or maybe not idk
#linear-algebra
dapper gorge
zinc timber
no, that doesn't mean the space is real
self adjoint operator are used extensively in quantum mechanics
dapper gorge
no obviously is not, but maybe the properties it would have it would resemble those of a real space
zinc timber
ex the Strum Liouvill operators are self adjoint so they have a spectral decomposition
dapper gorge
where the characteristic polynomial splits in R
I was asking because in the chapter I'm reading the author (it's friedbergs,.. book) didn't include an explicit theorem on when self-adjoint complex operators exist (it is inmediate from other results nontheless)
zinc timber
exists? like why shouldn't they exist?
wintry steppe
hey all im a bit confused on how to show this
i think orthogonality affects this somehow?
not sure how to go from there tho
viral magnet
what are you stuck on?
you have a biconditional statement. which direction have you tried first?
fringe fjord
Is y arbitrary or something you already have assumptions about?
viral magnet
arbitrary would be my interpretation
fringe fjord
In that case $\mathbf{y+X\beta}$ is \emph{itself} arbitrary, and writing the arbitrary N-vector that way seems to be a feint.
wintry steppe
you have a biconditional statement. which direction have you tried first?
well i think u need something with orthogonality
i think going from XT(y-xB) = 0 -> y-XB might be easier?
stoic pythonBOT
Troposphere
viral magnet
is there some theorem about range direct product null space equals the span
https://en.wikipedia.org/wiki/Rank–nullity_theorem
yea this one
fringe fjord
If we set z=y+Xbeta, then it should be clear (from the definition of matrix multiplication) that X^Tz=0 iff z is orthogonal to each row of X^T, and those rows are the columns of X, i.e. the generators of its range.
tired fossil
can anyone help me, i keep doing this and am not getting the same answer as the back of the book. 2.f)
this is the answer
gray dust
,w plot 5/(x^2+x-6)+1/(x^2-5x+6)-6/(x^2-9)
stoic pythonBOT
gray dust
in general dependency relations arent unique
u may get a different linear combo equaling 0
wintry steppe
hey all im super confusedo n what this is asking us
how do u even simplify something like that
fringe fjord
"simplify" seems to be a bit of a stretch -- it looks like it just wants you to write it out in components rather than matrix algebra.
Definitely looks sloppy of the problem setter to write $x_{i,1}$ in one line and then suddenly switch to $x_{1,i}$ in the next...
stoic pythonBOT
Troposphere
grand fog
hi guys, could someone tell me what this means?
im pretty sure it means for all real numbers
but im not sure
gray dust
belongs to the set of real numbers
grand fog
okay, thanks man!
wintry steppe
"simplify" seems to be a bit of a stretch -- it looks like it just wants you to ...
hmm i still am not sure how to set it up though
tired fossil
have a question as well. if the vector space is P2. What dimension would the subspace V={1-x,1-x^2,1-2x+x^2} have?
fringe fjord
Definitely more than 1, since your vectors are not all parallel.
Definitely at most 3, since that's how many vectors you have.
(I'm assuming you mean V is the span of {1-x,1-x^2,1-2x+x^2} instead of {1-x,1-x^2,1-2x+x^2} itself, which isn't even a subspace).
So it comes down to whether your set is linearly independent, or not.
wintry steppe
idk why i thought that was for what i posted but then i realized its not
fringe fjord
idk why i thought that was for what i posted but then i realized its not
Sorry. I'm afraid I don't have more intelligent responses about your problem.
wintry steppe
no worries haha im confused too
gray dust
@wintry steppe its not so much that tropo cant fully answer the problem
wintry steppe
oh like without giving it away?
gray dust
its that finding a basis of V takes a nontrivial amount of work
yeah so do it urself, recall how to find a basis of a space
then dimV is the size of ur basis
wintry steppe
right but if x = [x0 x1] how can u even multiply that with something infinite?
fringe fjord
I think X is N×2.
wintry steppe
so it would be 1 x 1, thus would it just be x0 + x1 + y?
fringe fjord
Note how in the definition of X that looks like a single column, the entries are actually x_i^T, where each x_i is a vector with two entries ...
fringe fjord
Just to add to the confusion, the boldface x_0 and x_1 with indices are defined to be the columns of X, so each of them contains elements from all he lightface x_i's.
wintry steppe
oh gosh
fringe fjord
(That's horrible notation but there you go ...)
wintry steppe
hmm yea i have no idea how to multiply those with y
so if its XTy then its
2 x N for XT and i think N x 1 for y?
so 2 x 1 for the result
would it just be x0y and x1y for both of the rows?
fringe fjord
My advice would be to write out the matrices and vectors with all the elements explicitly visible (which the question doesn't deign to do itself), to get a feeling for what's going on.
You can use "..." for the middle indices between 1 and N.
tired fossil
Im confused as to where to start on this question
so for example let A=[a,b//,c,d]
then a+c=b+d for the rule that they have equal column sums
but then where do i go?
fringe fjord
Check each of the parts of the definition of "subspace" in turn.
tired fossil
Check each of the parts of the definition of "subspace" in turn.
im assuming this is a reply to me
fringe fjord
Yes.
tired fossil
okay i will look
tired fossil
Yes.
It dows fit under the definition of a subspace
does*
but now, how must i find the dimension? That is where I am really confused
fringe fjord
I think the answer is intended to be something like "the dimension is such-and-such because here is a basis for the subspace with that many elements. I know it's linearly independent because [...] and I know it spans the entire subspace because [...]".
limber sierra
a hint for finding a basis: how many "degrees of freedom" do you have? that is to say, how many things can you "meaningfully change"?
fringe fjord
You're probably expected to puzzle out a basis by unsystematic exploration.
tired fossil
its not, there is not answer in the pack of the book for a, but for b it is the same thing with 3x3 matrix and it says the answer is 7
limber sierra
note that, if we pick specific values for 3 entries of the matrix, the 4th entry is "forced"
this might give you an idea of where to start looking
tired fossil
note that, if we pick specific values for 3 entries of the matrix, the 4th entry...
so like [1,2//c,6] for example?
limber sierra
right, the value of c is "forced" to be 7
theres no freedom
since we need 1 + c = 2 + 6
tired fossil
yes, that would make sense
limber sierra
so intuitively, theres only 3 meaningful "degrees of freedom", so we'd expect the dimension to be 3
tired fossil
this is b, and the answer for be is 7, Im not sure how they get it
limber sierra
to be clear, this is not a formal proof
its just a starting point
the logic for (b) is similar, we have 7 degrees of freedom instead of 3
by pidgeonhole, 7 is the minimal number such that an entire column of a 3x3 matrix is filled
limber sierra
construct a basis and show that it satisfies the definition of a basis
i.e.:
- The elements are linearly independent, and
- They span the entire set.
tired fossil
The basis would be 4
limber sierra
so start by picking 3 linearly independent matrices that satisfy the column-sum rule
what?
tired fossil
if im correct
limber sierra
do you know what a basis is?
tired fossil
sorry the max dimension
nocturne jewel
{4} spans the space \s
tired fossil
would be 4
limber sierra
okay, then try to construct a linearly independent set of 4 matrices that each satisfy the column sum thing
||you won't be able to||
tired fossil
a basis is a set of vectors in a vector space that satisfy 1. independent, 2 v=span of that set of vectors
If i am correct^
nocturne jewel
i.e.:
- The elements are linearly independent, and
- They span the entire set.
yes.
limber sierra
there's an equivalent characterization as the MAXIMAL linearly independent set
i.e.
