You need the inner product, yes.
#linear-algebra
wintry steppe
marble lance
I have no idea what you mean by the dual space then
If you say it consists of bounded linear functionals but you can't define bounded
Because you have no inner product
wintry steppe
I have no idea what you mean by the dual space then
Well, generally, the dual space is defined as the space of linear functions from the vector space to the underlying field.
When you have infinite dimensional spaces, it is restricted to bounded linear maps.
marble lance
Well, there are two kinds of dual spaces. Algebraic and continuous dual spaces. The former being like you said all linear functionals, the second being the continuous/bounded ones.
Riesz has to do with the continuous dual space
Which is why I thought you were talking about that
wintry steppe
Riesz has to do with the continuous dual space
You're right.
I was wondering if such an inner product based isomorphism existed for pseudo inner product spaces.
marble lance
But there is no continuous dual space
wintry steppe
Let's not worry about the infinite dimensional case right now.
Assume finite dimensions.
marble lance
Well, then I assume the answer is no. Because you won't be using the pseudo inner product at all when constructing the dual space.
So they shouldn't be connected.
wintry steppe
Well, then I assume the answer is no. Because you won't be using the pseudo inne...
Well, neither is an inner product.
marble lance
The inner product defines the continuity.
So it is what chooses which linear maps make it into the continuous dual space. It "chooses" the ones that have a riesz rep.
wintry steppe
So it is what chooses which linear maps make it into the continuous dual space. ...
No, I get that.
marble lance
The pseudo inner product will do nothing. So I mean maybe it's true, but it seems very unlikely.
wintry steppe
What about the musical isomorphisms?
marble lance
Yeah, you have gone out of the scope of my knowledge. Sorry
wintry steppe
LOL
quartz compass
what, you can't just learn this by skimming the wikipedia article @marble lance ? smh
marble lance
wintry steppe
And also help poor SK
what about the musical isomorphisms? the fact that they're even a thing is just non-degeneracy
quartz compass
it's just lowering and raising indices
wintry steppe
what about the musical isomorphisms? the fact that they're even a thing is just ...
Isn't this similar to Reisz?
perhaps it's similar, you're getting an isomorphism between a space and its algebraic dual
wintry steppe
it's just lowering and raising indices
Because you have an isomorphism between the vectors and covectors, defined using the pseudo inner product.
wintry steppe
perhaps it's similar, you're getting an isomorphism between a space and its *alg...
Yes.
I mean, aren't algebraic and continuous duals the same in finite dimensions?
that's true if you have an inner product, since then "continuous" makes sense. but right now we're only talking about psuedo inner products.
wintry steppe
that's true if you have an inner product, since then "continuous" makes sense. b...
Yes.
Are the musical isomorphisms, which are defined in terms of a pseudo inner product, actually isomorphisms?
yes
by non-degeneracy
why would you call them musical isomorphisms if they werent isomorphisms?
wintry steppe
why would you call them musical isomorphisms if they werent isomorphisms?
That's true.
quartz compass
maybe you should work out an explicit example of a pseudo inner product to see what happens
wintry steppe
Do you guys have any links to the proof of musical isomorphisms?
That'd help.
what do you mean? do you want to know why they're isomorphisms?
wintry steppe
what do you mean? do you want to know why they're isomorphisms?
Yes please.
what does "non-degenerate" mean for you?
wintry steppe
what does "non-degenerate" mean for you?
The non-degeneracy I am familiar with is defined in terms of the musical maps.
what is the precise definition of non-degeneracy that you know
wintry steppe
what is the precise definition of non-degeneracy that you know
A bilinear form is said to be non-degenerate if there exists the musical map
$\begin{eqnarray*}\flat : \Gamma(TM) &\rightarrow &\Gamma(T^M)\
X &\mapsto& \flat(X) := \flat(X)(Y)=g(X,Y)
\end{eqnarray}$
stoic pythonBOT
SK
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
wintry steppe
this map always exists. what do you want to say about it?
that it's an isomorphism?
wintry steppe
Do you guys have any links to the proof of musical isomorphisms?
if yes then the answer to this is "there is none" because you don't prove a definition
wintry steppe
that it's an isomorphism?
Yes.
wintry steppe
this map always exists. what do you want to say about it?
