i think?
#linear-algebra
dusky epoch
you're expanding over the FIRST column
if you were expanding over the second then you'd use the signs in the second column
ditto for the third
wintry steppe
i'm not sure if i agree
look at this
$\begin{pmatrix} 2 & 4 & - 3 \ 1 & 7 & 0 \ -1 & 5 & 2 \end{pmatrix}$
stoic pythonBOT
polynomial:
wintry steppe
using the trick above
dusky epoch
i am willing to bet you'll now show me something that falls victim to an arithmetic fuckup
dusky epoch
are you 101% sure there are no arithmetic fuckups in what you're about to send me
dusky epoch
ok what is it then
wintry steppe
$2(7\cdot2) - 1(4\cdot2 -(-3)\cdot5) - 1(3\cdot7)$
stoic pythonBOT
polynomial:
wintry steppe
$=-16$
stoic pythonBOT
polynomial:
stoic pythonBOT
Result:
-16
but the rest of the columns
dusky epoch
ok
dusky epoch
yes because you didn't use em
you expanded along the first
not the second
not the third
the first
wintry steppe
can you give an example of what you mean
dusky epoch
but IN PRINCIPLE it's possible to expand along any column
wintry steppe
when would you do it differently
dusky epoch
when convenient
wintry steppe
like..
dusky epoch
i could've expanded along row 2 for example because there's a zero there
and that means one less minor to worry about
wintry steppe
i don't know how to do what you are referring to
i only know of this expansion
i didn't even realize we'd be computing the same thing
if we "expanded differently"
dusky epoch
$-1 \begin{vmatrix} 4 & -3 \ 5 & 2 \end{vmatrix} + 7\begin{vmatrix} 2 & -3 \ -1 & 2 \end{vmatrix}$
stoic pythonBOT
Ann:
dusky epoch
this is what the expansion would be
wintry steppe
what happened to the last one
dusky epoch
it's zero
dusky epoch
laplace expansion
dusky epoch
in a linalg course ¯_(ツ)_/¯
wintry steppe
what is a minor
dusky epoch
the determinants of those struck-out matrices
wintry steppe
hmm
i thought i understood sarrus' rule
apparently not
😩
LA literally so annoyingly difficult
feels like it should be easy
What if you expand along the first column of a general 3x3 matrix and simplify the expression to see sarrus pop out
Might help you understand it
dusky epoch
matrix-bashing does have its hard aspects
wintry steppe
for a matrix to be not invertible
all columns have to be linearly dependent right
it can't just be 1 or 2 of them
(in a 3x3 matrix)
A matrix is invertible if and only if all columns are linearly independent
Not the other way around
so if 2 columns are linearly dependent out of 3, then it's not invertible
Yes, I think you meant invertible, not not invertible then?
Oh
Nvm I was the one who misread
Well actually the thing you said still wasn’t quite right
You don’t need all columns to be linearly dependent on each other but I think that was probably a typo since the other thing you said was fine
so my last statement is correct
Yes
ok
half storm
Yea the reasoning is because in order for a matrix that is invertible is equivalent to saying that it is both injective and surjective i.e. 1-1 and onto.
The part that is particularly important relating to linear independence/ dependence is the injectivity.
outer citrus
Quick question: is it correct to define the norm of a vector v as the square root of the inner product v·v ?
outer citrus
Thanks :))
dire bough
That's how you define the 2-norm of a vector, yes
half storm
Because if you construct a matix from a set of column vectors in say R^n then when you're asking whether or not the matrix is injective it's the same thing as asking whether or not the matrix equation Ax = 0 has nonzero solutions.
i.e. whether the columns or linearly dependent.
dire bough
But more broadly speaking, it depends on the vector space you're interested in. There's also the taxicab norm and the $p$-norm, which must be defined differently
stoic pythonBOT
owomorphism:
dire bough
And not all normed vector spaces are inner product spaces, so it'll break down in those
Just something to be aware of
half storm
It saysing that for any linear map T:V -> W , That the matrix represenation of the transpose of T is the transpose of the matrix representatio nof T.
T^t maps linear functionals on W into linear functionals on V.
So M(T)^T is not a linear functional.
M(T^t) maps carteisian products of what ever field V and W are over.
So $M(T^t) : F^n \rightarrow F^m$ where $n = dim(W) = dim(W^{}) $and $m = dim(V) = dim(V^{}).$
make sense?
stoic pythonBOT
JohntheDon:
ocean sequoia
im going to have to digest this a bit
ocean sequoia
so T^t put into a matrix is the same as putting M(T) then transposing it
half storm
Yes exactly.
half storm
Sure.
ocean sequoia
T^t maps linear functionals on W into linear functionals on V.
@half storm so if T is a linear functional on V then T^T is a linear functional on W?
i think this is where i am lost
i understand what a dual space/basis is i think but i dont understand how a T and T' are related
half storm
Not quite. If T is a linear map from V into W right. i.e. $T: V \rightarrow W$, then the tranpose of T, $T^t : W^{} \rightarrow V^{}$ meaning that the tranpsoe of T takes linear functionals on W and maps them into linear functionals on V. It maps the dual of W into the dual of V.
ocean sequoia
ok so wait i think i get that
stoic pythonBOT
JohntheDon:
ocean sequoia
T^t would be the same then as moving from W to V then appling the linear functional on V
half storm
Ummm, I'm a bit confused with the langauge that you've used. What I'm trying to say is that for any linear functional on W i.e. $g \in W^{*}$ $T^{t}(g) := g \circ T$ Note that $ g \circ T : V \rightarrow F$. That is $g \circ T$ is a linear functional on V.
stoic pythonBOT
JohntheDon:
half storm
So the transpose of T takes functionals and maps them into other functionals.
half storm
Cool cool.
ocean sequoia
sorry i was trying to phrase it differently thank you
half storm
No problem.
deep sluice
Hey guys, I am trying to do a problem for my homework and I am not understanding where this example for their numbers from
if someone understands this, could I get an explanation?
The S matrix is what I am confused by
ocean sequoia
it just means to draw a unit square
so [(1,0),(-1,0),(0,1),(0,-1)]
its not a linear transformation in this case
its jsut storing data
deep sluice
So are the numbers that are there not correect?