- The elements are linearly independent, and
- if we add another element from the set, NO MATTER WHAT ELEMENT WE ADD, the basis will no longer be linearly independent
this is what i was hinting at exploiting earlier with the "3 degrees of freedom" argument
fringe fjord
In that case just see how many independent matrices in the subgroup you can find.
At least that will be a start, and while looking for them you might notice some patterns.
tired fossil
like i know how to do this if it gives me a span with real numbers, im am just so lost when it is given arbitrarily
tired fossil
In that case just see how many independent matrices in the subgroup you can find...
these are arbitrary matrices?
fringe fjord
No, there are matrices with the specific property that the column sums are the same.
tired fossil
yes, so V={[a,b//c,d]|a+c=b+d} i understand that, but where do i go after
fringe fjord
It sounds like your psychological problem here is that you cannot immediately see how to get all the way to an answer. The way to overcome that is to start playing around with the situation, see what you can reach, see if you can find something that feels like it would be an ingredient in the answer.
Write down some example matrices!
See if you can find a way to generate matrices in the subgroup that you'll know will be linearly independent!
tired fossil
Are these matrices with real numbers?
fringe fjord
Yes.
(At least that ought to be the default assumption on this level when nothing else is specified in the problem).
tired fossil
okay
arbitrarily lets say [1,1//0,0],[0,0//1,1], [1,0//0,1], [0,1//1,0]
how is that a dimension of less than 3?
nvm i see it now, i used a matrix
dapper gorge
In general, when does an n-th root of a linear map or matrix (over an algebraically closed field) does not exist? A has an n-th root if A=B^n for some linear map or matrix B.
or do they always exist? I don't know, haven't thought about it. The existence of n-th roots when A is nonsingular is clear
Certainly, if n is big enough relative to the size of the matrix (dimension of the space), then if the matrix/map is singular, there is no n-th root for dimension reasons
you can say something similar from looking at the rank of the matrix/map.
probably not too interesting anyway
besides this trivial cases, I wonder if there are other limitations
zinc timber
they always do
ex take the diagonal matrix of one of roots of xⁿ-1
ok wait
I thought you are talking about nth root of I
errant palm
hello, i took a quiz and was given the solutions to the quiz, and I am confused about what i did wrong.
dusky epoch
well we can't say what you did wrong if all we see is your assertion in the answer that the (1,2) entry of A is 1
wintry steppe
wintry steppe
happens to everyone
mystic frigate
If you do the inner product of 1xn matrix with nx1 matrix and have an answer as a 1x1 matrix, are u allowed to just write the number like 7 or do you have to put it like [7]
wintry steppe
i think there's no harm in pretending 1 x 1 matrices and scalars are the same thing
exotic wedge
Any hints ?
slow scroll
Any hints ?
the kernel of pi is W
its kinda like a generalization of the first isomorphism theorem if you're familiar with that
exotic wedge
the kernel of pi is W
Yeah I showed that in the previous part but can't see how to use that
exotic wedge
its kinda like a generalization of the first isomorphism theorem if you're famil...
No not really
slow scroll
Okay, so lets let $\varphi : Maps(V/W, V') \to U \subset Maps(V,V')$ be the map in question (where $U$ is the subspace of maps $T$ such that $T(w) = 0$ for $w \in W$). Can you see that $\varphi(f) = f\circ \pi$ is well-defined?
stoic pythonBOT
kxrider
slow scroll
and for that matter, a linear map as well?
exotic wedge
Okay i think i kinda get it
Now we show injectivity and subjectivity right ?
Surjectivity*
slow scroll
yep
exotic wedge
Injectivity follows from the kernel ?
slow scroll
it would suffice to show that ker(phi) = {0} for injectivity, yes
exotic wedge
What about surjectivity ?
slow scroll
for surjectivity, you would just have to use the definition: that given a linear map g : V --> V' such that g(w) = 0 for all w in W, there is a linear map f : V/W --> V' such that g = phi(f) = f circ pi
slow scroll
npnp
ripe shale
For a particular map
I'm looking at the proof of part (b)
I came up with my own proof, which I think is a little cleaner. I wonder if it's valid
Suppose $Tu = Tv$, then $Tu- Tv = 0$, which means that $u-v \in$ null T. By 3.85, this implies, $u+$null T = v + null T. This shows that $\tilde{T}$ is injective
stoic pythonBOT
Minh Pham
ripe shale
slow scroll
Suppose $Tu = Tv$, then $Tu- Tv = 0$, which means that $u-v \in$ null T. By 3.8...
yes, this is correct
proper cradle
Ans is false, How to get it?
dusky epoch
try to think about what the eigenvalues of A can be
proper cradle
only 1 is possible i think
balmy sand
Hi, Can we say in a surjective function a co-domain sets and image sets are equal?
zinc timber
yes that's what being surjective means
balmy sand
@zinc timber Thank you.
heavy crown
I don't understand it why is the next statement true/correct?
Let A be normal matrix, A isn't the zero matrix.
It's impossible that: A^(2021) = 0
proper cradle
I don't understand it **why is the next statement true/correct**?
Let A be norm...
A is normal matrix then D=UAU*, where D is diagonal matric and U is unitary matrix. A=UDU* implies that A^2=UD^2U*, ... , A^(2021)= UD^(2021)U*, since D is non zeros because of non zero A, Hence A^(2021) is not Null matrix.
heavy crown
A is normal matrix then D=UAU*, where D is diagonal matric and U is unitary matr...
oh wow yes you're right, thank you :]
proper cradle
👍
dusky epoch
A is normal matrix then D=UAU*, where D is diagonal matric and U is unitary matr...
are you sure you didnt mean UDU*
as-is, $(UDU)^2 = UDU^2DU$
stoic pythonBOT
Ann
proper cradle
are you sure you didnt mean UDU*
yes yes sorry
@heavy crown please check the mistake, sorry
crisp pilot
I have absolutely no clue how to do part 3.b. I could do part a, but in part b I'm very confused as to how to use lagrange multipliers with matrices.
Can someone help me out ? Even tagging helpers didn't help in the help channel
nocturne jewel
so span is essentially c1v1 + ... + cpvp?
yes, span of a set of vectors is the set of linear combinations
zinc timber
wintry steppe
lagrange multipliers are linear algebra
zinc timber
lavish jewel
downgraded from ryu sama to ryu kisama
dusky epoch
zinc timber
cunning pier
I want to calculate the dimension of U
however the triplet (u1, u2, u3) is not a basis, because they are linearly dependent
Could someone give me input on this?
zinc timber
stick them up in a row-matrix and calculate the rref
cunning pier
rref?
dusky epoch
reduced row echelon form
cunning pier
ahh
heavy crown
anyone?
lavish jewel
the important parts are that it's a projection matrix for a projection p: R^2 -> R^2
heavy crown
the important parts are that it's a projection matrix for a projection p: R^2 ->...
yes, so we know it looks like this, but I got stuck in the next step
lavish jewel
to be honest more info is needed, since the 0 matrix works here
unless i'm missing something
heavy crown
it can't be the 0 matrix because it is the projection matrix
heavy crown
unless i'm missing something
yea I feel like more info is needed too but this is all the information given in the question, question from an exam from last year 😭
lavish jewel
doesn't the 0 matrix satisfy the definition of a projection mat?
it's idempotent
unless i'm missing something, the question is lacking the bit that it should project onto a dim 1 subspace
then you can use that info to find the orthogonal complement of the null space
heavy crown
doesn't the 0 matrix satisfy the definition of a projection mat?