How can I show that this map always exists?
ok, let me ask for clarification. what precisely do you mean by "this map always exists?"
i don't know what it means other than "this function is well-defined"
wintry steppe
ok, let me ask for clarification. what precisely do you mean by "this map always...
Let's say that I have a pseudo Riemannian manifold. How can I show that at each point, the pseudo inner product on the tangent space at that point induces an isomorphism between the tangent space and the cotangent space at that point?
quartz compass
by definition you have a nondegenerate metric tensor
otherwise it's not a pseudo riemannian manifold
wintry steppe
I feel like this is going in circles.
Okay, so, non-degeneracy of the pseudo inner product implies the existence of these musical isomorphisms, yes?
yes, by definition
quartz compass
this is good, it means you take your specific space, show it's pseudo riemannian, and everything else you get for free
wintry steppe
yes, by definition
Now, how does the non-degeneracy imply the existence of these isomorphisms?
what
I mean, if it's non-degenerate, we have these isomorphisms.
How do I show this?
it's by definition, i said
you defined it that way
they exist and are isomorphisms because you defined it that way
wintry steppe
it's by definition, i said
What about this definition of non-degeneracy?
that one's equivalent to yours from earlier, at least in finite dimensions
wintry steppe
that one's equivalent to yours from earlier, at least in finite dimensions
Okay, how do I show that it is equivalent?
I'm sorry for dragging this out for so long.
stoic pythonBOT
TTerra
wintry steppe
conversely, if flat is an isomorphism, then the kernel above is trivial, which implies the wikipedia definition of non degeneracy
so they are equivalent definitions, for finite-dimensional spaces
quartz compass
since this is the linear algebra channel, we can just think about it in terms of matrices
nondegeneracy of a metric tensor means the matrix we're using to represent it has nonzero determinant
so these musical isomorphisms are nothing more than just matrix multiplication of your vector by the metric tensor or its inverse depending on which way you're going
trying to bring this down to earth here, cause I feel like it's much clearer to see if you've worked out examples with literal numbers in number boxes
wintry steppe
literal numbers in number boxes
quartz compass
schuller called them I think 'cemetaries for numbers' or something like that if I remember lol
that's the person in @wintry steppe 's pfp I'm pretty sure
wintry steppe
that's the person in <@456226577798135808> 's pfp I'm pretty sure
Yes! 
lofty scroll
Can someone help me on the part where they show that beta is linearly independent? https://math.stackexchange.com/a/4199707/928469
wintry steppe
quartz compass
dont dox me bro 
lofty scroll
I understand that part where the two sums lie in the intersection $V_1 \cap V_2$, however I don't follow how that implies $\sum \alpha j v{i_j}=0$
stoic pythonBOT
tuturuuu
wintry steppe
gamma union {v_i's} is linearly independent
Let γ be a basis for V1∩V2. Complete it to a basis for V1 by adding vectors {vi}i∈I; also complete to a basis for V2 by adding vectors {wi}i∈I
by this part
lofty scroll
I had some help from someone and this is what they said
Am I right that $W_1$ is the span of ${v_i}$ and $W_2$ is the span of ${w_i}$? How does it follow that $W_1 \cap W_2={0}$?
stoic pythonBOT
tuturuuu
dire thunder
@lofty scroll the point is that if dim V_1 != dim V, then you can use it to generate basis for subspace that won't include V_1 and V_2 and provide direct sum
dire thunder
so ok
if we had only one space it would be trivial
we would just extend (v_1, ..., v_m) to basis of V and let W be space spanned by basis of V with vectors in basis of V_1 removed
however this fails in our case
dire thunder
@lofty scroll so point is to adjoining them correctly to some combination of extension of bases to get W
lofty scroll
Took me a long time but I somehow get it now, thank youu!
fast ice
hey guys i have a small question. i was watching ProfessorDaveExplain's video on eigenvalues and eigenvectors and i was solving this question. I got eigenvalues -2 and +2, which was correct... but the eigenvectors i got were [1,-1] and [1,-5] respectively. This answer is basically his answer but flipped. Is that fine or did i do something wrong?
feral jetty
Hi, I was reading wikipedia link on lagrange polynomials and lagrange interpolation: https://en.wikipedia.org/wiki/Lagrange_polynomial
Can someone explain how they computed the derivative here?