They were the ones that were given as part of an example
ocean sequoia
then they are mostly certainly right
deep sluice
hmmm, then I'm still not really understanding where the numbers came from that are being used
especially since there are 5 numbers in each of the two matrices
This is being done is MATLAB if that helps at all
ocean sequoia
unfortunately i don't know MATLAB but im pretty positive those are just coordinates
id put them into a matrix and i bet youll see it fall out
deep sluice
hmm that does seem to look a little like what it should be
Maybe they are just asking for some weird crap in a weird crap way
dire bough
What do you mean?
ocean sequoia
honestly im going to ask in a minute
i want to reread the section
sorry wasnt expecting such a fast response 🙂
ocean sequoia
Let $T \in L(V,V)$ and U be be a subspace of V such that $Tu \in U$ this means that there is some $Tu = \lambda u$ where u would be an eigenvector if we changed T but kept it as an operator for V would the eigenvector change?
stoic pythonBOT
brzig:
dire bough
haha thanks
I'm still not quite certain what you're asking
Are you asking whether $T$ defined on some subspace preserves its eigenvalues if you linearly extend it to the full space?
stoic pythonBOT
owomorphism:
ocean sequoia
yes
dire bough
Ah
I think so, yes
Almost by definition
It would be silly to extend an operator in such a way that it no longer behaved the same on the original space
Haha
ocean sequoia
fair enough lol thanks!
dire bough
half storm
I'm just trying to make sure that the argument that I'm making is correct.
So suppose that there exists a set of scalars in F $a_1, a_2, \dots a_n$ s.t. for any $p(x) \in P_n(x)$ $( a_0f_0 + a_1f_1 + \cdots + a_nf_n)p(x) = 0.$
Taking the hint and specifically considering $p(x) = (x - c_1)(x - c_2) \cdots (x - c_n).$ We have that
$( a_0f_0 + a_1f_1 + \cdots + a_nf_n)p(x) = 0 \implies a_0f_0 (p(x)) + a_1f_1(p(x)) + \cdots + a_nf_n(p(x)) = a_0p(c_0) + a_1p(c_1) + \cdots + a_n p(c_n) \implies a_0p(c_0) = 0 \implies a_0 = 0$
We can extend this argument for any of the scalars i.e. we can construct a polynomial $p(x) = (x - c_0)(x - c_1)(x-c_2) \dots (x - c_i) \cdots (x-c_n)$ and conclude that $a_i$ = 0. So we have that $ {f_0, f_1, \dots f_n } $ is a linearly independent set.
paper ether
\cdots
you may want to use \dots for lists and sequences and \cdots for multiplication
half storm
oh o.k.
stoic pythonBOT
JohntheDon:
half storm
<@&286206848099549185>
spice storm
To use the Best approximation theorem, the two vectors have to be Orthogonal?
outer citrus
Quick question: is it correct to define the norm of a vector v as the square root of the inner product v·v ?
Referring to my question before
I'm stuck on this problem
<x, y> = 2
<x, x> = 4 => norm(x) = sqrt(4) = 2
<y, y> = 8 => norm(y) = sqrt(8) = 2*sqrt(2)
Therefore using the angle between vectors formula:
cos(a) = 2/(4*sqrt(2))
spice storm
Do you know what is the inner product?
outer citrus
Therefore a = 0.78 rads, which is not an option
Do you know what is the inner product?
@spice storm
The definition is written in the exercise.
The inner product of x and y is the result of conjugating x and y by the Gramm matrix G
pallid rampart
,calc acos(2/(4*sqrt(2)))
stoic pythonBOT
Result:
1.2094292028882
Hmmmm
outer citrus
But then again, this is not an option either
@pallid rampart
It is, the last one
pallid rampart
There's an fourth option?
outer citrus
The third one
spice storm
option three @pallid rampart
pallid rampart
,calc 1.2pi69/180
stoic pythonBOT
Result:
1.4451326206513
That's not quite the answer either
outer citrus
,calc 360/(2*pi)1.2
stoic pythonBOT
Result:
47.746482927569
,calc 360/(2pi)1.2
stoic pythonBOT
Result:
47.746482927569
Hmmmm... I want to cry
spice storm
I would rather learn than crying. you are learning. Find out what you did wrong
is the second option is what I am getting
spice storm
cos(a) = 2/(4*sqrt(2)) I just typed it on my calcualtor.
wintry steppe
@outer citrus the thing is they arent using the regular definition of the inner product
The problem defines it differently so you should use that one
outer citrus
Where can I find a rigurous inner product definition?
spice storm
riguours inner product will be U *V = U^T *V
outer citrus
,calc acos(2/(4*sqrt(2)))
stoic pythonBOT
Result:
1.2094292028882
cos(a) = 2/(4*sqrt(2)) I just typed it on my calcualtor.
@spice storm
Isn't it 1.2?
wintry steppe
@spice storm look at the problem
outer citrus
riguours inner product will be U *V = U^T *V
@spice storm
But that's when the Gramm matrix is the identity, it's not that way otherwise
pallid rampart
Inner product is a function $\brk{\cdot,\cdot}:V\times V\to\bC$ such that for all $v,w,u\in V$ and $\alpha\in\bC$, $$\brk{v+w,u}=\brk{v,u}+\brk{w,u},$$ $$\brk{\alpha v,w}=\alpha\brk{v,w},$$ $$\brk{v,w}=\overline{\brk{w,v}},$$ and $$\brk{v,v}\geq0$$ with equality if and only if $v=0$. If $V$ is a real vector space then the condition is $$\brk{v,w}=\brk{w,v}$$
You can find this definition in any linear algebra book, Axler for example
stoic pythonBOT
Whoever:
spice storm
So I got -11 when I tried to find the orthogonality
patent cloak
Do semilinear operators correspond to a set of matrices or similar objects in the way that linear operators do with matrices?
Tell me if I'm interrupting the previous conversation 😁
median forum
probably asking if there is a set of objects naturally isomorphic to semilinear maps that can represent them well
void slate
Is this proof correct? I can send theorem 6.9 and 6.10 that I used for it
I have another longer proof for it but I noticed that I could possibly use these two theorems in this short proof instead
These are the two theorems I used
It’s only for 3i I haven’t started on 3ii
half storm
@void slate is this friedberg?
void slate
Yes
half storm
I learning out of friedberg too 😄
It's a very good book imo.