A projection matrix should have eigenvalues of 1 and 0.
If it is the zero matrix it means Av = cv, in our case Av = 0 = 0 * v. always equals to 0.
So it can't be that Av = 1 = 1*v (which we seek because we need the other eigenvalue).
heavy crown
then you can use that info to find the orthogonal complement of the null space
I have no clue :[ thanks for your help
lavish jewel
orthogonal complement
heavy crown
I know what you mean but I don't know where to begin with because there isn't any info there haha
lavish jewel
a zero matrix has eigenvalues 1 and 0
heavy crown
really hmm
wintry steppe
orthogonal complement
but i guess my question is what does that actually mean
lavish jewel
the image of the matrix X is a subspace spanned by its columns
you can take those columns and do gram schmidt to construct an orthonormal basis for this subspace
then, extend this to an orthonormal basis for all of R^n
the extra vectors you added are a basis for the orthogonal complement of X
any vector in (im X)^perp is orthogonal to any vector in (im X)
heavy crown
I understand you I think. We did it in the class in general case.
We said V=R2, U=sp{u}. Av=Pu(V). We look for finding A which is the projection matrix.
u= {(a, b)} is an orthogonal basis to U subspace. (No need for gram schmidt because it's dim1 so it's obviously orthogonal)
So using the projection formula. We found a general representation for that A matrix that meets those requirements.
heavy crown
yes, so we know it looks like this, but I got stuck in the next step
and what I wrote on the left side is the general representation we found
heavy crown
the important parts are that it's a projection matrix for a projection p: R^2 ->...
oh I understood how to answer it now
thank you)
lavish jewel
i really don't think your problem has a unique solution
there should be 2
the one listed there, which makes the projection matrix rank 1, and also the 0 matrix is valid
heavy crown
continuing from here, we find that a=-2b
Which means that we had u=sp{(a,b)} = sp{(-2b,b})
Now the solution is really simple because as we see In what we need to prove we need to multiply A by (-2,1)
Which belongs to u (the sp)
And according to the projection matrix Pu(v), if you put there v that belongs to u. So Pu(v)=v (=Av)
Which is the exacly the proof we're looking for
lavish jewel
seems wrong in the first step
wintry steppe
can you not just move them around bc of the commutative property of the dot product?
dusky epoch
this is not dot product this is matrix multiplication
lavish jewel
commutative property of dot product is <u,v> = <v, u> if it's over R
this means u^T v = v^T u
where u and v are vectors in R^n
that's not what you did
you said u^T v = v u^T, which is false
u^T v is a scalar, v u^T is a rank 1 matrix of size n x n
wintry steppe
hmm but
wintry steppe
this means u^T v = v^T u
if i did this
then im not sure how i would be able to get the transpose of A as well
lavish jewel
you didn't do that
you said u^T Av = Av u^T
substitute Av = w, because it's a vector
then you said u^T w = w u^T
when it should be w^T u
lavish jewel
that's what i did
just substitute back and use the properties of the transpose you were given
lavish jewel
what distributive property
wintry steppe
dot product
lavish jewel
no distribution is happening
dusky epoch
aren't yall overthinking this
stoic pythonBOT
Ann
lavish jewel
aren't yall overthinking this
yes
wintry steppe
sorry im just starting rn ;/
tough veldt
c = (c.n)n+(c.m)m
For unit basis {n, m} this is true, right?
What's the result called, tryna look it up
grizzled sentinel
Write a monomial and a trinomial with the variable x of your choice. The monomial must have a negative coefficient.
how do i do this
lavish jewel
Write a monomial and a trinomial with the variable x of your choice. The monomia...
please check out #prealg-and-algebra
lavish jewel
**c** = (**c**.**n**)**n**+(**c**.**m**)**m**
For unit basis {**n**, **m**} thi...
by unit basis you mean orthonormal?
lavish jewel
well, unit usually refers to normal here, from norm. the ortho part is orthogonality (confusing nomenclature, i know). n is orthogonal to m?
tough veldt
no
lavish jewel
then it is not necessarily true
take for example a basis [1,0] and [1/sqrt(2), 1/sqrt(2)]
and a vector c = [1,0]
the procedure yields [1,0] + [1/sqrt(2), 0] instead of [1,0]
tough veldt
@carmine echo ^ nope.
Urgh
Ok, so basically the problem is this
We are given 2 non-coincident planes in R^3
r.n = a
r.m = b
And want to find the equation of the line of intersection
r = k(nxm) + c
This is the progress, but stuck on finding c
lavish jewel
what's r here
tough veldt
the variable
lavish jewel
you can write this as a matrix vector equation and just solve that
[n,m]^T r = [a,b]^T
lavish jewel
oh oops, lemme fix
tough veldt
I was expecting something nice kinda
lavish jewel
well, yes
the matrix is 2 x 3, so it has a nontrivial kernel
a null space of dimension at least 1
meaning the planes intersect at at least a line
or at most a plane
tough veldt
Without going into coordinates I can't seem to write c
lavish jewel
solving this will give you the parametric equation of a line
tough veldt
in terms of the other vectors explicitly?
carmine echo
**c** = (**c**.**n**)**n**+(**c**.**m**)**m**
For unit basis {**n**, **m**} thi...
uhh, this was for R² though... for R^3 you can go for c = (c.n)n + (c.m)m + (c.mxn)(mxn) m,n,mxn have hats plz understand🤦♂️
m, n linearly independent
carmine echo
Yeah, xm + yn would still cover the plane containing vectors m and n, while mxn covers the 3D being perpendicular to both vectors
tough veldt
**r**.**n** = a
**r**.**m** = b
Ok for now lets assume n, m are unit (a and b can just be rewritten)
c.n + (c.m)(c.n) = a
This aint necessarily true is it?
carmine echo
r.n = a => c.n = a then? hmm?
lavish jewel
hmm?
tough veldt
r.n = a => c.n = a then? hmm?
yes
carmine echo
**c**.**n** + (**c.m**)(**c.n**) = a
if this true then c.n +(c.m)(c.n) = a + (c.m)(c.n) = a => (c.m)(c.n) = 0?
tough veldt
IF it is true
This is what I get after plugging your formula for c into the 1st plane equation
unless ab = 0, it aint true.
But you're close.