It seems that l(x) and l'(x) are the same?
tidal wharf
they used lots of product rule
but l' is different from l
like l(x_j) = 0
$\ell'(x) = \sum_{i=0}^k \prod_{j\ne i}(x-x_j)$
stoic pythonBOT
waschbaerrr
feral jetty
$\ell'(x) = \sum_{i=0}^k \prod_{j\ne i}(x-x_j)$
How did you work that out so quickly?
Is it something obvious with the product rule?
tidal wharf
I used this formula https://en.wikipedia.org/wiki/Product_rule#Product_of_more_than_two_factors or multivariable chain rule
the one that says $\frac{d}{dx}\left[\prod_{i=1}^kf_i(x)\right]=\sum_{i=1}^k\left(\left(\frac{d}{dx}f_i(x)\right)\prod_{j\ne i}f_j(x)\right)$
stoic pythonBOT
waschbaerrr
feral jetty
the one that says $\frac{d}{dx}\left[\prod_{i=1}^kf_i(x)\right]=\sum_{i=1}^k\lef...
oh thank you very much
glacial mango
hey guys i have a small question. i was watching ProfessorDaveExplain's video on...
So I think [1,-1] is fine because it's a scaled version of [-1,1], but I don't think [1,-5] is OK because it's not a scaled version of [-5,1]. Although it is still a basis for the eigenspace, I think it would correspond to a different eigenvalue.
You should get a second opinion though.
feral jetty
I used this formula https://en.wikipedia.org/wiki/Product_rule#Product_of_more_t...
Why does the sum disappear when I evaluate it at a point?
tidal wharf
Why does the sum disappear when I evaluate it at a point?
each term in the sum is a product of some (x-x_k)'s, so if you plug in a particular x_j, all the terms that have a factor (x-x_j) will vanish
and there is only one term in the sum that does not have a factor (x-x_j)
feral jetty
and there is only one term in the sum that does not have a factor (x-x_j)
oh, so when the sum is at i=0, the only way for the product to be non-zero is if we are evaluating at x_i
because that is where j != i
okay got it, thank you
tidal wharf
hey guys i have a small question. i was watching ProfessorDaveExplain's video on...
as @glacial mango said [1,-1] is fine. But [1,-5] is not an eigenvector
glacial mango
Do matrix multiplication and scalar multiplication commute with vectors? i.e. Is $Ac\vec{x_1} = cA\vec{x_1}$?
stoic pythonBOT
modus ponens
tidal wharf
yes you can use matrix multiplication formula to check
raven parrot
Hello, I realized that in ALA (Applied linear algebra), something called finite value decomposition is taught. I looked it up and had no clue how it works. Can someone give me insight on how it works?
twilit anvil
Do matrix multiplication and scalar multiplication commute with vectors? i.e. Is...
can one not easily see the property from A being a linear map ? (unless you have not talked about that yet, which is very possible, apologies if this is the case)
leaden tide
can one not easily see the property from A being a linear map ? (unless you have...
Well, it's like answering "well, doesn't that follow from f being a linear map" to "why does f(x+y)=f(x)+f(y)?"
turbid trellis
Can a vector be part of a span of a linear combination if there are free variables in the reduced echelon form ?
The book I'm reading is unclear about this
twilit anvil
Well, it's like answering "well, doesn't that follow from f being a linear map" ...
thats very true, not sure what i was thinking
leaden tide
Can a vector be part of a span of a linear combination if there are free variabl...
I'm unsure what you mean by this
turbid trellis
I'm learning about when a linear combination is equal to a vector
In a way that the vector is in the span of the linear combination
The book teaches me to create a system of linear equations from it, and to solve it
leaden tide
The vocabulary is not quite this
Namely, the span of a set of vectors is all the possible linear combinations made from that set
turbid trellis
And my question was, if there are infinite solutions to the system, will this say that it is part of the span ?
leaden tide
So, "span of the linear combination" sounds redundant
turbid trellis
Span of the set of vectors then
lavish jewel
yes, having at least one solution means it's spanned by the columns of the matrix
leaden tide
that's probably right
turbid trellis
Okay and how about a single row which contains zero's ?
