and the second line kinda gives like 0,3/2 and 1/2
#linear-algebra
old flame
By the way @native rampart , heres my another attempt for q13. Since $\forall U \subset V$, where $\dim U = \dim V -1$, it is invariant under T, which means that $\forall u \in U$, every $u$ is an eigenvector of $T_{U}$. Therefore, by question 12, T is the scalar multiple of the identity.
stoic pythonBOT
Otoro:
native rampart
@native rampart Is there any hints on how to start with (I-AB) is invertible, not sure how to continue with this first step
@old flame assume (I-AB) is invertible and multiply rhs with (I-BA)
You will get I
old flame
do we expand the brackets ? cause those are operators though
native rampart
Yes
old flame
but how do you know AB equals BA
native rampart
They are not
old flame
ok, if we expanded the brackets, we get I - BA - AB + AB*BA right ?
native rampart
No
You are not multiplying I-BA and I-AB
You are multiplying (I-BA) with that expression on the rhs
old flame
Im sorry what do you mean ?
glad sky
Hi
native rampart
halcyon pollen
@limber sierra he is implementing the lines by y = 0 for the second line man so which one is right?
native rampart
You are doing
$(I-BA) (I+B(I-BA)^{-1}A)$
stoic pythonBOT
DrunkenDrake:
native rampart
That will be I,if you expand
limber sierra
i have no clue what youre asking
you mean he's setting y = 0 and solving for x?
that's fine too
the point is just to find 2 points on the line
once you have 2 points on a linear equation, you're done.
whatever technique you want to use to find those points works
old flame
But techically, aren't we just taking that expression and multiplying both sides on the left by (I-BA) ? however, why are we allowed to take the result and work with it ? I thought we have to start from I-AB
native rampart
You are just taking this random expression and and if you multiply with (I-BA) , you happen to get I.(Whether you do expression* (I-BA) or (I-BA)* expression)
So,This expression should be (I-BA)^{-1}
*the term inside inverse should (I-AB),typo
old flame
so you mean the same thing happens whether it is AB or BA
native rampart
Not sure, what you mean by that
Yea,inverse of (I-AB) will be obtained similarly (Just replacing A with B and B with A,in all places)
native rampart
Yea,inverse of (I-AB) will be obtained similarly (Just replacing A with B and B with A,in all places in the expression)
If you mean this,sure
golden drum
By the way @native rampart , heres my another attempt for q13. Since $\forall U \subset V$, where $\dim U = \dim V -1$, it is invariant under T, which means that $\forall u \in U$, every $u$ is an eigenvector of $T_{U}$. Therefore, by question 12, T is the scalar multiple of the identity.
@old flame
What invariant means?
old flame
If you mean this,sure
@native rampart yes
@golden drum Invariant means that for $u \in U$, $Tu \in U$ , for $T \in L(U)$ right ? Or for restricted operator on an subspace, if the subspace is invariant under the operator
native rampart
By the way @native rampart , heres my another attempt for q13. Since $\forall U \subset V$, where $\dim U = \dim V -1$, it is invariant under T, which means that $\forall u \in U$, every $u$ is an eigenvector of $T_{U}$. Therefore, by question 12, T is the scalar multiple of the identity.
That need not be true. Take V to be $R^3$ ,
U to be space spanned by { (1,0,0),(0,1,0) }
Let $ T_{U} (1,0,0)=(1,2,0) and T_{U}(0,1,0)=(2,1,0) $
stoic pythonBOT
Otoro:
old flame
so the problem lies with operator ?
stoic pythonBOT
DrunkenDrake:
native rampart
Actually,Show that if a basis vector is not eigen,Some subspace is not invariant under T
old flame
may I ask what is the explicit error Ive done in my proof ? I want to understand what I got wrong
native rampart
You haven't shown how that implies u is an eigenvector of T(Which is not obvious,at all)
old flame
sure I will try that approach ty
native rampart
So,Your proof is not quite complete
halcyon pollen
what is linear combination of columns ?
old flame
btw, @native rampart Im still a bit unsure of how to expand for your q8. So far Ive got, $(I-BA)(I+B(I-AB)^{-1}A)=I-BA +(I-BA)B(I-AB)^{-1}A$, Im not sure how to continue
stoic pythonBOT
Otoro:
golden drum
Sum of columns scalated matrices
golden drum
Where your question come from?
halcyon pollen
i am looking at the mit open courseware in youtube
studying that cant understand what is linear combinations
native rampart
Do
$(I-BA)B(I-AB)^{-1}A
B(I-AB)^{-1}A -BAB(I-AB)^{-1}A
=B((I-AB)^{-1}-AB(1-AB)^{-1})A=BA$
stoic pythonBOT
DrunkenDrake:
old flame
ah okay I see it thanks
old flame
ok heres my trial for question 13 then
Let $(u_1,...,u_{dim V -1})$ be a basis in $U_{\alpha}$, where $U_{\alpha} \subset V$. Let T be an operator that is invariant $\forall U \subset V$, where $\dim U = \dim V -1$. Suppose $\exists j \in {1,...,\dim V-1}$ such that $T(u_j) \notin U_j$. Then pick $b \in U$, written as a linear combination, $b=a_1u_1,...,a_{\dim V-1}u_{\dim V-1}$ for $a_1,...,a_{\dim V-1} \in F$. Then $T(b)=a_1Tu_1+...+a_jTu_j+...+a_{\dim V-1}Tu_{\dim V-1}$. Since $Tu_1 \in U_1$,...,$Tu_{\dim V-1} \in U_{\dim V-1}$, but $Tu_j \notin U_j$, so T is not invariant on $U_{\alpha}$. Therefore, contradiction. Thus all basis vector has to be an eigenvector, which implies hat every $u \in U$ is an eigenvector $\forall U$ with $\dim U=\dim V -1$. Thus by Q12, T is a scalar multiple of the identity.
stoic pythonBOT
Otoro:
native rampart
So,you are taking different vector spaces and considering a space such that T(uj) is not in that space?
old flame
yeah, so T is not invariant in that
tepid violet
Sometimes i see functions like that arcsin(sin(x)), sin(cos(x)), cos(cos(x)) What sense to use nested trigonometry?
native rampart
Well, It's better to convey that as:
Consider $u_i$ as a member of space spanned by
{$u_1,u_2,...u_i,...u_{n-1}$} $\implies
T(u_i)=a_1u_1+a_2u_2...a_{n-1}u_{n-1}$
Now see $u_i$ as a member of {$ u_1,u_3,....u_i...u_n $}
Implies this subspace is not invariant under T,unless $a_2$=0. Repeating this process(seeing $u_i$ as a part of some subspace) gives us that $u_i$ has to be an eigenvector
stoic pythonBOT
DrunkenDrake:
old flame
are you saying that u_i is a vector that after applying T would be a vector that does not a have a projection on some subspaces ?
half forge
can someone help me with letter a
im confused on the solutions
didi they take the deriivative in yellow?
gray dust
yes
native rampart
are you saying that u_i is a vector that after applying T would be a vector that does not a have a projection on some subspaces ?