But no, this feels like an impossible task with n, m not necessarily orthogonal
There's no way this formula for c doesn't involve a and b
You can't have c inside the formula
carmine echo
For $\bR^3, , \vec{c}=(\vec{c}\cdot \hat{m})\hat{m}+(\vec{c}\cdot \hat{n})\hat{n}+(\vec{c}\cdot \hat{m\times n})\hat{m\times n}$, is false?
tough veldt
I'm quite sure it is.
stoic pythonBOT
Ansh
carmine echo
hmmm
tough veldt
If m and n arent perpendicular you encounter problems. Just dot that formula with n
And you get c.n = (c.m)(m.n) + (c.n)
We require (c.m)(m.n) = 0 for this to be true
And similarly, we require (c.n)(m.n) = 0
Hence c.m = c.n = 0 or m.n = 0
carmine echo
okay yeah right 🤦♂️ I need m.n = 0 for this
tough veldt
this feels like idk. Super annoying we can't get a closed form in terms of a b n m
carmine echo
in which case you can just pick 3 non-independent vectors and replace mxn with that one... mxn is a make do in case you got two perpendicular vectors m and n
tough veldt
c = (c.n)n + (c.m)m + ???
c = (c.n)n + (c.m)m + ???
c.n = (c.n) + (c.m)(n.m) + ???.n
c.m = (c.n)(n.m) + (c.m) + ???.m
carmine echo
not n, m... another vector say p or something
carmine echo
condition is am + bn + cp = 0 has no soln.
tough veldt
-c.m = p.n/(n.m)
-c.n = p.m/(n.m)
Find p so this is true
-(c.m)(n.m) = p.n
-(c.n)(n.m) = p.m
carmine echo
~~-c.m = p.n/(n.m)
-c.n = p.m/(n.m)~~
Huh?
c.n = (c.n) + (c.m)(n.m) + (c.p)(p.n)
-(c.m) = (c.p)(p.n)/(n.m)
carmine echo
Seriously? you got the first equation wrong from the start
carmine echo
c.n = (c.m)(m.n) + (c.n) + (c.p)(p.n)
c.m = (c.m) + (c.n)(n.m) + (c.p)(p.m)
c.p = (c.m)(m.p) + (c.n)(n.p) + (c.p)
🤦♂️
tough veldt
0 = (c.m)(n.m) + p.n
0 = (c.n)(n.m) + p.m
carmine echo
it's (c.p)(p.n)
tough veldt
man im going to #latex-testing sorry for hogging channel
tough veldt
it's (c.p)(p.n)
what why
carmine echo
not only p.n
carmine echo
c.n = (c.m)(m.n) + (c.n) + (c.p)(p.n)
c.m = (c.m) + (c.n)(n.m) + (c.p)(p.m)
c.p ...
because look at this
c = (c.m)m + (c.n)n + (c.p)p
tough veldt
yeah im kinda done
haughty berry
What are you guys proving?
tough veldt
Given
r.n = a
r.m = b
non-coincident planes (treat n, m as unit vectors if you want)
Find the vector equation of the intersection
Progress:
r = k(nxm) + c
Finding c seems to be hard/not possible in closed form
carmine echo
Also, given 3 non-independent m, n, p, is it right to write c = (c.m)m + (c.n)n + (c.p)p ? m, n, p are unit vectors
haughty berry
**r**.**n** = a
**r**.**m** = b
Sorry for being dumb, but what’s the plane here?
haughty berry
What’re n,m?
carmine echo
n, m are normal vectors and r is (x,y,z) 
carmine echo
yes
tough veldt
n m a b constant
wait wait wait, this looks helpful
c is the a in that post at the bottom
haughty berry
Isn’t the intersection just
[ {an + bm + c(n\times m)\vert c\in\bR} ]
?
stoic pythonBOT
Slurp
haughty berry
Or am I pulling a stupid
tough veldt
if m and n arent orthogonal
tough veldt
orthogonal?
tough veldt
m is orthogonal to its plane
n is orthogonal to its plane
They aren't necessarily orthogonal to each other
Ok yeah, that answer makes sense
We kinda need convert a, b in the plane equation
Basically, it's a pain in the current form
(r-a).n = 0
(r-b).m = 0
This would make it much more doable
fringe fjord
A straightforward way to proceed would be to set v=(x,y) and start calculating.
heavy crown
A straightforward way to proceed would be to set v=(x,y) and start calculating.
it should be simple but I don't get it still sorrry
lavish jewel
you could also check the eigenvalues of the matrix A and see if you can diagonalize it
fringe fjord
There's not much to get, just unfold the definitions and start simplifying.
A third way would be to notice that sqrt(1/10)A is orthogonal.
heavy crown
A third way would be to notice that sqrt(1/10)A is orthogonal.
you mean orthonormal, I understand it this way more
heavy crown
There's not much to get, just unfold the definitions and start simplifying.
but idk can you show me how in this way?
fringe fjord
Yes, that's a more logical name for the property. Unfortunately "orthogonal" is kind of traditional.
I'm not going to do your homework for you.
heavy crown
it's not my homework, i'm preparing for my exam hahaha
lavish jewel
tropo's suggestion builds up on the definition you already wrote
all you have to do is capitalize from associativity
fringe fjord
where "capitalize from associativity" really means "remember your high-school algebra".
heavy crown
but idk can you show me how in this way?
sorry. I mean I literally forgot how to open <Av,Av> ( not has to do with remembering algebra , crying from shame haha)
if v=(x,y) then how do I proceed.
I just need to see one example of this to understand..
fringe fjord
Look, if you don't even care enough to write the definition of A instead of the letter A, and write (x,y) instead of v, and then start applying the definition of matrix multiplication, then why should we do it for you?
vestal spire
There's clearly a huge conceptual misunderstanding here that has nothing to do with the problem at hand
heavy crown
Look, if you don't even care enough to write the definition of A instead of the ...
okay that made me understand now better haha sorry for being dumb.
I literally didn't know what to do but now I made it
thank you!
subtle gust
the question is solve for X
i set up a system of 6 equations in 6 variables
and solved it
everything checks out
except the element in the 2rd row 3rd col 🙂
[-4 9 -4]
[-3 5.5 -4.5]
that's the matrix i found
when you multiply the 2 matrices you don't get 10 in the 2nd row 3rd col u get 3/2
lavish jewel
first thing that comes to mind is to check your arithmetic, or try doing it with a different method and see if you catch where you made a mistake
fringe fjord
Obviously something must have gone wrong along the way. But there's a trick you can use to avoid having to solve 6 equations.
subtle gust
first thing that comes to mind is to check your arithmetic, or try doing it with...
i used an online gaussian elimination calculator.......
subtle gust
Obviously something must have gone wrong along the way. But there's a trick you ...
huh
how
lavish jewel
you might've input the matrix wrong
the easiest way here seems to invert the 3x3 mat
you can do that by augmenting it and then doing GJ on that
at least off the top of my head
maybe tropo has a more clever idea
subtle gust
the easiest way here seems to invert the 3x3 mat
i don't think we learned that in class hmmmm
we still haven't covered how to find the inverse of a matrix
fringe fjord
First transpose your equation so it looks like A X^T = B for some concrete 3×3 matrix A and 3×2 matrix B (which I'm giving names jsut because I don't want to repeat their values over and over). Then if only you had A^-1, you could multiply with that from the left and get A^-1 A X^T = A^-1 B, that is X^T = A^-1 B. Now, applying a matrix from the left amounts to doing row operations, so what you can get the effect of multiplying by A^-1 by forming the block matrix [ A B ] and doing Gaussian elimination on that until reduced row echelon form. Then the A has become I and B has become A^-1 B.
wintry steppe
Is this the correct way to check vector space axioms? I haven't done it in a long time
I didn't check all of them because I want to make sure my work so far is correct, maybe I'm making a silly mistake on each proof.
subtle gust
First transpose your equation so it looks like A X^T = B for some concrete 3×3 m...
i'm so confused rn
it's def something that we still haven't covered
idk why it's in the hw tho
lavish jewel
tropo's suggestion is equivalent to doing gaussian elimination twice on two small systems
lavish jewel
if you're careful, you can do this for two separate systems in 3 variables
subtle gust
tropo's suggestion is equivalent to doing gaussian elimination twice on two smal...