leaden tide
idk why i took the question, I haven't eaten in 24 hours
turbid trellis
I have questions which have that and are marked as 'not a linear combination of' which have a row of zeroes
lavish jewel
go eat 
leaden tide
don't worry too much about me, i'm not that hungry despite all likelihood
lavish jewel
if it's marked as "not a linear combination of", it means that there is no x such that Ax = b
twilit anvil
seriously? if i havent ate anything in 24 hours, id would eat my dummit&foote here in front of me
turbid trellis
if it's marked as "not a linear combination of", it means that there is no x suc...
ax+b is handled in the next chapter
I haven't had it yet
lavish jewel
you're already doing it, but ok
you have your vectors and you put them in a matrix and rref
you have a row of all 0s, meaning there are linearly dependent vectors
in this case, it can be that there are vectors that are not in the span of the set you were given
for example if i give you the vectors (1,0,0) and (0,1,0), and i ask you if (0,0,1) is in their span
lavish jewel
no
you could have 1, 0, or infinitely many ways of producing the vector from a given set
you have to check each time
if you rref the 2 vectors i gave you above, the last row is (0,0)
turbid trellis
I mean a row of zeroes in the reduced echelon form
lavish jewel
still
turbid trellis
All questions which had that are marked as not a linear combination of, and I was thinking if this somehow is a rule
That there only can be one answer
lavish jewel
it depends on the number of vectors
and which vectors they are
there is no rule that will tell you immediately
it also depends on which vector they ask you about
turbid trellis
3x4 augmented matrix, row 3 is all zeroes in the reduced echelon form
lavish jewel
do you know what the null space is?
lavish jewel
well, for square matrices (before augmenting), having a row of all zeros means you only have 2 linearly independent vectors in the matrix columns
this means you will either have infinitely many solutions, or no solutions
which one it is depends on which vector you are given
turbid trellis
Yeah I understand because 0 != 1
But I was asking for the reduced echelon form of the augmented matrix
lavish jewel
i don't know what you mean by that
turbid trellis
Nevermind then I do
However is there some rule for only one solution if checking if something is a linear combination ?
lavish jewel
you rref the augmented matrix and you get an identity and an extra column of something else (doesn't matter what it is)
that's the only case where you have exactly 1 solution
turbid trellis
I'll just send an image
lavish jewel
just read what i said. if you don't get a full identity matrix, you don't have 1 solution
turbid trellis
Yes I understand, but the book I'm reading gives some strange answer to question
An matrix which has infinite solutions
augmented matrix*
lavish jewel
aight, send the image
turbid trellis
13
As far as my calculations go, this has infinite solutions when checking if it's an linear combination of it
However it's marked as not a combination of
Or am I being a idiot at the moment ? I'm kinda overwhelmed with a lot of theory haha
Oh my god I found the mistake
Sorry for wasting your time I'm being an absolute idiot
I was solving the matrix the coefficient matrix
stoic pythonBOT
zealous junco
i think this is a property?
$\sum_i^D u_i\cdot u_i^\top = I$ if $u_i$ are some orthobasis
stoic pythonBOT
Anticipation
zealous junco
idk whether its true and if so how to prove it
lavish jewel
hmm only if you have as many vectors as the dim of the full space
wintry steppe
Does the positive definiteness of an inner product induce non-degeneracy? Or are they independent?
positive-definiteness implies non-degeneracy
stoic pythonBOT
TTerra
wintry steppe
not conversely, though
wintry steppe
not conversely, though
Thanks.
stoic pythonBOT
phao
wintry steppe
positive-definiteness implies non-degeneracy
How about positive semi-definiteness?
zealous junco
by defition, if its pos semi-def but not pos def, then u got an x so that x^tAx = 0
silent sandal
Hi.
Is there a "standard" way to take a vector norm on $\bR^n$ and extend it to $\bC^n$ such that the induced matrix norms coincide?
I mean, say I have a norm $N$ in $\bR^n$. I'd like a norm $M$ in $\bC^n$. Such that
$$\sup\limits_{N(x)=1, x \in \bR^n} N(Bx) = \sup\limits_{M(x)=1, x \in \bC^n} M(Bx)$$
and also that for all $x \in \bR^n$, $N(x) = M(X)$.