@old flame yes
The subspace spanned by{u_1,u_2,...,u_(i-1),u_(i+1)....u_n}
old flame
So you're saying that $u_i$ has to be spanned by all the basis vectors, cause if not then $\exists k$ where $T(u_k)$ not in the span, which means T is not invariant right ?
stoic pythonBOT
Otoro:
native rampart
I am saying T(u_i) should be ku_i for some k
old flame
So in simpler terms. We let $u_i$ be a vector spanned by $u_1,...u_n$, where a projection of $u_i$ does not exists for some subspaces. Then T would not be invariant, so the projection of $u_i$ must exists under all subspaces. Hence $u_i$ is an eigenvector ? Am I right
stoic pythonBOT
Otoro:
native rampart
u_i is a basis vector
wintry steppe
can anyone help me solve this problem? I'm completely lost
you can start by looking at the third given equation
arctic karma
you can turn it into an augmented matrix so you can visualize it better
wintry steppe
that's not really needed here but if it helps them
big hint: ||work upwards||
like this?
i guess you meant x_2 instead of x^2 in the second part
you can only raise indices like that once you've taken a differential geometry course 
oh yeah, my writing is messy lol
that aside
you got the right answer
any system of equations in a "triangular" form like this can be solved using a similar method
triangular form?
I'm kinda new to the class so I'm not really familiar with the terms
like the non-zero coefficients (i filled in the zero ones) form a triangle
there is a more precise definition
if you write the system in matrix form as Ax = b, then we'll say the system is upper triangular if every element below the diagonal of A is zero (if you don't know what this is, don't worry too much about it. you'll probably learn it soon)
this sort of "working up" is a valuable technique
"up" as in we started from the bottom equation and found variables solutions moving up one equation at a time
spiral star
@old flame i'd like to suggest an alternative for the first part of your proof of Q13. an indirect approach via linear dependence is straight forward imo.
radiant topaz
can someone help me solve this? i dont even know what the problem is asking for or how to interpret the augmented matrices "containing A in row echelon form"
i tried reducing both the given matrix at the top to see if it matched any of the answers
and the answers to see if they matched given
and they didnt
so im clearly going about it in the wrong way
can someone please point me in the right direction
wintry steppe
what is it mean when they say:
You can subtract any multiple of one equation from another.
That is, you can replace any equation with that equation less a
multiple of any other equation.
?
spiral star
@radiant topaz they followed the row elimination algorithm step by step and finished after row echelon form was achieved. no further manipulations were done. you should get the bottom matrix as the answer
the algorithm runs column by column
radiant topaz
i dont think we've ever done reducing matrices through manipulation of columns
spiral star
no thats not what i mean
you manipulate the rows but it creates pivots column by column
you shouldnt do try to do any further manipulations
you are done after only a few steps
spiral star
yes
first get rid of everything non-zero before the first pivot
then do the same with the next pivot
until you are done
spiral star
yes
the algorithm follows the pattern
like, after you fix everything below the first pivot you should get
then your normalize the 2nd pivot
and then you fix everything below that
and normalize the last pivot
and you get
spiral star
this is all the algorithm would do
you can do more transformations but they would not be part of the algorithm
radiant topaz
yeah i just didnt understand exactly what it was asking for at first
i was just using an online calculator so
spiral star
they wanted you to play computer
radiant topaz
it was going too far as well
yeah
usually they're way more specific about whether or not its acceptable to use technology to solve
i guess i probably should've tried manually once i decided i wasnt getting the right answer from symbolab
spiral star
online calculators usually give you rref instead of ref
radiant topaz
symbolab gives a choice between the two
spiral star
i see
i mean the row space is preserved, so you could always take an rref or whatever and do operations to get back into the ref lol
radiant topaz
too much effort at that point tbh
spiral star
definitely
matrix computations are boring af anyway
completely understandable if you wanted to do them with a computer
radiant topaz
yeah we use technology for most of that stuff now
while i have you here, any chance i can ask another question? im not sure if it falls entirely under the branch of linear algebra but i can explain the process to get to the stage where i have an augmented matrix
i just dont know how to go from there
spiral star
i dont really know what you mean
spiral star
uh i dont remember the page rank algorithm and i kinda dont wanna look it up rn
radiant topaz
thats fine, hopefully someone knows
my process so far was based on a video that they included to help us work through it
which basically like
setting up each of the equations first based on how they're connected without taking into account the amount of connections
so like
x1 = x4
x2 = x1 + x3
x3 = x1 + x2 + x5
x4 = x1 + x5 + x6
x5 = x2 + x6
x6 = x4 + x5
and then after, adding in division for the amount of connections
for example, x4 has two other outward connections, so any instance of x4 on the RHS will be become x4/2
wintry steppe
I have the answer if you want it
radiant topaz
sure
i already worked through most of it, i just couldnt work out how to do the final part of question b
wintry steppe
I'll just dm it to you
radiant topaz
like i got it into an augmented matrix and reduced it
wintry steppe
cause it's like 6 pictures
radiant topaz
sure
but to keep explaining for anyone else, i got it to an augmented matrix which was
uh hold on i'll have 2 take a pic of it
this
radiant topaz
reduces to this though
in the example that was given there was a row of 0s, which was used as a free variable
old flame
@wintry steppe Let x3 be t
radiant topaz
and then other solutions were found in terms of that
old flame
Then u have X2 in terms of t and then X1 in terms of t as well
radiant topaz
<@&286206848099549185> apologies for the ping, have waited 15 minutes since my last message
i started describing the problem and how we've been taught to solve it so far starting from this message: https://discordapp.com/channels/268882317391429632/540211747613704221/762608174288470026
wintry steppe
I tried entering it into my homework and it says, "variable t is not defined in this context"
old flame
because your website says use s1 or s2 i guess
@spiral star thank you very much, this seems like a method very similar to one I saw before, using the linearity
so are you saying that since T is invariant in all subspace U, then the transformation of a vector $\sum v_i$ equals to the sum of the transformation of each basis vector, and this implies that the individual eigenvalues are the same as the eigenvalue for the sum and thus all the eigenvalues are the same ?