I haven't gotten to inner product spaces yet though.
void slate
Nice!! I was able to pick up the fourth edition paperback for 8 dollars at half price books
Tho I found the PDF for the 4th and 5th online it doesn’t hurt to have a physical
half storm
i'm excited 
Yea I have physical copies of some books that have bad photocopies
or non-pdfs
wintry steppe
friedberg's jordan forms chapter is kinda ehhhh in my opinion since he just says "you can find a jordan basis using cleverness" instead of actually giving a method for finding them
half storm
Lol it's always nice to have algorithms.
wintry steppe
anyways, i won't disturb any more, it seems like noobiebooby had a question up
half storm
I'm also reading a book after this called "The Linear Algebra that a beginning graduate student ought to know"
So hopefully that has a better jordan forms chapter
void slate
I ended up staring at the dot diagram for three hours trying to figure out how it related Jordan blocks and I ended up figuring it out. I was stuck on finding the Jordan canonical basis but I ended up understanding how to solve for the basis in most cases
slate mauve
what's the difference between a unit vector and a basis vector?
pallid rampart
A unit vector is a vector with length 1 when length is properly defined. A basis vector is a vector in a basis
slate mauve
okay thanks
so are $\hat x, \hat y, and \hat z$, as commonly used in $\mathbb{R}^3$ space, unit vectors or basis vectors?
stoic pythonBOT
matt_:
pallid rampart
Any vector can be part of a basis
I haven't seen people use x hat y hat and z hat, but I have seen i hat, j hat, and k hat for the standard basis vectors
gray dust
i've seen hat x/y/z and e_x/e_y/e_z, they're all the same
void slate
The writing above the question is my proof for 3i how do I go about showing 3ii? I know a basis is orthonormal if it’s made up of unit vectors. But do I just make up my own transformation matrix with respect to the basis? How does that work?
I know my notation is sloppy😭 but I just need help with 3ii. Would it be too weak to say “theorem 6.10 requires a basis to be orthonormal, thus if the basis is not, the equality cannot be proved”
glass meteor
what did i get wrong here, working out cross product
dusky epoch
the j component should be flipped
it's $-\begin{vmatrix} 6 & 7 \ 2 & -7 \end{vmatrix}$
stoic pythonBOT
Ann:
glass meteor
would that be just this one or is it always like that?
dusky epoch
always bc this is how determinant expansion works
glass meteor
awe okay thank you. first time looking at it, trying to teach myself, thanks 🙂
void slate
Quick question what does ||v1||^2 again when referring to a vector?
pallid rampart
use \|\| \|\|
void slate
||v1||^2
wintry steppe
the norm of the vector v_1, squared
or v_1 inner product v_1 (if you don't know what that is, then v_1 dot v_1)
void slate
I have v1= [1 i 0] and then it says the norm of the vector squared is equal to 2
Shouldn’t it be 0?
glass meteor
do i need to write it a different way?
wintry steppe
oh, it's complex
gray dust
recall the standard inner product on C^3
void slate
Like so right?
gray dust
for future reference, the standard inner product on C^n is
for all $\mathbf x=(x_1,\ldots,x_n),\mathbf y=(y_1,\ldots,y_n)\in\bC^n$
$$\ip{\mathbf x}{\mathbf y}:=x_1\cjg y_1+\ldots+x_n\cjg y_n$$
stoic pythonBOT
RokettoJanpu:
void slate
Ah ok thank you both!!! I appreciate it(:
wintry steppe
and one thing to note about the thing rokabe just posted
if all of the components of y are real (so imaginary part = 0), then you get something that looks like the dot product on R^n
which is what i thought you were working with before you showed what v_1 was
gray dust
this inner product is conjugate-symmetric, NOT symmetric like inner products on real vector spaces
ie for all $x,y\in\bC^n$, $\ip{y}{x}=\cjg{\ip{x}{y}}$
stoic pythonBOT
RokettoJanpu:
wintry steppe
how did you get an \ip command like that? what's the code?
gray dust
note it's also conjugate linear wrt the 2nd argument ie
$$\ip{x}{cy+z}=\cjg c\ip{x}{y}+\ip{x}{z}$$
stoic pythonBOT
RokettoJanpu:
void slate
Oh wait! Now that you mention that Rokabe, if I have <f,g>= integral of 0 to 1 f(t)g(t)dt where V is the collection of continuous functions from [0,1) to Complex. Is this conjugate symmetric?
gray dust
you seek $\ip{f}{g}:=\int_0^1f(t)\cjg{g(t)}\dd t$ instead
stoic pythonBOT
RokettoJanpu:
void slate
Is that where I do the 1/2pi integral 0 to 2pi thing?
gray dust
idk where 1/2pi is from
void slate
Bc so far I proved the other parts of inner product spaces but the one that confuses me is showing <f,cg>=(conj.c)<f,g>
gray dust
how did you get an \ip
btw \ip is already in physics package
void slate
Ignore the 1/2pi it’s something from the textbook that looked familiar I thought might’ve been what you were getting at
gray dust
<f,cg>=c*<f,g> is from (ab)*=a*b*
void slate
Does it still work if f,g are in the complex? I know that when you needed to take * of c in reals taking the * of others didn’t do anything but now that they’re in complex how would that work? If I do <f,cg>*= I’m not quite sure what I’d get. Sorry if the answers obvious and I can’t see it 😅
gray dust
does what still work
the codomain of f & g is C so their images lie in C so the conjugate of their images is well defined
void slate
gray dust
can you post full context
void slate
I have to show that it is a inner product space but the only part I’m having trouble with is showing <f,cg>= (c bar) <f,g)
Is that enough? I can write it out by hand if you’d like! I dont wanna be too confusing
gray dust
were you specifically asked to show that V, with this defn of inner product, is a complex inner product space?
void slate
Yes it says : “Decide whether each of the following is an inner product space. Justify your answers”
gray dust
decide but not prove
void slate
Yes decide
gray dust
is V a vector space over C or R?
void slate
It says collection of continuous functions from [0,1) to C
forest quail
if i have the polynomials p1=1-2t and p2=-3+t+4t^2, can i represent this in a matrix like this: \begin{bmatrix}1&-2&0\-3&1&4\end{bmatrix}\begin{bmatrix}1\t\t^2\end{bmatrix}
gray dust
what i'm asking is, is V's field of scalars R or C?
stoic pythonBOT
soaringbear:
Compile Error! Click the 
reaction for details. (You may edit your message)
forest quail
or are the polynomials supposed to be columns? im confused
void slate
I’m not sure? The problem not instructions specifically say that anywhere. The theorem I’ve been using says c is in the field so I’d assume it’s C?
gray dust
ok take C and you want to see if V, together with that defn of inner product, is a complex inner product space or not
does <f,cg>=c*<f,g> for all vectors f,g and scalar c?
gray dust
plug into the inner product defn, <f,cg>=?
void slate
Integral(f(t)cg(t))dt= Int. f(t)cg(t) ?