I imagine that if this norm $N$ comes from an inner product, then this is true. I'm wondering about the general case.
stoic pythonBOT
phao
wintry steppe
by defition, if its pos semi-def but not pos def, then u got an x so that x^tAx ...
What's A?
The pseudo inner product?
zealous junco
yea
wintry steppe
yea
I was asking about non-degeneracy.
Can I show non-degeneracy from positive semi definiteness?
zealous junco
oh oops
wintry steppe
but that's what anticipation did
if g(x, x) = 0 for some non-zero x (which may happen if you are working with a positive semidefinite tensor) then g cannot be non-degenerate
wintry steppe
if g(x, x) = 0 for some non-zero x (which may happen if you are working with a p...
Wait, so pseudo inner products aren't non-degenerate?
a psuedo inner product is non-degenerate by definition. i am answering your question (or rather expanding on what anticipation said) of whether positive semi-definiteness implies non-degeneracy
wintry steppe
a psuedo inner product is non-degenerate by definition. i am answering your ques...
Okay, so positive semi definiteness doesn't imply non-degeneracy; pseudo inner products have to additionally satisfy non-degeneracy together with being positive semi definite.
zealous junco
tbh i was unsure because i don't know about bilinear forms so I thought what you asked may have been out of scope lol
wintry steppe
psuedo inner products are not required to satisfy positive semi definiteness
only non degeneracy
nothing else
So, how do we actually pseudo inner products?
Symmetric bilinear forms that are non-degenerate?
yes
lucid glacier
Wait are they only nondegerate or anisotropic
fast ice
can someone help me? I know that a) is correct because of the pivot columns, but somehow b) and d) are also correct? how does that work?
tidal wharf
If the columns of R form a basis for the column space of R, then the corresponding columns of A form a basis for the column space of A
(since row operations don't change dependency of columns)
stoic pythonBOT
tuturuuu
lofty scroll
Is it true that if $span(\beta)\bigcap span(\gamma)={0}$, then the union of $\beta$ and $\gamma$ is linearly independent?
stoic pythonBOT
tuturuuu
wintry steppe
should be
if $\beta={v_i}$ and $\gamma={w_j}$, and if the spans intersect only at $0$, then, if $\sum_i a_iv_i + \sum_j b_jw_j=0$, $$\sum_i a_iv_j = -\sum_j b_jw_j.$$ but this sum is simultaneously in the span of $\beta$ and of $\gamma$, so by linear independence of each, the coefficients vanish
lucid glacier
In general if 2 nonzero subspaces have trivial intersection a union of independent sets from both would he independent
stoic pythonBOT
TTerra
lofty scroll
Thank youuu!, that was the justification I tried
fast ice
If the columns of R form a basis for the column space of R, then the correspondi...
this would explain a) being correct, but how does it explain b and d also being correct?
lucid glacier
I don't think it was ever stated only 1 option is corrext
It said to state which are correct and which are incorrect
fast ice
yea, the answer sheet says a b and d are correct but i only understand how a is correct
lucid glacier
It also explains how b and d are correct, since the 2nd, 3rd and 4th columns still form a linearly independent subset of size 3
Since
$$\begin{pmatrix}1\0\0 \end{pmatrix}, \begin{pmatrix}0\1\0 \end{pmatrix}, \begin{pmatrix}1\0\1 \end{pmatrix}$$
Is linearly independent (The corresponding columns in R)
stoic pythonBOT
ShiN
fast ice
OOH i got it
thanks for the help
i just assumed that i only needed to look at the pivots in the reduced matrix and decide the basis from that
i didnt think to look for other combinations of independent columns
verbal shard
wintry steppe
small glen
Can I get some advice on how to solve this question? No answer needed.