yeah when i did that i got a def wrong answer
lavish jewel
instead of one huge system with 6 vars
wintry steppe
y-xB must be in the kernel for X and the kernel = range X ^ orthogonal
but idk how beyond that to actually 'prove it'
subtle gust
when i separated the 6 equations into 2 systems
lavish jewel
well, i can't say anything other than check your arithmetic :x
subtle gust
i got a whole def matrix
subtle gust
well, i can't say anything other than check your arithmetic :x
again, i used an online calculator lol
lavish jewel
you evidently did something wrong though
subtle gust
i'll try again and check idk
subtle gust
i'll type the augmented matrices for the 2 systems here
lavish jewel
notice that the part circled in red is precisely the definition of being in the orthogonal complement of range X
it might be easier to see if you substitute X^T with something else
wintry steppe
but i think the direction here is to do something with the kernel space no?
since we know that
lets say z = y - xb
then we have X^T z = 0
thus we must be able to say that z is in the kernel space no?
wintry steppe
hmm im not sure yet since i didnt start the proof but
that was my thinking behind this proof
lavish jewel
what you said is the same thing i said
subtle gust
\begin{bmatrix}
-1 & 2 & 3\
0 & b & c
\end{bmatrix}
wintry steppe
what you said is the same thing i said
oh sorry lmao
lemme try to write it out and see if i can try it myself
fringe fjord
notice that the part circled in red is precisely the definition of being in the ...
I think that should have been X at the end, not X^T, right?
subtle gust
\begin{bmatrix}
-1 & 1 & 3 & 1
0 & 1 & 1 & 5
1 & 0 $ -1 & 0
\end{bmatrix}
lavish jewel
I think that should have been X at the end, not X^T, right?
yes, my bad. forgot about the transpose
fringe fjord
It may help to be explicitly aware that the kernel of a matrix is the same as the orthogonal complement of its row space.
thorn yacht
Would this be an acceptable argument for showing that a matrix with orthonormal columnvectors is orthogonal? Let Q be the matrix with orthonormal columnvectors, then:
$$(QQ^){i,j} = \sum{r=1}^{n} Q_{i,r}Q_{r,j}^{} = \sum_{r=1}^{n} Q_{r,i}^{} Q_{r,j}^{} = \sum_{r=1}^{n} (Q_{r,j} Q_{r,i})^* $$
$$= <Q_{,i} , Q_{,j}> = \delta_{i,j}$$
and likewise for $Q^{*}Q$.
stoic pythonBOT
Kaishin
subtle gust
this is the augmented matrix of the first system
the augmented matrix for the seconds system
wintry steppe
http://mathb.in/69867
Is this the correct way to check vector space axioms? I haven't done it in a long time
I didn't check all of them because I want to make sure my work so far is correct, maybe I'm making a silly mistake on each proof.
Reposting my question from earlier as it got buried
subtle walrus
Note that f(x)+g(x)=g(x)+f(x) as they are elements of R.
why isn't the proof of commutativity done here?
(same for associativity)
you do a lot of unnecessary work
wintry steppe
Like what?
wintry steppe
Yeah isn't that what I do?
subtle walrus
sure, but why do you think that (f+g)(x) = f(x)+g(x) = g(x)+f(x) = (g+f)(x) does not suffice
you write a lot more
subtle walrus
you did that in (1)?
for both f+g and g+f
(as it holds for f and g by assumption)
jaunty needle
as I show that the bijections=U between the sets $L(\mathbb{R}^{m},\mathbb{R}^{n}) \ y \ M_{n x m}(\mathbb{R})$ and $\mathbb{R}^{nm} $ are isomorphic
stoic pythonBOT
alexix21
wintry steppe
Oh okay, so I should've just stated by (1), the rest follows
subtle walrus
i mean, you have f, g satisfying that property by assumption, f+g and g+f as well by (1) and you only have to show f+g = g+f
i am not quite sure why you want to show more than that
you already showed "additive closure" before, and you show it again
there is no reason to even mention it
wintry steppe
Yeah okay
subtle walrus
in (5) you don't mention that g is in V
since you did everything else in a lot of detail, it seems like you forgot to check
subtle walrus
yes, you show that such a g exist as a function R -> R
but you do not confirm that it is in V
wintry steppe
I see what you mean so I should add a line like g(x) = 0 = g(1 - x), so g in V
subtle walrus
i miss a "clearly g \in V" after defining g
yes, exactly
everything else looks good to me
wintry steppe
Thanks for the feedback
subtle walrus
maybe a thing to mention
if you know that the functions R -> R are a vector space already, you can save yourself a lot of work
since then you only have to show that V is a subspace and you get commutativity/associativity for free
wintry steppe
I see, yeah
subtle walrus
(also distributivity and the compatability thing)
wintry steppe
Distributivity is compatibility isn't it?
wintry steppe
Oh okay yeah
dapper gorge
Would this be an acceptable argument for showing that a matrix with orthonormal ...
It seems like you are assuming that $Q_{ri}$ is real? You can just show that the rows are orthogonal, it follows that the columns are also orthogonal (you can apply the same result to $Q^t$). From the second equality, note that $Q^*{rj}=\overline{Q{rj}}^t=\overline{Q_{jr}}$. And so the sum equals $\sum_{r=1}^n Q_{ir}\overline{Q_{jr}}$ whcich is the dot product of rows $i$ and $j$.
stoic pythonBOT
Croqueta
dapper gorge
To bring this to nicer forms, is the usual way of doing it by multiplying the 1-degree terms by another variable, say z, and then appplying Jacobi's theorem or principal axis theorem (not sure how you call it) and fix the new variable z, so that z^2 is just a constant ?
wintry steppe
I've finished most of the theorem, can someone verify that I'm doing the correct ideas?
(I could list axioms used, but I'd need to include a screenshot of which axiom is which, but it's just the standard vector space axioms)
native ore
im not understanding this question
Pn is the vector space containing all polynomials of the nth degree
one of these are "no" but I dont understand why
wintry steppe
@native ore Have you gone through the axioms for each one?
nocturne jewel
They got help already
tired fossil
hey guys general question here, a linearly independent set written;a1v1+...akvk=0, can it have a non trivial solution?
so it must only be a trivial solution?
wintry steppe
We just say that if not all of the c_i are zero then it's a nontrivial solution. However, because the set is independent, you can't get 0 without all of the c_i being zero
tired fossil
makes sense, i read something but I had a feeling is wrong, thanks
wintry steppe
What did you read?
tired fossil
also,
how would i evaluate to see if 2.f) is linearly dependent
how do i make a system of equations to evaluate? the fact that it is in the denominator is messing with me
wintry steppe
What do you mean system of equations? There is no system right now, just use the definition and make an equation, and see what happens then
Also I haven't seen that notation before. Does that mean functions with domain [0,1] to F?
nocturne jewel
how do i make a system of equations to evaluate? the fact that it is in the deno...
the definition of linear (in)dependence
nocturne jewel
dont write $\vec{0}$
stoic pythonBOT
Mosh
nocturne jewel
it implies coordinate vectors
wintry steppe
Just multiply through by the denominators like you normally would?
nocturne jewel
however, write it all as 1 rational function
wintry steppe
Just keep in mind the domain restrictions
But they're easy to factor so that shouldn't be a problem
nocturne jewel
write it as $\frac{P(x)}{Q(x)}=0$
stoic pythonBOT
Mosh
wintry steppe
Or just P(x) = 0 if you remember the domain restrictions for where it is undefined
tired fossil
so add them all together first?
nocturne jewel
yes
tired fossil
OK, why would you do that out of curiousity
nocturne jewel
want a sum of fractions = 0
and easier to solve that when you have fraction = 0
cause then it becomes numerator = 0
ie you get rid of the rational expressions and end with a polynomial
nocturne jewel
sure, could've been simpler
cause factors are repeated I believe
however, you now have a quartic = 0
whose coefficients are functions of a1,a2 and a3
tired fossil
which cannot be independent
wintry steppe
what in the world
http://mathb.in/69868
I've finished most of the theorem, can someone verify that I'm doing the correct ideas?