Wait
My stars are gone
=f*(t)cg(t)
It’s supposed to be f star c star g star
void slate
Oh yes
gray dust
h(3)=?
void slate
9
gray dust
h(2e^x)=?
void slate
(2e^x)^2=4e^2x
gray dust
let $p,q\in V,~\ip{p}{q}=,?$
stoic pythonBOT
RokettoJanpu:
void slate
p1q1+...+pnqn?
gray dust
no the pic you posted
void slate
Oh, integral 0 to 1 p(t)q(t) dt
gray dust
let c be a scalar. <p,cq>=?
void slate
integral 0 to 1 p(t)cq(t) dt ?
gray dust
see any way to show if this equals c*<p,q> or not?
void slate
I mean, would I just take the star of c ? Or would I have to take the star of the whole thing and then factor out c star and then take the star again of only the functions?
Man you’ve been so patient, thank you
gray dust
the goal is see if you can reach one side of <p,cq>=c*<p,q> from the other
gray dust
idk what you wrote
void slate
Did the picture not come up? Or what I wrote is just not valid💀
gray dust
latter
void slate
Oh, welp, I’m out of ideas
gray dust
idk what you did after line 1
and i'll emphasize, the aim is show "<p,cq>=c*<p,q> for all p,q,c" is either true or false
void slate
Ok so ima try again I have <p,cq> => <p,cq> star = integral 0 to 1 p star c star q star => c star (integral 0 to 1 p star q star) => c star <p star, q star>
So, then it’s false?
gray dust
<p,cq> = <p,cq> star
no
void slate
gray dust
let c be a scalar. <p,cq>=?
integral 0 to 1 p(t)cq(t) dt
start from here, and recall you can "pull out" constants in integrals
gray dust
finish off
void slate
How? (c<p,q>) star => c star <p star, q star> = c star <p,q> ??
I’m about to just stop existing
gray dust
i think you have tunnel vision
you're still under the impression you're doing a proof rather than seeing if smth is true OR possibly false
DECIDE if V is an inner product space, not PROVE
glass meteor
sorry its a mess, but where did i go wrong. cause its miles off the correct answer
void slate
since c<p,q> =/= c star <p,q> it is not an inner product space?
glass meteor
answer was
gray dust
it's very important to know the difference of "true for all x" vs "true for some but not all x"
a necessary condition for V to be an inner product space is <p,cq>=c*<p,q> for all vectors p,q and scalar c
i can easily pick specific p,q,c where the equality holds but it should also be just as easy to pick out other p,q,c where it doesn't hold, so "<p,cq>=c*<p,q> for all vectors p,q and scalar c" is false, so V fails to be an inner product space
void slate
So if c=2i then c star is -2i which won’t be equal?
Ima just hit the sack. Thank you again for being patient. I really appreciate it you sticking it out for so long! As well as guiding me instead of giving me the answer right away, really makes me think more and more
gray dust
@void slate rethink in the morning: again, you either have <p,cq>=c*<p,q> true for all possible p,q,c or not
is c<p,q>=c*<p,q> true for all p,q,c? i can pick specific examples to test by letting p,q be such that <p,q>!=0, so i'm looking at c=c*. then at this point it should be obvious it's not true for all c, so c<p,q>=c*<p,q> is not true for all p,q,c
old flame
Quick question here, if "$F^{n}$ is a vector space over $F$, where $F$ is a field", what does over mean ? Is it sort of like induction, where if $F$ is then $F^{n}$ holds ?
dusky epoch
no
stoic pythonBOT
Otoro:
dusky epoch
it means that the set F^n admits the structure of a vector space whose scalar field is F
old flame
What so far I know that a vector space has to satisfy those properties (i.e commutativity.....) And permits scalar multiplication and addition
dusky epoch
it should be a list of axioms which make reference to two sets:
the vector space proper (i.e. the set whose elements are called vectors and is probably denoted V)
and the scalar field (i.e. the set whose elements are called scalars and is probably denoted F or K)
when we say "V is a vector space over F" we mean that there is a definition of addition and scaling on V, with the scalars coming from F, which satisfies all these axioms
the reason for this generalization is that, for example, we want to talk about real vector spaces and complex vector spaces on the same footing without having to prove the same thing twice (once for the real case and once for the complex) when the proof can carry over word-for-word
and if you go into other disciplines you might encounter spaces over weirder fields like GF(2)
old flame
Ah alright
dusky epoch
basically this is abstracting away from requiring the scalars to always be real numbers is what i'm saying
old flame
So the vectors are the same dimensions as the space, while the scalars are also from the space that it's over
old flame
Wait, so a vector field V over F just means that V abides all the axioms while also having scalars from F that satisfy the axioms as well ?
dusky epoch
V and F aren't in isolation from one another
gray dust
V, drawing scalars from F and considered along with given definitions of vector addition & scalar multiplication, satisfies the vector space axioms
old flame
So it's just that as V itself is already a vector space, when instead it's scalars are taken from different space call F and this also satisfies the axioms, we call this vector space V
over F ? Is this correct
gray dust
most of this is nonsense
F^n is a vector space over F, where F is a field, what does over mean ?
it means that the set F^n admits the structure of a vector space whose scalar field is F
when we say "V is a vector space over F" we mean that there is a definition of addition and scaling on V, with the scalars coming from F, which satisfies all these axioms
you were already given a precise meaning of "V is a vector space over F"
old flame
Ok thanks
subtle walrus
what have you tried?
dusky epoch
@supple hemlock so what have you tried
supple hemlock
Umm, Let A = [aI, 0, Ib, 0]
Basically identity matrix
but I multiplied by constants
supple hemlock
The only commutable matrices are identity matrices, no?
dusky epoch
you're not looking for matrices that commute with everything
you're only looking for matrices which commute with B specifically
supple hemlock
Ohhh
So first, I solve for AB = BA
which is AB - BA = 0?
then find all answers?
supple hemlock
What would be better wording?
dusky epoch
well you're looking for a basis of S
so maybe getting a more concrete description of what matrices are in S would be helpful
so maybe write $A = \bmqty{a & b \ c & d}$ and then write out $AB=BA$ as a system of equations in $a, b, c$ and $d$ and do something to it
stoic pythonBOT
Ann:
dusky epoch
you're overthinking it already
write out what AB and BA are, in terms of a, b, c and d
get a system of four equations in four unknowns
surely you must know a thing or two about those
dusky epoch
they come in many shapes and sizes but rly like. don't overthink it
supple hemlock
I can see that.