stoic pythonBOT
Otoro:
spiral star
well in the first part we have proved that every vector is an eigenvector
we just didnt confirm that there is only one eigenvalue
if every vector is an eigenvector, then the vector you get by taking the sum of all basis vectors is also an eigenvector
and each of the basis vector themselves are also eigenvectors
and then you show that all of them have the same eigenvalue
the second part doesnt really need invariance anymore
it just works with the fact that every vector is an eigenvector
and you get the eigenvector property from the first part
@old flame does that make sense or do you want me to unwrap it more?
old flame
ohhhh
so its because we have proven $(v,Tv)$ is linearly dependent, then it means we can write this as a linear combination, $av+bTv=0 \Rightarrow Tv=(-b/a)v$, which implies that every v is an eigenvector, since all vectors in U could be represented as this, so do the basis vectors ? So this is pretty much the most important step right ? Since everything follows through this
stoic pythonBOT
Otoro:
native rampart
Yes
old flame
ahhhhh, got it, thank you @native rampart and @spiral star
wintry steppe
Ohh thank you didn’t realize that
spiral star
@old flame technically you have to be careful with your argument because you could have divided by zero here
but in general you can say (v,w) is linearly dependent iff v = a*w or w = b*v for some scalars a,b
you can always express one in terms of the other
well i guess if you assume that v is not zero then you can make the claim
because if you assume (v, Tv) is independent, then neither v nor Tv can be zero. but the thing is that you dont need to construct that scalar, you just know that some scalar will work because a vector is the span of another
old flame
oh, so I guess just the idea of dependence is all that matters, since explicitly writing it may cause problems if details is left out
spiral star
all you need is Tv = lambda v for some lambda
which is equivalent to Tv being in span(v)
which is perfectly fine even if Tv = 0 so the kernel doesnt get in the way
thats what you get from linear dependence here
native rampart
Is it true,If (Tv,v) are linearly dependent ,then Tv=av if we take the vector space to be over a finite field,for some a?
spiral star
why would it not be
spiral star
i guess we can confirm it by writing it out but i dont immediately see any argument that shouldnt hold in finite fields
the initial assumption in my proof was v != 0
radiant topaz
rip my question from a while ago :(
spiral star
when v = 0 then (Tv, v) are always dependent, and in particular Tv = 0 by linearity
when v != 0 and (Tv, v) is dependent, then then Tv is in span(v)
because if we take out Tv then (v) is independent
so Tv = cv for some constant c in any field
we can always generate a basis for span(Tv, v) by removing vectors until we are left with an independent set
that should work in all fields afaik
and removing Tv always works if v != 0
removing v doesnt always work because v could be in ker(T)
but given v != 0 and (Tv, v) dependent, then Tv = cv for some c
native rampart
If the group was finite,would that change anything?
the linear dependence condition (for vectors x,y) says non zero c1,c2 exist such that c1x+c2y=0 I was thinking something like c1=0 and c2y still being zero,which doesn't give us a relation between x and y
spiral star
if c2 y = 0 then c2 = 0 or y = 0. that holds in all fields including finite
native rampart
(nvm field definitions)
So,They behave exactly the same as vector spaces over R or C when talking about linear independence?
spiral star
yea
like in terms of choosing a basis and linear independence, sure
you get into trouble when you add multiples of vectors tho
for example, if your field has characteristic 2, and you have some basis vector v, then v+v = 0
so this is kinda annoying sometimes
but in terms of choosing a basis and showing linear independence they behave the same
chilly solstice
What does the following mean:
$A$ is a product of elementary matrices?
Let us take $A=\left[\begin{array}{ll}2 & 1 \ 1 & 2\end{array}\right]$ as an example.
stoic pythonBOT
Schuams:
native rampart
Isn't that only for invertible matrices?
Btw,Do you know gaussian elimination?
An elementary matrix is a matrix E,such that B=(EA) is also the matrix obtained by applying some elementary row operation on A
native rampart
This is equivalent to saying,You can apply a bunch of row operations on an invertible matrix to get I
chilly solstice
So If E represents e.g. multiplication of first row with 2, then E is an elementary matrix.
Rihgt?
native rampart
Yes
chilly solstice
Thanks DD!
chilly solstice
Is this also called the max norm on R^n ? :
soft burrow
yes
soft burrow
if your book defines it previously, use that definition
magic acorn
for question 2 a idk what they're asking they are telling me that i should find the sum of a matrix that equals to 1 but how do i do that if they give me a whole probability table
spiral star
@magic acorn maybe you are overthinking it. its like a classic high school probability question. you are supposed to find the probabilities of transitioning from square 1 to any square
magic acorn
yea cause im looking at this question and im like how does this relate to linear algebra
magic acorn
they're asking me to find the sum of 1 with probabilities of 1 2 and 4 but they already gave the chances for them
probably will get into that for question b
spiral star
well the real transition probabilities differ slightly from the table
because if you would end up in 3, you automatically get back to 1
like, let me give you an example i guess
for the transition 1 -> 1
you need to roll either a 4 or a 2
4 gets you back to 1 immediately, 2 would get you onto 3 but then you automatically go back to 1
so the probability for 1 -> 1 is 1/8 + 2/8
because thats the probability of rolling either a 2 or a 4
so i would expect this to be the answer
P(1 -> 1) = 3/8
P(1 -> 2) = 1/8
P(1 -> 3) = 0
P(1 -> 4) = 4/8
unless i misread the task
magic acorn
no no you read it right, so for square 4 why would it be 4/8 again
spiral star
because if you are on square 1 you need to roll a 3 so 1 + 3 = 4
and the probability of rolling a 3 is 4/8
magic acorn
oh ic ic
spiral star
:)
magic acorn
alright thanks really appreciate it
spiral star
im pretty sure this is heading towards a transition matrix
magic acorn
yea b and c looking rough rn
lemme try them first
then if i need help ill come back
thanks again
spiral star
np
stuck stratus
Prove that a nilpotent matrix A (i.e. that a certain power of A yields the zero matrix)
merely has the eigenvalue zero
how would I prove this?
spiral star
suppose A^n = 0
now take an eigenvector
and apply A n times to it
@stuck stratus
spiral star
what about the zero vector
stuck stratus
A n times a vector is the zerovector right?
stuck stratus
I saw something online
spiral star
note: by definition, an eigenvector cannot be the zero vector
stuck stratus
😶
spiral star
and short as well
stuck stratus
It said that: take A^k = 0
then A^k v = 0
so λ^k v = 0
then λ^k = 0 so λ=0
spiral star
yes
stuck stratus
is that the proof you wanted to show me as well
spiral star
yes
stuck stratus
nice
and
Prove that an eigenvalue of an idempotent matrix A (i.e. A^2 = A) must be 0 or 1
stuck stratus
We know that A^2 - A = 0
Now take an eigenvector x for λ, an eigenvalue of A
Then (A^2-A)x = 0
So (λ^2-λ)x=0
And x is not the zerovector so λ^2-λ = 0. So λ = 0 or 1
spiral star
yea
stuck stratus
:)
If v is an eigenvector of the matrix A, then v is an eigenvector of A + cI as well for every scalar c.