(I could list axioms used, but I'd need to include a screenshot of which axiom is which, but it's just the standard vector space axioms)
Reposting from above since it got buried
tired fossil
so i get a solution of a_3(-5/6. -1/6,1)
so how do i use that to answer the linear combination?
nocturne jewel
you have infinitely many options for how to pick the scalars
pick 1 that isn't the 0 vector
exotic wedge
Can i get some help on this, i might forgotten a lot about dual spaces
wintry steppe
with response to this
i have written up this proof
do you have any thoughts on this?
i meant to write kernel of x transpose but besides that
warm dome
Has anyone seen the new Matrix movie? It wasn't as good as the old ones, but apparently WB is using it to launch a whole cinematic universe of linear algebra-themed action movies. Personally I'm really looking forward to the upcoming Quentin Tarantino-directed The Tensor Product. Michael Bay's Gaussian Elimination, slated for 2024, also doesn't look terrible, and I'm normally not a fan of his work.
exotic wedge
If we have a infinite dimensional space V, can we find a linear transformation from V to a subspace W in V ?
exotic wedge
Can i get some help on this, i might forgotten a lot about dual spaces
I am trying to solve this and thought maybe if we find such a mapping then we can take the function on W and construct a function on V, then we prove i is surjective
functional*
dusky epoch
surjective maybe
what i gave is surjective onto {0} lol
also i think you might want to refresh yourself on the definition of adjoint maps
wintry steppe
http://mathb.in/69868
I've finished most of the theorem, can someone verify that...
bumping for 3rd time
jaunty needle
what is basis space of linear functions ℒ ( ( ℝn, ℝm)
slow scroll
think about matrices. These are mxn matrices
i.e. how would you construct a basis for the space of mxn matrices?
exotic wedge
Any hint on this ?
wintry steppe
carefully write out what it means for i^\vee to be surjective
(if you haven't, figure out what i^\vee is in the first place)
exotic wedge
(if you haven't, figure out what i^\vee is in the first place)
should we find some projection from V to W to find i* ?
wintry steppe
i don't really know what you mean
well, maybe i do. it'd work, but it doesn't really address either of the things i wrote, at least not directly
exotic wedge
so u say there is a simpler way of finding i*? cuz at the first i was using basis to find this projection but i was pointed out that I shouldn't assume finite dimensions
exotic wedge
I think i roughly i understand dual spaces and transformations, could clarify more about finding i and will go back and refresh my understanding ?
wintry steppe
if f is a linear functional on V, what is i*f?
exotic wedge
if f is a linear functional on V, what is i*f?
a linear functional on W ?
wintry steppe
which one?
exotic wedge
not sure
wintry steppe
you have to review the definitions in your book/notes/lecture, then
exotic wedge
yeah I should review this topic. Thanks a lot
exotic wedge
you have to review the definitions in your book/notes/lecture, then
I am sorry, could u clarify what does "i given by the inclusion of a subspace" mean?
wintry steppe
it's the map from W to V which sends every element w of W to w.
stoic pythonBOT
TTerra
wintry steppe
this maybe looks a little silly
but it is important
this is a linear transformation, so you can talk about its adjoint linear transformation
exotic wedge
so i just reviewed the definition
if we have v' in V'
then i guess i' v' should also be v' ?
because <iw, v'> = <w, i' v' >
wintry steppe
well, you have to be a bit careful about the domain of v'
v' is defined on V, but i'v' is only going to be defined on W
exotic wedge
well, you have to be a bit careful about the domain of v'
yeah i was gonna say it should restricted to W right ?
exotic wedge
I think surjectivity should follow from this nicely even though i can't see immediately how
wintry steppe
there are a few ways to show i* is surjective
thinking concretely, you want to show that every linear functional on W extends to one on V
right?
exotic wedge
sure
wintry steppe
since restricting the extension is just gonna give you your original functional back, and that says it's in the image of i*
dusky epoch
hahn banach 
wintry steppe
lmao
a more abstract way to do it is to use the fact that surjectivity is equivalent to having a right inverse
passing to adjoint transformations switches around compositions, so you'll want to find a left inverse to i. but that's injectivity, so...
figuring out how to do it in terms of extending functionals is probably the point of the problem
~~or you can use hahn banach ~~
exotic wedge
since restricting the extension is just gonna give you your original functional ...
Isn't the intuition given here is sufficient? or should we construct the right inverse ?
wintry steppe
you should show how to construct an extension to V of any linear functional on W
maybe best to ignore the abstract stuff i said about finding right inverses, for now
i only brought it up because you mentioned projections of V onto W earlier, and that's where that comes in
well
exotic wedge
you should show how to construct an extension to V of any linear functional on W
but can i just do a silly extension? like just let it be a constant function on V\W ?
wintry steppe
eh, both ways are fine. either showing you can extend functionals, or showing that you can find a projection. they're basically the same problem
why would that be linear?
you gotta extend your functional W -> F to a functional V -> F, so it's gotta be linear
exotic wedge
was thinking the identity function but not necessarily linear aswell
wintry steppe
it's something to think about
exotic wedge
it's something to think about
can't think of something tbh
exotic wedge
it might help to play around with bases of W and V
how to assume basis if it is not finite dimensional ?
wintry steppe
hmmm
well, bases still exist
you can still find a basis of W and extend it to a basis of V, for example
something something axiom of choice
maybe that'll help
wintry steppe
bumping for 3rd time
bumping for 4th time
exotic wedge
you can still find a basis of W and extend it to a basis of V, for example
so the extension just puts zero on all other coordinates that's not in W ?
wintry steppe
i think you've got the right idea
exotic wedge
Oh okay that's nice, thanks a lot. Is it okay to ask for a hint for the second part? or did I ask too many questions ?
wintry steppe
for that one i recommend slowly unpacking the definitions
you want to relate functionals on V/W to functionals on V, somehow
(since one side of the isomorphism is (V/W)* and the other side lives in V*)
exotic wedge
so Ker(i') is just the functionals on V that are 0 for all w in W right ?
wintry steppe
righgt
exotic wedge
all i think about is that pi from V to V/W is zero for all w in W but this is probably irrelevant
wintry steppe
the quotient map might be relevant
exotic wedge
hmm
wintry steppe
is this channel free
big hint: it might be a good idea to think about the quotient map and its adjoint
exotic wedge
okay so in a previous part
viral magnet
is this channel free
exotic wedge
I proved that L(V, V') and L(V/W, V') are isomorphic for another V' vector space
wintry steppe
<#488120190538743810>
sorry i didnt wanna interrupt a convo but its a linear algebra related one
exotic wedge
does it make sense to just apply this with V' = R ?
wintry steppe
can you post a screenshot of that part?