Lemme try solving the eqn, brb
Another Q: Most efficient way would be finding all solutions to Det(P) = 0, and your answer will be anything BUT those solutions?
dusky epoch
what's capital P
supple hemlock
Woops, S. My bad.
dusky epoch
that's even worse now bc S is a set and not a matrix
if you asked whether you could answer this by taking the complement of the solution set of $$\begin{vmatrix} t & t & 0 \ t & t & 1 \ 1 & t & t \end{vmatrix} = 0,$$ then yes. you can.
stoic pythonBOT
Ann:
dusky epoch
you want to show that there EXISTS a matrix A such that L(A) = p
supple hemlock
This would exist if b + c = a1, right?
dusky epoch
what?
dusky epoch
why did you say b+c = a_2 though
dusky epoch
so you need to show that no matter the values of a_0, a_1 and a_2 the system a = a_0; b+c = a_1; d = a_2 will always have a solution
dusky epoch
you're overthinking it again
dusky epoch
the answer is obvious
a and d are fixed anyway, and setting b = a_1 and c = 0 (or the other way around) will do the trick
dusky epoch
one of
supple hemlock
Also, part c, do I just use my answer found in part b, and find the null space for it?
dusky epoch
,,,,,,,,,,,
i really really REALLY don't like being asked questions like "do i do this?" or "do i use X?" or "do i make a Y?"
supple hemlock
Oh, my apologies
one of
@dusky epoch How would I go about finding other answers?
(Part B again)
supple hemlock
Oh, just show it's onto
Lol, I need a break
Thank you very much, and have a good day (or night if it's night time there)!
supple hemlock
@dusky epoch, I have 1 more question.
Part C
A bit lost w/how to approach it actually
since it's asking for nullspace of L, not A.
dusky epoch
Ker(L) consists of those matrices which get sent by L to the zero polynomial
supple hemlock
The only matrix I can think of is the 2x2 0 matrix.
The only matrix I can think of*
Is there any way to re-formulate the problem into a matrix?
dusky epoch
what happens when you apply $L$ to the matrix $\mat{0 & -1 \ 1 & 0}$?
stoic pythonBOT
Ann:
supple hemlock
0 😄
I found 3 matrices only
would that be my basis then?
$\mat{0 & -1 \ 1 & 0}$, $\mat{0 & 1 \ -1 & 0}$, $\mat{0 & 0 \ 0 & 0}$
stoic pythonBOT
Octoc00pler:
$\mat{0 & -1 \ 1 & 0}$, $\mat{0 & 1 \ -1 & 0}$, $\mat{0 & 0 \ 0 & 0}$
```Compile error! Output:
! Undefined control sequence.
<recently read> \mat
l.54 $\mat
{0 & -1 \ 1 & 0}$, $\mat{0 & 1 \ -1 & 0}$, $\mat{0 & 0 \ 0 & 0}$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
$\mat{0 & -1 \ 1 & 0}$ $\mat{0 & 1 \ -1 & 0}$ $\mat{0 & 0 \ 0 & 0}$
stoic pythonBOT
Octoc00pler:
Compile Error! Click the reaction for details. (You may edit your message)
dusky epoch
- you're trying to use my custom latex macros rn
- no
and your kernel doesn't consist of those three only, anyway
supple hemlock
@dusky epoch, how would I expand my answer to find the other matrices
would the rest just be multiples of the column rows of (0,1)^T and (1,0)^T?
dusky epoch
"the column rows"
supple hemlock
Yep.
half storm
@supple hemlock she’s saying that what you said is a contradiction”
wintry steppe
@supple hemlock when does $a+(b+c)x+dx^2 = 0$
stoic pythonBOT
robo™:
forest quail
can someone help me with this question? i don't really understand what i'm supposed to solve for
wintry steppe
@forest quail well p(0) = 3, p(-3) = 9, p(2) = -1
forest quail
am i solving for the polynomial p?
half storm
Yup,
wintry steppe
x_1 + 3x_2 + 2x_3 - 2x_4 = 0
-5x_3 + 5x_4 = 0
--
How do I solve for this in terms of r, and k ?
I set x3 = r and x4 = k.
But then I'm stuck because the first equation has all four variables in it.
half storm
I'm just looking at this system of equations and you can't really et x_3 and x_4 as two different variables because it's clear from the second equation that they are necessarily equal.
half storm
That last equation $ -5 x_3 + 5 x_4 = 0 \implies x_3 = x_4$ Can you see why?
stoic pythonBOT
JohntheDon:
wintry steppe
Yes because if you bring x_4 to the other side then you divide by -5 you get x_3 = x_4
half storm
Yup.
wintry steppe
My mistake is that I have to set variables for the columns of the rref which don't have pivots
I just picked 2 random columns instead
ocean sequoia
When it’s said that’d U is a sub space of V and U is invariant under T in L(V) does it imply that U would be invariant under S in L(V)
I’m thinking no but I’m not sure
wintry steppe
you're gonna have to give some conditions on S here
median forum
I think theyre asking if a subspace is invariant under a linear transformation, till be under another
well, not necessarily ofc
wintry steppe
yeah that's why i think they left out a condition on S
median forum
probably
ocean sequoia
What do you mean my condition
median forum
it doesnt change anything
ocean sequoia
But I’m pretty sure> well, not necessarily ofc
@median forum answers the question
median forum
maybe the question youre meaning to ask is if there are non trivial subspaces that are invariant over any linear transformation
in any case, gotta dip
ocean sequoia
No worries thanks
median forum
anytime 
ocean sequoia
^
forest quail
i'm still confused, would the solution be in form of a vector or a polynomial? because there aren't any variables like t, t^2 in here (resending the pic)
wintry steppe
polynomial
which you can identify with a vector if you'd like (if you don't know what i mean by this, ignore it completely)
forest quail
oh i get what you mean, but how would i know which value (3, 9, 1) corresponds to whatever variable degree (t, t^2, etc)?
wintry steppe
polynomial
@wintry steppe
yes?
lol
i was responding to soaringbear's q
oh 😦
anyways
@forest quail you're being asked to find a polynomial p of degree 2 that satisfies T(p) = (3,9,-1). can you write out what this means in terms of the definition of the linear transformation T?
something like p(0) = ....., p(-3) = ...., and so forth
but how would i know which value (3, 9, 1) corresponds to whatever variable degree (t, t^2, etc)
i have no clue what this means. im just working with what's in the picture
forest quail
yeah it would mean p(0)=3, p(-3)=9, and p(2)=-1 right?
gray dust
yes
forest quail
how would i be able to solve for the polynomial p with this info though?
wintry steppe
one method is lagrange interpolation
forest quail
oh i don't think our class has learned that yet
is there an easier way to solve it? this was a hw question from the vector spaces/linear transformations section
supple hemlock
@supple hemlock when does $a+(b+c)x+dx^2 = 0$
@wintry steppe When b = c
stoic pythonBOT
Octoc00pler:
wintry steppe
i learned about lagrange interpolation before linear transformations, but ymmv
grizzled tide
hello
for single variable decomposition
why is the orthonormal eigenbasis of the symmetric matrix A^tA necessarily the set of vectors such that application of L(x) = Ax to the set returns another set of orthonormal eigenvectors
median forum
@wintry steppe thanks 
spice storm
Need help: why am I wrong?