can you explain this one?
spiral star
that's a proof
stuck stratus
how do you know that
stuck stratus
yeah but you cant prove with an example
spiral star
its not an example
you prove it in general
for any matrix A and any scalar c
and any eigenvector of A
stuck stratus
oh wait
spiral star
like, where did you get the idea with the example from
stuck stratus
I get it
spiral star
(A + cI)v = ??
stuck stratus
λv
spiral star
more than that
spiral star
you forgot the cI
stuck stratus
(λ+cI)v
spiral star
(A + cI)v = Av + cIv = λv + cv = (λ+c)v
so (A + cI)v = (λ+c)v shows that v is an eigenvector of (A + cI)
spiral star
you had nothing to do but apply the matrix to the eigenvector
all those proofs are oneliners lol
stuck stratus
If v is an eigenvector of the invertible matrix A, then cv is an eigenvector of A−1 for
every scalar c != 0
and this last one?
spiral star
well guess what you can do here :p
stuck stratus
are the eigenvalues f an invert. matrix the same as those of the inverse?
spiral star
how about you figure that out by doing the same thing again that you have been doing for the last two proofs
If Av = cv for c != 0, then what happens when you apply A^-1 to both sides
stuck stratus
v= A^-1 cv
stuck stratus
divide both sides by c
spiral star
yes, and you can do that because c != 0
stuck stratus
then A^-1 v = 1/c v
stuck stratus
how?
spiral star
v= A^-1 cv was good already
stuck stratus
oh
spiral star
it shows that cv is an eigenvector
stuck stratus
with eigenvalue 1
stuck stratus
1/c
spiral star
yes
stuck stratus
🎊
spiral star
so now you can say something about the non-zero eigenvalues
stuck stratus
they're all the inverse of the eigenvalues of A
spiral star
if c is an eigenvalue of A, then 1/c is an eigenvalue of A^-1
stuck stratus
yep
spiral star
and if v is an eigenvector of A, then v is also an eigenvector of A^-1
modern quarry
Find a,
does this even have a solution?
i have a vague memory of something about "if the numbers of variables are more than numbers of equations, its not solvable"
true?
dusky epoch
a isn't an unknown
it's a parameter
as far as i can see from the layout of this system
vernal pebble
you have two variables and two equations
dusky epoch
but also i'd appreciate if you posted the exact problem statement @modern quarry
it could clear this up
modern quarry
it is find the value of a
vernal pebble
i think they just want you to find a
dusky epoch
find the value of a under what condition?
modern quarry
no conditions given
dusky epoch
do you have the ENTIRE problem statement
if not, i'll assume your task is to find a such that x=-1, y=7 is a solution of the system
vernal pebble
i'm going to assume the system is consistent
dusky epoch
don't guess, let zakuto provide the info i'm asking for
modern quarry
"for what values of a does the equationsystem lack solutions"
dusky epoch
OH
vernal pebble
ok so it wants you to find a so that the system is inconsistent
dusky epoch
WELL WHY DIDNT YOU SAY SO
vernal pebble
that is important information you shouldn't leave out LOL
modern quarry
sorry xD
vernal pebble
@modern quarry first make an augmented matrix out of the system of equations
then turn it into reduced echelon form
vernal pebble
true but it always works ;P
dusky epoch
you can just isolate y in the second equation and substitute the result into the first, obtaining a linear equation in x with a nonzero constant term
vernal pebble
^
dusky epoch
then require the coefficient of x to be zero so that you get the required lack of solutions
vernal pebble
yeah that's probably a smarter way to do it @modern quarry
vernal pebble
a just can't be a certain real number or else the system is inconsistent
dusky epoch
you will have at most one value of a here
vernal pebble
your answer will be akin to a = x where x is some real number
dusky epoch
no
on two accounts
first off x is taken
and second, the answer will be a = (some number)
dusky epoch
yes
vernal pebble
thanks!
dusky epoch
do you have a fetish for multiples of three or for unsimplified fractions?
also, $ax - 6x \neq -5ax$
stoic pythonBOT
Ann:
modern quarry
right
dusky epoch
but also what stopped you from simplifying $\frac{12-9x}{3}$ to $4-3x$
stoic pythonBOT
Ann:
modern quarry
dusky epoch
lines 4 and 5 why
there was no need to isolate a
once you have the equation (a-6)x = -2, you should think back to your goal: to find which values of a make this equation (and hence the system) have no solutions
normally you would solve this equation for x by dividing both sides by (a-6). where could things go wrong?
modern quarry
when a = 6?
dusky epoch
exactly
modern quarry
Thanks alot! <3<3
sonic shale
hey guys quick question
take matrix A
1 0 k
0 1 m
1 0 n
What values of k m and n would guarantee a consistent linear system Ax = B for whatever B
wouldn't the answer be correct as long as k and n are equal or am i missing somethin
slate haven
This translates to: Let phi be a surjective map and let U be a linear subspace of W. Show the equivalence. U0 is the annihilator and phi* is the dual map of phi. I have already shown that the first set is a subset of the second but i am struggling with showing the other direction. So far my approach has been to start with a psi in (phi^-1(U))^0 and i have to find a function f in U^0 such that psi = phi*(f). this is what i've been trying the last 20 to 30 minutes without any success. I would be very glad if someone could help.
slate haven
so that is
$U^0 = {f \in W^* \mid f(u) = 0 \forall u \in U}$ and $\phi^* \colon W^* \to V^* , \alpha \mapsto \alpha \circ \phi$
stoic pythonBOT
LyricalMiracle:
gaunt field
half storm
wheat ermine
They busy
soft burrow
that's not a reason to spam the other channels
gritty frigate
Hey. If a system of equations is 3x2 and it is has only one solution for x and y. What is the value of the determinant of the augmented matrix ?