it doesn't seem right to me as you've stated it
exotic wedge
wintry steppe
yeah, you can apply this
exotic wedge
yeah it makes total sense
wintry steppe
you got it
wintry steppe
wintry steppe
hey all im trying to solve this https://cdn.discordapp.com/attachments/495064710278938662/936132911088939098/Screen_Shot_2022-01-26_at_11.37.31_PM.png
this is the matrix im trying to solve it with
but i have no idea how they work together
peak plinth
How would I find RREF of Augmented matrix from a solution set
dusky epoch
did you mean eigenvalues?
also this isn't true in general unless you say something else about A
$A = \bmqty{1&0&0\ 0&2&0 \ 0&0&3}, t_1 = 1$ and $t_1 = 2$ does not satisfy $(A-t_1I)(A-t_2I) = 0$
stoic pythonBOT
Ann
dusky epoch
@teal flare
teal flare
@dusky epoch
dusky epoch
well why didn't you say so
teal flare
Sorry, could you help me?
dusky epoch
anyway in that case this is just cayley-hamilton
dusky epoch
plugging a matrix into its own characteristic polynomial gives 0
though if you don't know this theorem then there are other ways to explain why your result holds
teal flare
I'll try to understand that, but if you have others, I'd like to hear them, if you don't mind.
dusky epoch
how familiar are you with the concept of diagonalization
lavish jewel
in that case, notice that your matrix has 2 distinct eigenvalues and is diagonalizable
so let's say A = QDQ^-1
and notice also that QQ^-1 = I
that can be substituted into (A - t1 I) (A - t2 I)
as (QDQ^-1 - t1 QQ-1)(QDQ^-1 - t2 QQ-1)
then notice that the Q and Q^-1 can be factored from each term, leaving you with
Q(D - t1 I) Q^-1 Q(D - t2 I) Q^-1
and the Q^-1 Q in the middle can be replaced with another I
so now we have Q(D - t1 I)(D - t2 I)Q^-1
then recall that D will have as diagonal entries t1 and t2
so D - t1 I is a matrix of the form [0, 0; 0, t2 - t1]
similarly for D - t2 I, where we get [t1 - t2, 0; 0, 0]
so the product of the 2 inner terms is 0
teal flare
I see. Thank you.
lavish jewel
Q 0 Q^-1 = 0 (where 0 is the 0 matrix)
this holds more generally for matrices that are not diagonalizable, but you'd have to look the cayley hamilton theorem as ann said
dusky epoch
in that case, notice that your matrix has 2 distinct eigenvalues and is diagonal...
nobody said t1 ≠ t2 though
lavish jewel
good catch
i saw your example and was like "ah, t1 = 1 and t2 = 2" lol
the argument still works under the stronger assumption that the matrix isn't defective
otherwise i guess they'd need jcf
thorn yacht
Could someone help me with these two problems?
I'm having some difficulty writing stuff out... it's as if it makes sense, but I can't structure my thought process.
zinc timber
what is P(R)?
thorn yacht
Polynomials with coefficients in $\mathbb{R}$.
stoic pythonBOT
Kaishin
zinc timber
and ' stands for dual?
thorn yacht
Yes
I think defining an isomorphism between them would probably be the easiest way, but I'm not sure how. Perhaps if we define
$T \colon (P(\mathbb{R}))' \to \mathbb{R}^{\infty}$
as
$$T ( \phi) = (\phi (1), \phi (x), \phi(x^2), \ldots)$$
maybe?
stoic pythonBOT
Kaishin
thorn yacht
I think that should be an isomorphism; not sure...
proper cradle
Which map preserve area? From following.
thorn yacht
Which map preserve area? From following.
I'm just curious: are area preserving maps the same as isometries?
proper cradle
Idk
dusky epoch
I'm just curious: are area preserving maps the same as isometries?
no
a map can preserve areas without being an isometry
thorn yacht
I see
zinc timber
**Kaishin**
don't think this will work because dual space of P(R) is not that simple
I don't know what's the dual space of P(R) is tho
thorn yacht
I'm not sure here, but I don't think the first one preserves, as it sends the triangle with corners (0,0),(1,0),(0,1) to (1,0),(6,3),(3,1), I think.
2 sends same points to (0,0),(1,1),(1,1), so th area is 0, so also not preserving, I think.
Don't know about 3rd. @proper cradle
zinc timber
Which map preserve area? From following.
something something || determinant ||
thorn yacht
The set of all linear functions from P(R) to R, but it's infinite dimensional, so we can't do the simple ol' basis switcheroony
thorn yacht
something something || determinant ||
AH, of course. Forgot about the geometric meaning of this.
proper cradle
something something || determinant ||
But they are not linear transformations so how to find det
zinc timber
But they are not linear transformations so how to find det
jacobian
thorn yacht
also P(R)' is complete (wrt induced norm) where as P(R) is not
Sorry, I don't know what induced norm and complete mean.
zinc timber
But they are not linear transformations so how to find det
also are these TIFR prev year questions?
proper cradle
also are these TIFR prev year questions?
Yep
zinc timber
proper cradle
Yep
zinc timber
and?
proper cradle
If they independent of variable then they preserve areas?
zinc timber
No, do you know the geometrical meaning of the jacobian?
proper cradle
No
zinc timber
in simple words jacobian at point x_0 measures the infinitesimal distortion of volume (area) at point x0
proper cradle
alright then?
wintry steppe
bumping for 4th time
bumping for 5th time
subtle walrus
it works
in (c) you don't mention that additive inverses are unique (which you need)
wintry steppe
Why do I need to do that?
dusky epoch
(-a)x is the additive inverse of ax
you can't speak of "the" additive inverse until you know additive inverses are unique
subtle walrus
in (d) everything after "Now assume x \neq O" is unnecessary and you also messed it up a bit
wintry steppe
Oh okay, yes I do know that, just wondered why
subtle walrus
everything else is correct as far as i can tell
wintry steppe
in (d) everything after "Now assume x \neq O" is unnecessary and you also messed...
I need to prove both cases don't I though?
wintry steppe
there are cases when it doesn't
So (d) is incorrect?
dusky epoch
no, you don't. you have proved a ≠ 0 implies x = O, which is equivalent to (a = 0 or x = O)
subtle walrus
your two cases are a=0 (in which case you are done) and a\neq0
dusky epoch
by material implication
or that
the part that starts with "assume x ≠ O" can just be erased
subtle walrus
alternatively everything before that (but you would have to fix the stuff that comes after)
wintry steppe
No after rereading it I understand now what you mean
subtle walrus
(this would correspond to the cases x = O or x \neq O)
wintry steppe
We work under the assumption that $ax = 0$. Assume that $a \neq 0$. Then $ax = O$ and $a$ is an element of some field, so $a^{-1}$ exists. Therefore, $a^{-1}(ax) = a^{-1}O = O = (a^{-1}a)x = x$. Therefore, $x = O$. Notice that if $a = 0$, then by (a) we are done.
stoic pythonBOT
n/c
dusky epoch
also
wintry steppe
I think that's good enough for (d) then
dusky epoch
a is an element of some field
this is unnecessarily vague
a is an element of the base field of your vector space
which you don't give a name, but it's not "some field"
dusky epoch
if you want, you can give the base field of your vector space a name at the beginning (usually it's called F or K) and refer to it by name when needed
wintry steppe
Like let V be a vector space over the field F?