I asked a question yesterday to see if the the vectors have to be orthagonal but no one answered.
pallid rampart
I assume the method you used was $$\frac{z^Tu}{u^Tu}u+\frac{z^Tv}{v^Tv}v$$ right?
stoic pythonBOT
Whoever:
pallid rampart
Ok
Yes for this to work you need the vectors u and v to be orthogonal
Hence why I told you to use gram schmidt's first to create an orthonormal basis
spice storm
But I never learned gram schmidt's till today....
grizzled tide
you could have done orthogonal projection
concatenate u and v into a matrix A and then compute |b - Ax|
spice storm
But I still can't since prof said the vectors have to be zero in order to find the approxpation
grizzled tide
the vectors have to be zero?
grizzled tide
no the dot product of the vectors have to be zero not the vectors themselves necessarily
also i'm pretty sure you can do that
grizzled tide
you have to find z-parallel onto the image of A
spice storm
you are confusing me.
grizzled tide
the image of A is a two dimensional plane in a 4d space
spice storm
I did U^TV then I got -11. But I didn't know U^TV had to be zero to find the Best Approximation
forest quail
i found the solution to the problem i was working on earlier, how did they get the matrix [c, 9a-3b+c, ...]?
half ice
Np, feel free to ask if you need anything else!
wintry steppe
@grizzled tide its because they are an eigenbasis
Meaning they satisfy Ax = λx.
So if you apply the linear transformation A onto any of its eigenvectors, you will get a scalar multiple of the input
forest quail
to solve for basis of a kernel, i would first need matrix A, set it to equal 0, rewrite it to rref, rewrite to parametric vector form, then write solutions as a basis right?
but i'm confused on how i would get matrix A from the info provided
half ice
Since T is a linear transformation, there's a matrix representation of it. You're interested in that matrix
An easy way to find it, is to check T(1,0,0), T(0,1,0) and T(0,0,1)
half storm
I have a question, do you even need to resort to the matrix representation of the transformation?
half ice
No, you don't! This is straightforward by finding what maps to 0
forest quail
to check T(1, 0, 0), would i be plugging 1, 0, 0 as values of x in polynomial p?
half ice
So the function is T: P2 → R³
That is, you feed it a quadratic and it gives you back a point in 3D space
When I say (1,0,0)
That refers to the polynomial 1x² + 0x + 0
forest quail
ohh that makes sense
how would i find the transformation matrix from those polynomials?
half ice
Wait, now that I'm rereading it... why do they give you a function?
p = -2x + 3
I'm going to pretend that isn't there! Rofl
forest quail
it was continued from another problem
half ice
Okay, so first question, what's
T(x²)?
forest quail
oh do you not have to plug it into p?
half ice
x² itself can be represented by (1,0,0) in P2
T(x²) is the transformation applied to that
forest quail
oh i see
half ice
This happens to be the first column of the matrix representation
T(x) is the second
T(1) is the third
forest quail
oh yeah that makes sense
half ice
And so we have our matrix
|0 0 1|
|9 -3 1|
|4 2 1|
forest quail
so now i would set that matrix to equal the zero vector and solve?
forest quail
ok!! thank you so much for helping
torpid horizon
anyone here knowledgeable in comp sci
limber sierra
wintry steppe
Let's say I have vectors in R2: v1, v2, v3, ..., vn. I want to find a rotation that will minimize how far (where distance can be defined suitably to make the problem easier) the vectors are from either the horizontal or the vertical. How do I do this, short of just testing angles?
wintry steppe
<@&286206848099549185>
half ice
@wintry steppe
This sounds ripe for a least squares regression haha
And the problem is differentiable so that's lit
Let θn be the angle of each vector, an is the length. The distance away from the angle r is:
an sin(θn - r)
I care to minimize the sum of the squares:
Σ an² sin²(θn - r) summed over n
I can control r, which represents our rotation. Cuz r is differentiable it's enough to take the derivative in terms of r and set it to 0:
Σ an² sin(2θn - 2r) = 0
Solving that for r gives the rotation
Natural question,"how"
Iunno.
Wait no I know exactly how. Sum and difference formula ftw
Σ a²sin(2θ)cos(2r) - a²cos(2θ)sin(2r) = 0
cos(2r) Σ a²sin(2θ) = sin(2r) Σ a²cos(2θ)
tan(2r) = Σ a²sin(2θ) / Σ a²cos(2θ)
Awesome
wintry steppe
hmm I'll probably just use the minimization objective and gradient descent
weary isle
Could anyone explain https://en.wikipedia.org/wiki/Cramer's_rule#Proof ?
limber sierra
is there anything in specific you dont understand?
weary isle
Both https://math.stackexchange.com/questions/1269854/cramers-rule-proof-question and https://math.stackexchange.com/questions/1941590/how-does-cramers-rule-work are confusing to me...
limber sierra
what in specific is holding you up about this proof?
do you see how we derive the second equation from the first? do you see how that second equation tells us that Delta x = Delta_1? do you see how that proves Cramer's rule?
(at least in the two variable case, but it generalizes as one would expect)
weary isle
@limber sierra why add (0 1) to the right ? What is the purpose of doing so ?
limber sierra
that's how cramer's rule works?
weary isle
???
limber sierra
are you familiar with the statement of cramers rule
weary isle
cramer rule does not add (0 1)
limber sierra
in the equation Ax = b, cramer's rule says "replace a column from A with b"
in other words, going from $\begin{pmatrix}a& b\c& d\end{pmatrix}$ to $\begin{pmatrix}p& b\q&d\end{pmatrix}$
weary isle
okay
stoic pythonBOT
Namington:
limber sierra
(if you chose the left column, that is)
weary isle
but what does it mean by delta(x) = delta(1) ?
limber sierra
do you see how $\begin{pmatrix}p&b\q&d\end{pmatrix} = \begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x&0\y&1\end{pmatrix}$?
stoic pythonBOT
Namington:
weary isle
yes
limber sierra
cool
weary isle
@limber sierra I am asking about delta(x) = delta(1)
limber sierra
delta here is det(A) where A is this matrix
delta_1 is det(A_1) where A_1 is A but with the 1st column replaced with b
weary isle
@limber sierra but why delta(x) = delta(1) ?
limber sierra
similarly, delta_2 would be det(A_2) where A_2 is A but with the second column replaced
first, its not delta(x)
it's delta * x
x is a vector
the vector {x, y}
weary isle
?