What I was thinking was that, if the system has a solution then two of the equations have to be the same
Otherwise because of Rouché–Frobenius and Determinant the system would not have just one solution
But the remaining system is now 2x3, and it does not have determinant defined
In that case, Det(A) = 0 ? Being A a 2x3 matrix ?
dawn remnant
yeah, it must mean that one of the rows can be made to be 0
a solution to a 3x2 system is an intersection of 3 lines on a plane, which can generally be an empty set, but here it's a single point. If it has exactly one solution, then one of the lines must be expressible as a linear combination of the other two
keen zinc
Hi guys how would you do the part that's highlighted? I tried to set r(t1) equal to r(t2) and solve it on Mathematica but apparently that doesnt work for some reason...
midnight hedge
Can someone explain how they got the vectors for the eigenvalue bases?
spiral star
@slate haven bitteschön :) hoffe das ist verständlich
radiant topaz
apologies for the long problem but was trying to work this out yesterday and didnt get very far
im just struggling to get the complete system of equations
someone confirmed that the text is a little unclear and that each car uses 3 power units + has 2 spares
as opposed to just each car having 1 power unit + 2 spares
i started doing this
I1 + T1 + E1 = 10
I2 + T2 + E2 = 20
I1 + I2 = 10
E1 + E2 = 10
T1 + T2 = 10
but im not sure if its correct
midnight hedge
nvm i figured out my own question
hearty pulsar
I would help if I knew
Time to translate
Give the dot product of CA.BD using L and l
I got L^2-l^2
b) i using points K and H as intermediate points write BD as the sum of 3 vectors
I got BD=BH+HK+KD
ii. Recalculate CA.BD substituting BD from i
So I end up with CA.BH+CA.HK+ÇA.KD and then im not sure where to go from there
That’s as far as I got but iii. Asks to find the distance between H and K using L and l for length and height
agile tangle
Anyone can explain what means an eigenvalue of 0 for a matrix?
midnight hedge
@agile tangle it means Ax = 0x
so Ax = 0 basically
a eigen vector multiplied by A yields a zero vector
lucid cedar
Im trying to answer this question and came up with
-1 3
3 -1
for A.
Im just trying to make sure I have the right idea before I do a ton of these for this assignment
Wait I just realized I dont understand how to do B.
gray dust
recall the defn of a matrix of a map wrt bases of the domain/codomain
lucid cedar
hmmm
I think the columns are just the Return of T when given that element of the basis of V
but when examining w1 and w2 I think I am now dealing with rows
but the matrix
1 1
-1 1
does not fit T
gray dust
what's V, what's return
lucid cedar
uhh sorry im just talking in general with the notation axler usually uses
our class usually puts V the domain at the top and the columns are then filled out by the values obtained from T(v)
but we never talked about the domain of the range
gray dust
what's V, what's v, what's "domain of the range"
lucid cedar
T is in L(V,W)
V = F^2 W = F^2
gray dust
beside all that, you should look back to the definition of basis, then the definition of the matrix of a linear map with respect to given bases of the map's domain & codomain
lucid cedar
axler doesnt really talk about those terms :/
ill just google it
okay
im looking at a definition and i think i see what im supposed to do
gray dust
yes
lucid cedar
okay
so
when they say that (1,1) (-1,1) will be the basis of the domain. I think I know what to do with that information. I plug those into the formula like so:
Im taking some liberties but I hope u understand what i mean. I take the basis plug them into T and generate columns
but when it says that it is also the co domain
I suspect that means I should set it up as coefficients to those calculations
gray dust
do you know what codomain means
lucid cedar
so with the actual calculations done it should look like this:
It is the domain of the space that T goes to
If T : V -> W
then codomain is W?
gray dust
yes
1st step to building the matrix is getting the images of the basis vectors under T, T(w1) & T(w2) are?
gray dust
we call f(x) the image of x under f
gray dust
the given map
lucid cedar
right
gray dust
{w1,w2} are also given as a basis of the codomain. get the coords of the images wrt {w1,w2}
lucid cedar
uhhh
I dont really undersand
I already input w1 and w2 into T so im just not sure what im supposed to be doing witht hem for the codomain
gray dust
let B={b1,...,bn} be a basis of an n-dim vector space V and let x in V. there exist unique scalars c1,...,cn where x=c1b1+...+cnbn. we call (c1,...,cn) the coords of x wrt B
gray dust
the 1st col of the matrix is the coords of T(w1) wrt {w1,w2}
lucid cedar
so is what ur asking for just T(w1) and T(w2)
im having trouble following, Im really trying to understand
I answered all of axlers questions no problem, but my professors questions dont follow the book
gray dust
1st step to building the matrix is getting the images of the basis vectors under T, T(w1) & T(w2) are?
{w1,w2} are also given as a basis of the codomain. step 2, get the coords of the images wrt {w1,w2}
get the coords of T(w1) wrt {w1,w2}
fallen karma
I'm working out of axler myself I like it
gray dust
get the coords of T(w1) wrt {w1,w2}
did you do this
lucid cedar
the 1st col of the matrix is the coords of T(w1) wrt {w1,w2}
is the first column not the answer to that?
wait
those columns are actually the answer to T(w1) and T(w2)
i didnt just like make up those numbers
is that the issue here?
gray dust
no you're far off
gray dust
so?
lucid cedar
gray dust
what're the coords
lucid cedar
(2,2)
gray dust
did you see the defn of coords wrt a basis above?
@wheat ermine use #prealg-and-algebra
wheat ermine
I did no one answered
gray dust
this channel's not for prealgebra
wheat ermine
Its been 10’
lucid cedar
This isnt a paid service. people dont have to respond. These people do it because they are kind enough to
its late people prolly arnt awake
just try again tomorrow
wheat ermine
Lol people are so mean
lucid cedar
brother wut
wheat ermine
I know this isnt a paid service
gray dust
there are rules. stick to the right channels
wheat ermine
wtf.. ok
Im tryna solve something easy u would have helped me by now but u choose to fight over a channel
This is insane
gray dust
let's speed up. here's the defn of coords wrt a basis
let B={b_1,...,b_n} be a basis of an n-dim vector space V and let x in V. there exist unique scalars c_1,...,c_n where x=c_1b_1+...+c_nb_n. we call (c_1,...,c_n) the coords of x wrt B
get the coords of T(w1) wrt {w1,w2}
use the defn. {w1,w2} is a basis of F^2 so there exist unique c1,c2 where T(w1)=c1w1+c2w2. get c1,c2
lucid cedar
ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
(2,2) = c1(1,1) + c2(-1,1) c1 = 2 c2 = 0
gray dust
the coords of T(w1) wrt {w1,w2} are (2,0). get the coords of T(w2) wrt {w1,w2}
lucid cedar
(0,-4)
gray dust
the 1st col of the matrix is the coords of T(w1) wrt {w1,w2}
gray dust
a self esteem boost
gray dust
you're welcome
lucid cedar
So quick question
if the domain and the co domain are just (1,0),(0,1) then does it end up working out how I thought it would
where u just put the vectors into T and make that the column
Thats how axler does all the matrices he builds in the text and im trying to find the situation where that is in fact how it works
gray dust
you mean BASIS of the dom/codom
gray dust
what're the coords of (x,y) wrt that basis
lucid cedar
it does end up working out
that way
T((1,0)) = c1(1,0) + c2(0,1)
-1 0
and the other column works out the same way
interesting
gray dust
what're the coords of (x,y) wrt that basis
you didn't answer
lucid cedar
(-1,0) ^ (0,-1)
gray dust
idk wym
lucid cedar
T((1,0)) = c1(1,0) + c2(0,1)
(c1,c2) = (-1,0) = T(1,0)
and the second column is almost exactly the same
gray dust
i didn't ask about T
lucid cedar
uhoh
gray dust
for any (x,y) in F^2, what're its coords wrt {(1,0),(0,1)}
gray dust
for a) the 1st col of the matrix is the coords of T(w1) wrt this which is just T(w1)
gray dust
so reread the defn of the matrix of a map, then this is it summarized https://ptb.discordapp.com/channels/268882317391429632/486844875632017408/750109864876834857
lucid cedar
Im glad im not the only one that asked atleast
gray dust
the 1st part of the message is important. 2nd part no. and no shame in asking
lucid cedar
Thanks again. Im sure ill be back before the semester is over
gray dust
no prob
molten hill
So I'm a bit confused about how the set containing just the zero vector $Z = {\vec{0}}$ is linearly dependent.
