the free variables
how does that look
vernal yew
no idea
lavish jewel
just write the vector [x1, x2 ,x3]
but use only free variables
x2 and x3 are already free
replace x1
vernal yew
[-3/2x2 + 5/2x3 , x2 , x3 ] ?
lavish jewel
mhm
vernal yew
now what
lavish jewel
now notice some terms depend on x2, others depend on x3
split this into a sum of vectors
have one vector depend ONLY on x2
the other, ONLY on x3
i.e. write it as a linear combination
the only thing you're doing is writing linear combinations
all of the problems we have done so far have been about that
writing linear combinations
vernal yew
i dunno
lavish jewel
still wrong
$\begin{bmatrix} -3/2 x_2 + 5/2 x_3 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} -3/2 x_2 \ x_2 \ 0 \end{bmatrix} + \begin{bmatrix} 5/2 x_3 \ 0 \ x_3 \end{bmatrix}$
stoic pythonBOT
Edd
vernal yew
doesnt that = x1 [ 0 , 0 ,0 ] + x2[ -3/2 , 1 , 0] + x3[ 5/2 , 0 , 1 ]
lavish jewel
yes
lavish jewel
sure
vernal yew
haha
lavish jewel
and that's the solution, too
vernal yew
okay, uve been most helpful. another?
lavish jewel
but do you understand what you did?
x1 [ 0 , 0 ,0 ] + x2[ -3/2 , 1 , 0] + x3[ 5/2 , 0 , 1 ] what does this mean?
vernal yew
sort of. let me try again with different one
lavish jewel
no
you have to explain what this means
there's no point in doing another if you don't understand what these things mean
vernal yew
it means that x1 is dependent on x2 and x3
vernal yew
what is the basis of the null space of the given stuff
lavish jewel
and what did you do to show that this is case?
vernal yew
umm , row reduced, and factored the nontrivial solution?
then took the linearly independent factored vectors
and boom basis of the nullspace?
lavish jewel
x1 [ 0 , 0 ,0 ] + x2[ -3/2 , 1 , 0] + x3[ 5/2 , 0 , 1 ] what does this mean?
what i mean is: you have written every single vector x that satisfies Mx = 0 as a linear combination of these vectors
vernal yew
ummm what
lavish jewel
you realize x1, x2 and x3 are arbitrary scalars?
vernal yew
yeah, u said so
lavish jewel
what you wrote is an arbitrary linear combination
i.e. the span of a set of vectors
i.e. a vector space
vernal yew
alright
lavish jewel
you directly constructed a vector space and its basis
vernal yew
cool
lavish jewel
the vector space was the null space
lavish jewel
yes. so again, you found a general linear combination
same as all the other questions
vernal yew
all right
lavish jewel
given {[1,1,0], [1,0,0], [0,1,0]}, place these vectors as columns of a matrix M. find the dimension and an orthonormal basis for the column and null space of this matrix.
vernal yew
i dont understand the question. can you break it into multiple parts? u want an orthonormal basis and the nullspace and you want to know the dimensions of each?
column space = range
so you want dimension and basis for orthonormal, range and null space?
lavish jewel
i want an orthonormal basis for the column space, and an orthonormal basis for the null space
plush canyon
lavish jewel
also write the general form of the vectors in the null and col space
plush canyon
Can someone help me with this if u got the time
lavish jewel
we wrote 2 different solutions for this when you asked yesterday
let me find the link
lavish jewel
starting at your message here
ryu gives a solution using minimal polys
my solution involves moving the I to the RHS and factoring out A
this gives you -A(A+2I) = I, meaning -(A+2I) is the inverse of A
lavish jewel
now consider what A + 2I does to the eigenvalues of A, and what A^-1 does to the eigenvalues of A
i was, but it was in the spirit of showing that your problem did have a sol
plush canyon
ok
lavish jewel
you can take it from there, i think
plush canyon
If you have 2I it double matrix A
lavish jewel
it adds 2 to the eigenvalues of A
you can check yourself that A + cI has the same eigenvectors as A, and the eigenvalues are those of A + c
plush canyon
ah ok ok
My probkem is why is this a 5 mark question
thats soo much for something simple
vernal yew
halfway there
lavish jewel
wdym so simple 😛 you're asking for a reason
plush canyon
well not simple
but in the context of the solotion I htought there would be more to it
lavish jewel
it's not so straightforward, you need like 3 key observations
regardless of whether you do it how i did or you use the char and min polys like ryu did
lavish jewel
characteristic
plush canyon
ok
lavish jewel
he relied on cayley-hamilton and properties of the minimal polynomial
(he didn't explain them, but mentioned them briefly, followed by "blah blah")
plush canyon
this gives you -A(A+2I) = I, meaning -(A+2I) is the inverse of A
I get this
Now u just want me to solve for A
and that would give me -1
lavish jewel
wdym solve A
we never find A
we don't even know what size it is
that's the whole point of the problem
plush canyon
Oh yeah I'm dumbass.
so how do u know the eigen value is -1
I'm unsure/condused now
lavish jewel
i use 3 key observations. the first is that, rearranging and factoring, we can show A is invertible
plush canyon
yes
lavish jewel
this means the eigenvalues are nonzero
plush canyon
yes I understand that
lavish jewel
then we observe the inverse of A has ith eigenvalue 1/lambda_i, where lambda_i is the ith eigenvalue of the original matrix A
furthermore, we know that A + 2I has ith eigenvalue lambda_i + 2
plush canyon
wait
lavish jewel
and the - in front makes this negative