limber sierra
$\Delta x = \det(A) \cdot x$ and $\Delta_1 = \det(A_1)$ where $A_1 = A$ but with the 1st column replaced with $b$
ugh
stoic pythonBOT
Namington:
limber sierra
dividing both sides by delta gives
$x = \frac{\Delta_1}{\Delta} = \frac{\det(A_1)}{\det(A)}$
stoic pythonBOT
Namington:
limber sierra
which is cramer's rule
since x here is part of the solution vector {x, y}
ok let me clarify since
im abusing the fuck out of notation here
i should say that
the solution vector is MADE up of entries
delta_n / delta
as appropriate
limber sierra
the last equation is Delta x = Delta_1
this is already determinants
since Delta is just notation for the determinant of A
and Delta_1 is just notation for the determiannt of A_1, which is what you get when you take the matrix A, and replace the first column with the vector {p, q}
and x here is the first entry of the solution vector {x, y}
weary isle
@limber sierra I mean how to take the determinant of this last equation ???
stoic pythonBOT
Namington:
weary isle
but what about the left side of the expression ?
limber sierra
can you take the determinant of $\begin{pmatrix}a&b\c&d\end{pmatrix}$ and of $\begin{pmatrix}x&0\y&1\end{pmatrix}$?
stoic pythonBOT
Namington:
stoic pythonBOT
Namington:
limber sierra
taking determinants and applying this rule gives you the result
$\det\begin{pmatrix}a&b\c&d\end{pmatrix} = \Delta$
stoic pythonBOT
Namington:
limber sierra
and $\det\begin{pmatrix}p&b\q&d\end{pmatrix} = \Delta_1$
stoic pythonBOT
Namington:
limber sierra
and clearly $\det\begin{pmatrix}x&0\y&1\end{pmatrix} = x\cdot 1 - y \cdot 0 = x$
stoic pythonBOT
Namington:
limber sierra
so
$\det\left(\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x&0\y&1\end{pmatrix}\right) = \det\left(\begin{pmatrix}a&b\c&d\end{pmatrix}\right)\cdot \det\left(\begin{pmatrix}x&0\y&1\end{pmatrix}\right) = \Delta \cdot x$
stoic pythonBOT
Namington:
limber sierra
that accounts for the left hand side
whereas the right hand side is just Delta_1 as established
so Delta * x = Delta_1
i.e.
det({a, b; c,d}) x = det({p, b; q, d})
limber sierra
thats just the statement of cramers rule
not a proof
maybe youre confusing "what is cramer's rule?" and "prove cramer's rule"
those are two different questions
your links answer the second quesstion
but that image is just answering the first.
limber sierra
thats an alternate proof just from algebraic manipulations
note here that when we write (b, a_2, a_3), we're actually writing the columns of a matrix
b is the first column, a_2 the second, a_3 the third
so |b a_2 a_3| is the determinant of a matrix with first column b, second column a_2, third column a_3
i.e. the determinant of the matrix A but with the first column replaced with b
from there it just does algebra
using the fact that the determinant is "multilinear on columns"
this is a piece of terminology jargon
weary isle
limber sierra
if youre not super comfortable with linearity it might be weird
weary isle
how to obtain the second line of equation from first line ?
limber sierra
bleh writing this is a pain in the ass
you just break it up using multilinearity
are you familiar with the determinant property that
$\det\begin{pmatrix}a+b & c & d\end{pmatrix} = \det\begin{pmatrix}a & c& d\end{pmatrix} + \det\begin{pmatrix}b&c&d\end{pmatrix}$?
stoic pythonBOT
Namington:
limber sierra
where here a, b, c, d are column vectors
from there it uses the related property that $\det\begin{pmatrix}xa&c&d\end{pmatrix} = x\det\begin{pmatrix}a&c&d\end{pmatrix}$
stoic pythonBOT
Namington:
limber sierra
so basically, "first" the author "splits up" the nasty determinant based on addition
then the author factors out x_1 from the first determinant, x_2 from the second, and x_3 from the third
but he combines this into one step
as for why this stuff goes awway
again its another determinant property
when multiple columns are the same, the determinant must be 0
so we have x_2 * 0 + x_3 * 0
which is of course
0
by the way: i'm talking about columns here, but you may be familiar with these in terms of rows of a matrix, not columns
because det(A) = det(A^T), all statements about rows apply to columns
and vice versa
so everything that holds for rows holds for columns
weary isle
x_2 * 0 + x_3 * 0 ???
weary isle
huh ?
limber sierra
again since multiple columns are the same
if youre not convinced:
recall that det(A) = det(A^T)
weary isle
Why det(a_2 a_2 a_3) must be 0 ?
limber sierra
also recall that det(A) = 0 iff A has a zero row
so take the transpose
det((a_2 a_2 a_3)^T)
so a_2, a_2, a_3 are the rows
subtract the first row from the second
you have a zero row
boom, determinant of 0
if this:
det(A) = 0 iff A has a zero row
is confusing you, remember that a matrix is invertible iff its RREF is an identity matrix
but here we've shown the RREF is not the identity matrix
so the matrix cant be invertible
hence its determinant must be 0
since a noninvertible matrix has 0 determinant
there's like a billion ways to prove all these facts
i've just given a couple
but they're honestly basic facts you should know if you want to work with the determinant in a proofs context
they're very handy
and the reason why the determinant is so useful for proofs
weary isle
$\det\begin{pmatrix}a+b & c & d\end{pmatrix} = \det\begin{pmatrix}a & c& d\end{pmatrix} + \det\begin{pmatrix}b&c&d\end{pmatrix}$
stoic pythonBOT
promach:
limber sierra
the idea is very similar
det(x_1a_1 + x_2a_2 + x_3a_3 , a_2 , a_3) = det(x_1a_1 , a_2 , a_3) + det(x_2a_2 , a_2 , a_3) + det(x_3a_3 , a_2 , a_3) = x_1det(a_1 , a_2 , a_3) + x_2det(a_2, a_2, a_3) + x_3det(a_3, a_2, a_3)
as i just explained
they just combined these two steps into one.
weary isle
wait, why det(x_1a_1 + x_2a_2 + x_3a_3 , a_2 , a_3) = det(x_1a_1 , a_2 , a_3) + det(x_2a_2 , a_2 , a_3) + det(x_3a_3 , a_2 , a_3) ?