\medskip
For a set of vectors, $S$, to be linearly dependent, there must exist some $\vec{v}\in S$ that can be expressed as a linear combination of the other terms in the set. But how can this definition apply to a set that only has one term? How can $\vec{0}$ be expressed as a linear combination of the other terms'' in $Z$ if there \textit{aren't any} other terms'' in $Z$?
stoic pythonBOT
Isaiah:
dusky epoch
i mean, a linear combination of nothing is just the zero vector
just like a sum of nothing is 0 and a product of nothing is 1
wheat ermine
can u check for intervals on a number line if u have 3 numbers on the line?
or u need a table
golden drum
So I'm a bit confused about how the set containing just the zero vector $Z = {\vec{0}}$ is linearly dependent.
\medskip
For a set of vectors, $S$, to be linearly dependent, there must exist some $\vec{v}\in S$ that can be expressed as a linear combination of the other terms in the set. But how can this definition apply to a set that only has one term? How can $\vec{0}$ be expressed as a linear combination of the
other terms'' in $Z$ if there \textit{aren't any}other terms'' in $Z$?
@molten hill
A set is linear dependent, if exist a non trivial linear combination that goes to 0.
Since the unique element of S is 0
1•0 = 0, and 1 ≠ 0, then, you find a non trivial linear combination.
(The trivial one is 0•0)
golden drum
If the set is linear dependent by 1.
Then there exists a1,..., an not all 0, such that, a1v1 + ... + anvn = 0
By this, you can take vj such that aj ≠ 0, then
ajvj = -a1v1 + ... + -anvn.
You can divide by aj, because is a field and get vj
How can I write in LaTeX here?
I think the two are equivalent, but, is tricky to think in other terms, when, you have only one term
gray dust
$\LaTeX$
stoic pythonBOT
RokettoJanpu:
golden drum
I see, thanks
gray dust
have fun
golden drum
@molten hill
molten hill
@golden drum it looks like you didn’t even read what I wrote
You basically just said the same thing again
I think I’ve figured it out though, the second definition there isn’t fully rigorous and only applies to sets with length > 1
golden drum
You ask if the second is derived from 1
The tricky thing is "Other terms"
See this
In Definition part
Actually, you need the linear combination to be non zero, because, always exists a linear combination that goes to 0, $a_i = 0, \forall i$
stoic pythonBOT
Enigsis:
chilly solstice
Quick question. If I have a $n\times n$ matrix $A$ and $\eta := \max_{i}(-a_{ii}) > 0$, can I conclude from this that $a_{ii} < 0 $ or $a_{ii} \le 0 $ ?
slate haven
@spiral star thank you for that detailed solution ^^. Now that i read your solution i feel like i have been very close 😂 what bothered me yesterday was that i wasnt sure whether the map f was well defined, since i thought that even though phi is surjective there will be a preimage but i wasnt sure whether there will be exactly one. The "well-definedness" can be justified via "Prinzip der Linearen Fortsetzung" which you more or less used in the step where you defined f, right ?
oops should have been 😄 not 😂 discord is turning ':'D' into the laughing smiley lol @spiral star
stoic pythonBOT
Schuams:
chilly solstice
I do not know how to solve $\max_i(-a_{ii}) >0 $ wrt. $a_{ii}$ lol
stoic pythonBOT
Schuams:
dusky epoch
no you can't. take A = diag(-1, 3, 4)
max(-a_ii) = 1 but not all diagonal entries are nonpositive
chilly solstice
thanks!
chilly solstice
Here is a slight modification of my previous question.
Let $A$ be a $n\times n$ matrix and $\eta := \max_{i}(-a_{ii}) > 0$. Can I conclude from $1+\eta^{-1}a_{ii} \ge 0$ that $a_{ii}<0$ ?
stoic pythonBOT
Schuams:
chilly solstice
If $||\cdot||$ is matrix norm, how does this then follow $||P^n|| \le ||P||^n$ ?
stoic pythonBOT
Schuams:
dusky epoch
define matrix norm?
do you require the norm to be submultiplicative, i.e. $|AB| \leq |A| \cdot |B|$?
stoic pythonBOT
Ann:
chilly solstice
yes! @dusky epoch
Yes, | | A B | | = | | A | | · | | B | | is one of the proberties of the matrix nomr
shrewd mortar
Yeah lol
So just induct
Like this
||P^n|| = ||P * P^(n-1)|| <= ||P||||P^(n-1)|| and then repeat that with ||P^(n-1)||
like this
stoic pythonBOT
Archsys:
gritty frigate
There is any problem if I switch columns on a parameter matrix ?
gritty frigate
Well, a have a system with some parameters
I need to find the numbers that allow all possible system states
spiral star
@slate haven yes will use the fact that we can fully determine a linear map by knowing the images of a basis. that's quite common when you argue in a finite dimensional setting. maybe i can clarify the rest by unwrapping my proof from earlier even further
slate haven
@spiral star i think i understood your first proof already for the most part but now i definitely do understand it 🙂 thanks a lot for your very thorough proof, i appreciate it.
gritty frigate
@dusky epoch
dusky epoch
ok so alpha is a parameter?
well, the parameter, as far as i can tell
what are you asked to do with this system?