I just don't want F to be interpreted as R or C
dusky epoch
it won't be unless you explicitly say you're only considering real or complex vector spaces
(or context indicates as much)
subtle walrus
Like let V be a vector space over the field F?
this or "let V be an F-vector space"
proper cradle
How to solve this?
wintry steppe
is this a true/false?
a hint: if you could find a proper non zero invariant subspace, that corresponds to a non-trivial factor of the characteristic polynomial
proper cradle
Its true
in particular for identity map which one is invariant?
got it span by eigen vectors
gray dust
in particular for identity map which one is invariant?
every subspace is invariant wrt identity
wintry steppe
I did several "versions" of proving something is a subspace. I need feedback on which one is the best, or which parts of each version are the best and more correct
(I know I didn't compute dim S)
proper cradle
every subspace is invariant wrt identity
ohh yes, what happen if some linear transformation have only one eigen value with AM and GM equal are n(dim of v) then there exist any proper invariant subspace for this transformation?
gray dust
ohh yes, what happen if some linear transformation have only one eigen value wi...
if GM=dimV then ur map is a multiple of the identity
proper cradle
then eigen space with respect to this eigen value will be V right, so it is not proper
gritty swift
doing all these exercises from axler seems like a waste of time, i know how to do the computation so its reasonable to skip these right? or is there some reason axler is including tons of very duplicate looking exercises
proper cradle
since it is bijective, then ker is zero and range V, so my question is then which subspace W is invariant for this such transformation?
which is non-zero proper
proper cradle
doing all these exercises from axler seems like a waste of time, i know how to d...
in every exercise, U is changing if you are comfortable with finding the basis of U in every ex then it is fine to leave
gray dust
then eigen space with respect to this eigen value will be V right, so it is not ...
yes but any proper subspace of V is invariant
see msg about identity map
proper cradle
but for every map T, any subspace of V is not invariant under T
heavy crown
Anyone?
Let A be a normal matrix of order n x n. A has exactly 2 different eigenvalues: b, c.
Show that if b=-c, then A² is a a diagonal matrix.
--
We know that: A² = PD²P^(-1) = P * {{c^2, 0} ,{0, c^2}} * P^(-1)
How do I show it's diagonal matrix the result?
zinc timber
D² is constant multiple of identity
dusky epoch
this is P * c^2 I * P^-1
gray dust
but for every map T, any subspace of V is not invariant under T
im talking about ur example (GM=dimV), not in general
wintry steppe
http://mathb.in/69873
bump
heavy crown
oh you're right thank you two ryu and ann
proper cradle
I did several "versions" of proving something is a subspace. I need feedback on ...
All are good
wintry steppe
@proper cradle Thanks but surely one is better than the other?
heavy crown
Anyone?
A and B are matrices of order n x n that have the same characteristic polynomial. Show that A-B is invertible.
Means we want to do find |A-B| != 0 ? but how
slow scroll
If A = B this is not true though
heavy crown
If A = B this is not true though
That's the full question(from an exam)
A and B are matrices of order n x n that have the same characteristic polynomial. Circle the correct statement (1) or (2):
(1) A is invertible if and only if B is invertible.
(2) A - B isn't invertible.
I didn't know how to answer both options lol and I thought the answer is (1) is the correct answer.
So assumed the (2) is wrong = A-B must be invertible(?) thus my question above
heavy crown
If A = B this is not true though
yea youre right hm
slow scroll
nvm about the thing i deleted
heavy crown
yea I'm thinking hmm
slow scroll
okay the first one is correct
because if their char polys are the same, A has 0 as an eigenvalue iff B has 0 as an eigenvalue @heavy crown
heavy crown
oo you're right damn. haha
so both share the same eigenvalues then neceserrialy both don't have the 0 eigenvalue so both are invertible if one is
slow scroll
yep
heavy crown
so how to prove A-B is invertible? haha
slow scroll
is that supposed to be true?
it could be sometimes true, sometimes false or something like that
heavy crown
it could be sometimes true, sometimes false or something like that
yea it makes sense to me thinking about it more this way but
if we ticked (1) as correct, so (2) must incorrect and (2) says A-B isn't invertible.
thus A-B is always invertible somehow
or maybe this exam is broken haha
slow scroll
that's not how negating statements work
heavy crown
aha yea I guess it just means then that it could be invertible or not
slow scroll
the negation of the second statement is that there exists A and B with the same char poly such that A - B is invertible
slow scroll
npnp
keen bronze
If I have a linear transfomation and it's matrix T, is the Im(S) the vector space generated by the vectors I get from applying the transformation to each generator of S?
wintry steppe
If we define L(S) to denote the subspace spanned by a subset S of a linear space V, then how can we define the set S? Is it just some basis vectors of V, or scalar multiples of?
Because if S subset T subset V and T is a subspace of V, then L(S) subset T, but how do you justify that unless T is a set of (even more) linear combinations as well
halcyon spindle
Because if S subset T subset V and T is a subspace of V, then L(S) subset T, but...
Since T is a subspace of V, we have T is closed under scalar multiplication and vector addition. So taking a subset S of T we have L(S) is a subset of T. wdym by how do we define S? If we have generating system in V we can define S as the span of that system. Is that what you mean?
wintry steppe
@halcyon spindle I meant can we explicitly construct S? Because for example, given e_1,..,e_k basis vectors of S, $L(S) = { \sum_{i=1}^k \zeta_i e_i : \zeta_i \in K, e_i \in V}$ where $V$ is a v.sp. over the field $K$
stoic pythonBOT
n/c
halcyon spindle
hmm wait a second you have a basis for S, so you can construct S and S would be a subset of L(S).
wintry steppe
right
But that depends on what S is
Is S just some basis vectors, or scalar multiples of that are fixed?
tender zenith
The dimension of a subspace is at least one less than its parent space right?
wintry steppe
@tender zenith no
A subspace can be equal to the 'parent' space as well
@halcyon spindle I think we need to define S to be equal to the span of S, if we want S to be a subspace of V
gray dust
The dimension of a subspace is at least one less than its parent space right?
no, any space is a subspace of itself. but this is true for proper subspaces (assuming finite dimension)
wintry steppe
I don't understand. We define $S = {x_1,x_2,\dots,x_k} $ of $k$ independent elements from some vector space $V$. Then we let $L(S)$ to be the subspace spanned by S. So how can $L(S) = S$, ever?
stoic pythonBOT
n/c
gray dust
If I have a linear transfomation and it's matrix T, is the Im(S) the vector spac...
whats S, whats Im
wintry steppe
nvm didn't see who you tagged.
keen bronze
@gray dust this is the linear transformation
and this is S
and Im is the image, that is what we obtain when we apply the transformation to the space S
what I did was find the generators of S
and then applied f to every generator of S
and said that Im(S) was the span of the vectors that I obtained
Is that correct? If not, then how do I find the image of a subset given a transformation?
nocturne jewel
Image is the subspace of output vectors, yes
keen bronze
Thx it made sense in my head but I wasn't sure
gray dust
and Im is the image, that is what we obtain when we apply the transformation to ...
ic, Im(S) isnt the usual notation
if f is a function & A is a subset of the domain then f(A) denotes the image of A under f
so here we say f(S)
and indeed, if S is a subspace of the domain then f(S) is a subspace of the codomain
gray dust
If we define L(S) to denote the subspace spanned by a subset S of a linear space...
wdym how we define S?
wintry steppe
@gray dust I'm just trying to explicitly construct S
wintry steppe
I don't understand. We define $S = \{x_1,x_2,\dots,x_k\} $ of $k$ independent el...
here's my question
gray dust
who said L(S)=S?