@limber sierra
limber sierra
how do i answer this
these are just determinant properties
you can ask for a proof but then youre asking a different question
and i dont want to spend my night proving every single theorem from linear algebra
over and over again
weary isle
@limber sierra never mind, I will study this on my own
by the way, have you seen https://math.stackexchange.com/questions/1941590/how-does-cramers-rule-work#comment4024604_1941606 before ?
wintry steppe
If I am asked to determine the image of a matrix
Is it just to put it in rref and list the columns that don't have pivots ?
And those columns also form a base ?
Image is range I think
half storm
I mean if you just want to determine the image of a matrix, then you just want the set of all possible column vectors / n-tuples that you can achieve from the matrix product Ax
If you have a m x n matrix A right then , the left-multiplication transformation and $L_A : F^n \to F^m := Ax$ so you just perform the matrix multiplicaton.
stoic pythonBOT
JohntheDon:
half storm
Finding a basis is a different problem.
You know that the columns of the matrix span the image of the matrix so if you want to find a basis for it, then you can rref A. If there is a pivot in every column, then they are a linearly independent set.
If you find that this is not the case then the columns are linearly dependent and you can exclude one of them from the set.
supple hemlock
Hey
For 4c, how would I represent my answer?
I get [0,0,0,0]
[0, a, -a, 0]
[0,-a,a,0]
I just struggle with formalizing my basis
dusky epoch
i diagnose you with positive variable disease
dusky epoch
no, it would not
anyway, positive variable disease is when you make the implicit assumption that a variable without a negative sign in front of it always represents a positive value and one with a negative sign always represents a negative value
stoic pythonBOT
Ann:
supple hemlock
LOL ahh
So by making that assumption, I struggled w/compressing my solution into te kernel you gave me?
the*
dusky epoch
you struggled with...
algebra, really
basic algebra
point is, those who have the disease don't know they have it
supple hemlock
Welp, thank you for diagnosing me!
dusky epoch
anyway from here it should be obvious that ker(L) has dimension 1
and for your basis you only need one matrix, for example [0 1; -1 0]
ocean sequoia
im not following where the second step came from
umbral smelt
What's 5.12?
ocean sequoia
umbral smelt
Ah Axler
Okay, so from 5.12, what do you get after multiplying both sides with $\lambda_k$?
stoic pythonBOT
Wilston Lynx:
ocean sequoia
wouldnt you just get $\lambda^2_k v_k = \lambda_k (a_1 v_1 \lambda_1 +....+a_k-1 v_k-1 \lambda_k-1)$
stoic pythonBOT
brzig:
umbral smelt
Uh, how did you get $\lambda_k^2$ on the LHS?
stoic pythonBOT
Wilston Lynx:
umbral smelt
We only multiply $\lambda_k$ on each sides no?
stoic pythonBOT
Wilston Lynx:
umbral smelt
5.12 is, $v_k=a_1v_1+...+a_{k-1}v_{k-1}$
stoic pythonBOT
Wilston Lynx:
ocean sequoia
oh ok
umbral smelt
So you can get $$\lambda_k v_k=a_1\lambda_kv_1+...+a_{k-1}\lambda_kv_{k-1}$$
after multiplying with $\lambda_k$ on both sides
stoic pythonBOT
Wilston Lynx:
umbral smelt
Then you can subtract this equation with the one you get after applying T on both sides
umbral smelt
You're welcome!
ocean sequoia
wintry steppe
I'm wondering what my initial state matrix is in this case:
If the weather is good one day, there is a 10% chance it will be sunny the next day, 60% chance that the weather will
is rainy and a 30% chance (or rather risk) that it is snowy.
If the weather is rainy, there is a 50% chance that it will be sunny the next day, and 0% chance that the
weather remains rainy.
Finally, if it is snowy, there is a 25% chance that it will be sunny the next day and a 50% chance that the
weather is rainy.
What is the long term probability that the weather is good in this city?
From my understanding I have to use markov chains for this ? But no initial state matrix is given, so I'm not sure what to do.
tribal lodge
hi uh could someone walk me through this proof
I know theres a definition that says (AB)^T = B^T A^T but im not sure how that fits into this
& also if its a nxn matrix, AA^T = I, and A^T = A^-1
teal topaz
Remember that since $B^{-1}=B^T$, we have $B^TB=I$
stoic pythonBOT
Stract:
half storm
Think about what it means to take the transpose of the product of two matrices yea
teal topaz
Try working backwards a bit from the RHS
tribal lodge
wait so
B^T (A B^T) ^T
B^T (B A^T)
B^T B A^T
i A^T
A^-1 = A^T
A^-1 = A^-1
:o?
OH
yay : D
thank you 🙂
barren void
Should be all the linear polynomials such that T(p) = 0
I think that the kernel is trivial
Just the 0 polinomial
tribal lodge
a lil confused, what do you mean by all linear polynomials where T(p) = 0?
barren void
This transformation goes from the vector space of polynomials at most degree 1 to the vector space of 2x2 matrices
So, your kerner is a subspace of the vector space of linear polynomials
So, let p be a polynomial, p is in the kerner of T if T(p) = 0 matrix
half ice
@wintry steppe
Wait what? Why lol? I gave you a closed form. Maybe I wasn't clear about something?
tribal lodge
@barren void your 'kerner' is kernel yes?
Also, if P_1 and the M22 dont have the same dimensions, does this mean that T is "onto"?
forest quail
would this be false because if u and v are linearly dependent, it wouldn't be a plane?
slow scroll
yup
forest quail
ok tyy
slow scroll
npnp
barren void
@barren void your 'kerner' is kernel yes?
@tribal lodge yes sorry lol
Also, if P_1 and the M22 dont have the same dimensions, does this mean that T is "onto"?
@tribal lodge
For being onto it should be a surjective function
It means that the Image of T (also known as the Range of T) is all vector space of matrices 2x2
You should be able to tell if T is onto, i.e, surjective or not.
Its quite easy to tell think about what matrices you can obtain using this transformation
tribal lodge
@barren void
So this means that T is onto because the range of T is all vector sp. of 2x2?
Onto for matrix transformations say that: If A is a m x n matrix, and T(x) = Ax be the associated matrix transf, then
- T is onto
- T(x) = b has at least 1 solution for every b
- Ax = b is consistent for every b
- Columns of A span R^m
- A has a pivot in every row
- range of T has dimension m
wintry steppe
@wintry steppe Ax - 7x = Ax - 7Ix = (A - 7I)x
yeah
but why is 7x equal to 7I
it's not
7x = 7Ix because x = Ix, where I is the identity matrix
ohh ok
great
thanks
it's the x that becomes Ix , not 7 that becomes 7I
yeah
then you just pull x out the right
and you get a nice looking (A - 7I)x
Why is this "obviously" linearly dependent
Everytime I try to look it up online I don't get it lol