crisp rock
I think this is the room I want
I am in a linear systems course and we are doing Laplace transforms and using the program Matlab does anyone know how to use it for this topic
wintry steppe
help
plis
x-intercept is when the line hits the x-axis, i.e. when y = 0. y-intercept is when the line hits the y-axis, i.e. when x = 0
now solve
I need help in this one now Plis
dusky epoch
<@&268886789983436800> racial slurs
wintry steppe
wait who?
dusky epoch
@wintry steppe help does not mean doing it for you, and also this is the wrong channel
dusky epoch
wintry steppe
ok thanks
wintry steppe
nothing
jagged pendant
banned
dusky epoch
thank you
chilly solstice
<= not =
@dusky epoch I read it by mistake as "=", so that was why my life was so hard 😄
chilly solstice
A $n\times n$ matrix $A=(a_{ij})$ sub–intensity matrix has non–negative
off diagonal elements, negative diagonal elements and non–positive row sums. Precise definition
$$a_{i j} \geq 0, i \neq j \text { and } a_{i i} \leq-\sum_{j=1 \atop j \neq i}^{n} a_{i j}, i=1, \ldots, n$$
stoic pythonBOT
Schuams:
chilly solstice
The one million $ question. Is a sub-intensity matrix always diagonaziable?
midnight hedge
Can someone explain what happens to x1(t) and x2(t) as time goes to infinity?
How would i check what happens do I just plug in large values for t and see what happens to the graph?
dawn remnant
they go to 0, because there's an exponent there
How would i check what happens do I just plug in large values for t and see what happens to the graph?
uhh, no? You can just prove it using the definition of a limit.
midnight hedge
that's why I'm confused. In the book example how did they get a spiral?
its not the same but similar
so would it spiral to 0
dawn remnant
indeed
without the exponent it's the equations of something rotating around the origin(plot it in Desmos if you want)
the exponent makes it also decay exponentially towards 0.
midnight hedge
okay thanks
hollow finch
is there some sort of sense for an orthogonal matrix under a different inner product definition? or is that just silly?
cursive osprey
raw sand
do you know what the dot product = 0 tells us
raw sand
it meants they are perpendicular
so u and v are both perpendicular to w
if they are both perpendicular to the same vector what could you say about them?
open pumice
they are equal?
raw sand
they are parallel
and then w must be 0 to create a dot product of 0 with two vectors
so now you know w must be 0
so that narrows it down to c and d
limber sierra
if they are both perpendicular to the same vector what could you say about them
they are parallel
hold on
what?
consider the unit vectors in R^3
e_1, e_2 are both perpendicular to e_3, but they arent parallel
lucid cedar
Im assuming this room is open since last question was an hour ago. Im trying to answer this question:
I believe basis of the range of T is a linearly independent combination of A's columns
but the null space of T, Im not sure how to solve for that
oh wait am i solving for Ax=0?
gray dust
I believe basis the range of T is a linearly independent combination of A's columns
idk wym
oh wait am i solving for Ax=0?
ker(T) consists of all x where T(x)=0. the q wants a basis of ker(T) not ker(T) itself
lucid cedar
- I mean that the columns are vectors of T(ui) for some ui in this case it says standard basis so T(1,0,0) = (1,0,-1)
so those columns taken as a linear combination should have a basis in there somewhere
and 2. Wouldn't solving for T(x)=0 be a good start to solving the basis. I remember something about this from my sophomore linalg class but that was a while ago
gray dust
what's ui, and
so those columns taken as a linear combination should have a basis in there somewhere
your wording is all over the place, just state concisely how to get a basis of im(T)
sure solving T(x)=0 is a step in the right direction, just remember the q wants a basis of ker(T) not ker(T) itself
lucid cedar
Im sorry im a CS major. Much of my unis CS is spent avoiding rigorous math infavor of handwavy explanations. I am taking more math on the side because I wanted a better treatment of the material but clearly i am not there yet. I still dont quite speak the language
but let me try again
wintry steppe
don't be sorry for being in cs 
lucid cedar
Given T in L(V,W) and v1,...,vn is the basis of V: I am positing that I can find a basis of W by taking a linear combination of T(v1)+T(v2)+T(v3)+...T(vn). I am further positing that the columns of that matrix in the problem statement are T(v1)+T(v2)+T(v3) IE a linear combination of the columns will contain the basis of W (which is the Range of T).
i am not sorry for being in cs
i am sorry for being dumb
I guess what i said is not actually true. W could have a higher dimension than V
gray dust
your wording suggests you're not sure what linear combo means. the i'th col of A is T(i'th standard basis vector). T's codomain is a very distinct idea from T's image
lucid cedar
what about my wording suggest that.
its prolly the only thing im speaking about right now that im certain I understand
but regardless
this statement: "the i'th col of A is T(i'th standard basis vector)" is what im saying
are you saying thats not true? im not sure what ur getting at
my claim is that T(v1,v2,v3) (IE column 1,2, and 3) are elements of W and can be reduced (or expanded) to a basis
gray dust
you use linear combo outside its defn
i'd rather you use what's given rather than rename so many things. T is in L(R^3). refer to the standard basis of R^3 rather than arbitrary v1,v2,v3
my claim is that T(v1,v2,v3) (IE column 1,2, and 3) are elements of W and can be reduced (or expanded) to a basis
this is half ok. there's still this which you didn't address
T's codomain is a very distinct idea from T's image
you seek a basis of the image which generally is distinct from the codomain
lucid cedar
Yes
What i am positing
is that
the image of T with respect to the basis of its domain can be expanded or reduced to its codomain
Many of axlers results depend on letting the basis of W be T(v1,...,vn),u1,...,um where u's are vectors added if W is a larger dimension
If this isnt true then I have a very fundamental misunderstanding
This is the idea im talking about:
gray dust
the wording is making it hard for me to understand you. idk what reducing im(T) means
lucid cedar
oh I can answer that
axler uses expand and reduce as term to say that a set of vectors is an incomplete basis
so you can expand (1,0) to a basis
its missing (0,1)
and when he says reduce to a basis he means that there are too many vectors and some need to be removed
gray dust
when you say basis, always mention the vector space the set of vectors is a basis of
just basis by itself means nothing
gray dust
no
lucid cedar
okay
Im a bit frazzled im sorry. I feel like im trying to ask where the bath room is in a foreign country
gray dust
you have trouble using linalg terminology. i'd go back and review defns
my claim is that T(v1),T(v2),T(v3) (IE column 1,2, and 3) make up a spanning set of im(T) and can be reduced (or expanded) to a basis (of im(T))
here's an edit. this is the closest thing to a coherent statement
you probably know WHAT to do but have til now talked about it in a very roundabout way, abusing terminology along the way
it's nice you want to study linalg but at this point there's a prerequisite amount of prior ladr content to learn before continuing
polar sparrow
The reduced row echelon form of a singular matrix always has a row of zeros, right? I'm confusing myself
native rampart
Yes
gray dust
A is singular iff A's rows are linearly dependent iff a nontrivial linear combo of A's rows gives the 0 vector
polar sparrow
yeah that's good, my entire understanding of LA was just breaking down because of how my handbook was phrased
thanks
gray dust
no prob
lucid cedar
This is my attempt at finding the nullspace of this matrix
in order for that combo to equal (0,0,0) then x1,2,3 must all be 0
so I said that this was the basis
oh wiat
no
x1 x2 and x3 must all be 0
so the basis of the null space would be (0,0,0)
and the basis of the range is what i have above
4 hours to answer 1 question not bad
lucid cedar
i got it (1,-1,1) i had to get my notes from my first lin alg class. I dont see how that process fits into axlers content but whatever its done
thanks to everone that spent time helping me
fallow patio
is one of this incorrect or are both of them valid ways to complexify a real inner product space?
top one is from LADW and bottom one is from advanced linear algebra by Roman
dusky epoch
these are two different things
one talks about defining addition and the other talks about the inner product
fallow patio
aren't they the same? the sentences above both images explicitly say that it is about complexifying an inner product space
dusky epoch
oh wait nvm
i misread
can i have some more context for the first one
cause the inner product as defined there doesnt look conjugate-symmetric to me
severe cedar
Is there a proof that singular values are always positive or does this property just follow by definition?
waxen jacinth
How do I do this? I think I am missing something out or I'm going through a dumb minute. I assigned variables to the matrices and then isolated X. The other side with X isolated was B/(A-C), where A B C are the matrices in their corresponding order
dusky epoch
how do you divide by a matrix 
dusky epoch
no, how do you divide by a matrix
waxen jacinth
Oh ur not referring to me xd
dusky epoch
what does 1/M mean when M is a matrix
waxen jacinth
It's the inverse
dusky epoch
i am referring to you
so?
if it's the inverse then write it as the inverse
right before that you should've had (A-C)X = B
waxen jacinth
I did
waxen jacinth
What xd
dusky epoch
you should've had $X = (A-C)^{-1}B$
stoic pythonBOT
Ann:
dusky epoch
THEN WHY DID YOU WRITE IT AS DIVISION
waxen jacinth
Cuz I was in algebraic form
waxen jacinth
When I switched back I separated
dusky epoch
we dont divide by a matrix
waxen jacinth
Into 1/(A-C) * B
dusky epoch
THEN DONT FUCKING WRITE IT
dusky epoch
do not write nonsensical notation
waxen jacinth
SEcondly, the notation is useless if we both reach the same, wrong result
Because the operations are the same
And I'm just getting started on these
And it's not giving me a good result
So either help like a normal person or don't try at all, and avoid profanity
dusky epoch
then maybe you aren't calculating (A-C) or (A-C)^-1 correctly
also you will not tell me to avoid profanity
waxen jacinth
Yeah, that could be my problem
I subtracted first
Then found the inverse of the subtraction result
dusky epoch
yeah you aren't showing me any intermediate calculations so i cannot tell you whether or not you did anything wrong
waxen jacinth
So after isolating X, I got (A-C)^(-1) * B.
I did the subtraction of A-C
I then Inverted the result
Then multiplied the inversion by B
Those are the steps. I dont have anything written down as Im using a calculator
But that's basically wt Im doing
dusky epoch
??
waxen jacinth
Idk lol, just asking
As I said, just starting out
I was told that AB is different from BA
So Im just asking if there are any other such rules to follow
waxen jacinth
Yeah, but that is basic algebra right?
I'm confused xd
Ok I'll just write the entire expression and algorithm, and we'll see if there's fault in them
AX+B=CX
AX-CX=-B
(A-C)X=-B
X=(A-C)^(-1) * (-B)
pale orchid
-B
waxen jacinth
Right ye, But I did do that
K Ill try it one more time
Maybe I forgot it as I did here
pale orchid
yeah, it's missing in the last step
waxen jacinth
Oki, give me a min
fallow patio
can i have some more context for the first one
From my understanding, since the complexification is the sum of two real inner products, then it is a real number, so it is not affected by conjugation
wintry steppe
hint: if such a matrix P exists it's gonna be orthogonal no matter what
which reduces the problem to something you might be familiar with
dim venture
so existence will prove the matrix is orthogonal
wintry steppe
right
dim venture
not quite sure what you're hinting at
wintry steppe
have you seen changes of bases?
dim venture
are you talking about the transition matrix / change of basis?
wintry steppe
that yes
dim venture
ah okay alright
wintry steppe
one way to approach these problems where you're asked to show something exists is to see what kinds of properties that thing should have before you even start constructing it
dim venture
I was thinking of using this corollary
dim venture
so should I be showing that this matrix P is the conversion (not transition whoops) matrix? the way I thought of it previously was to consider the conversion matrix as P, but I didn't think it was sufficient because it didn't seem like I proved existence
wintry steppe
actually now that you've posted the second picture i think you should go through with that instead
using that gives you a really nice way to do this problem
gloomy hatch
wintry steppe
@dim venture you don't need any fancy reductions actually, you had the right idea
dusky epoch
the proper notation is ecks cube plus ecks squared minus four times ecks minus four
(/j)
wintry steppe
overthinking problems is my specialty, at least when they're not homework problems
try using that result to express the basis b_i in terms of the a_j
dim venture
ah yeah I know what you mean I thought I was really closing to proving it to. I just can't seem to formulate it into a proof. I had an idea with the corollary but it didn't end up working, though that might just be me not following through / making a mistake
wintry steppe
should always follow through, even if you think it might not work
you can still get something out of an attempt
(at least in my experience)
dim venture
laughs in part marks
wintry steppe
lol true
dim venture
alright, giving it another shot, will report back later
we know that the size of P is a nxn matrix right?
wintry steppe
yes, otherwise Pa_i = b_i doesn't make sense
dim venture
This is what I've got, which means I have all vectors from the b basis in terms of the vectors from the a basis
My idea to continue is to have a nxn matrix with each term as the coefficients from the expression of b_i in terms of a_1, ... , a_n
wintry steppe
yes this sounds like a good idea
dim venture
is that not P directly?
wintry steppe
you'll end up getting a change of basis matrix, and that has to be unique, so yes, it will be P
dim venture
is this a formal existence proof?
wintry steppe
of course
dim venture
so I just have to verify the multiplication holds and is true, then show orthogonality, and uniqueness?
(well as you said orthogonality is fine)
dim venture
okay, no problem