*Edit: To anyone seeing this question in the future, the answer is that v is not a linear combination of u1 and u2.
tidal wharf
set up a matrix for row reduction
north hedge
i would just do sim equations with the bottom 2 components an see if its consistent with the first
stoic pythonBOT
bobles
north hedge
and then just check the top
zealous junco
i dont get what they are doing here in the middle line
nvm they are just equal nothing going on
native rampart
How did a 5x6 matrix become a 5x5 in the last step
fleet orbit
How did a 5x6 matrix become a 5x5 in the last step
Because since a_5=1, we can move it over to the RHS no?
fleet orbit
I used a matrix calculator to reduce it
fast ice
stupid mistakes like that cost me hours
this basically sums up the whole topic of linear algebra 🥲
zealous junco
real beginner question, lets say i got this png image of plane
like 500x219 = 109500 pixels with rgb 3 channels
so id expect it to be 328500 bytes but its 190000 bytes, what kind of techniques is it using
lavish jewel
google says "png uses DEFLATE, a lossless data compression alg. using LZ77 and huffman coding"
zealous junco
oh thanks
lavish jewel
dictionary techniques and something like run length encoding
wintry steppe
✈️
silent sandal
Hi. I have a real $n \times n$ matriz $B$. I have a vector norm on
$\bR^n$ and I know, for the induced matriz norm, $||B|| < 1$. How do
I prove the spectral radius of $B$ is also smaller than 1. If all
the eigenvalues of $B$ are real, then this is simple. I'm wondering
what I can do in the case $B$ has complex eigenvalues.
stoic pythonBOT
phao
lavish jewel
my best guess is it depends on which induced norm it is
silent sandal
I've managed to show that $||B|| \geq \frac r {\sqrt 2}$, where $r$ is the spectral radius of $B$.
stoic pythonBOT
phao
nocturne jewel
I might be wrong but that means $\frac{r}{\sqrt{2}}\leq \norm{B}<1$
stoic pythonBOT
Mosh
nocturne jewel
so $r\leq \sqrt{2}$
stoic pythonBOT
Mosh
silent sandal
Right. I'd like to show that $r < 1$.
stoic pythonBOT
phao
silent sandal
Right. What that gives me is that if $\sqrt 2 ||B|| < 1$, then $r_B < 1$.
stoic pythonBOT
phao
silent sandal
Btw... idk if that can be proved.
lavish jewel
which induced norm are you working with? or do wanna do something general?
silent sandal
It should be general.
lavish jewel
not even a p-norm?
silent sandal
Right.
For example, for norms which come from an inner product (like the 2-norm) I know this is true. There is a way to take this into the complex numbers in such a way that I can translate results back to $\bR^n$
stoic pythonBOT
phao
lavish jewel
the only trick that comes to mind would be using an easy one like the 2 norm and then equivalence of norms stuff
silent sandal
Right... I'm starting to think this might be false tbh
lavish jewel
you could try to look for a counter example
zealous junco
**phao**
i think it is true right? for any induced norm $|Bv| \leq |B||v|$ so can't you just show that iterating this inequality you get $B^n v \to 0$ w.r.t whatever vector norm $|\cdot |$ your using but if the spectral radius was not smaller than 1 then there would be a $v$ so that $|B^nv|_2$ never diminishes, then use equivalence of norms to show its also supposed to go to 0 which is a contradiction?
stoic pythonBOT
Anticipation
lavish jewel
found this on wiki
they don't show a proof there tho
so let r = 1. if the induced norm on your matrix B is <= 1, then so is the radius
more or less like what anticipation was saying, but without requiring the norm to go to 0 as you increase the exponent
@silent sandal does this do the trick?
fast ice
can someone explain this answer? they assume that T^2 * X = 0, but if you multiply the same transformation matrix by any vector and always get 0, how can that be one to one?
lavish jewel
that's the point.
north hedge
can someone explain this answer? they assume that T^2 * X = 0, but if you multip...
its not for all x
fast ice
how tho? one to one means that you can always get back to the original vector by looking at the result of the transformation, but if you always have 0, and its the same transformation matrix, then how can you differentiate between different original vectors
lavish jewel
they show that the only way you get T^2x = 0 is if x = 0
north hedge
so it has trivial kernel and is hence 1 to 1 (injective)
lavish jewel
they pick a generic x, and then show that this x can only be the 0 vector
stoic pythonBOT
bobles
fast ice
hmm
another question
oh wait i think i gotit
so basically if were assuming that T is 1 to 1, then it would make sense that 0 is a valid outcome for some vector x times the transformation matrix
lavish jewel
yes, for the 0 vector
lavish jewel
only for the 0 vector
fast ice
thanks for the help
silent sandal
@lavish jewel @zealous junco thank you
silent sandal
i think it is true right? for any induced norm $\|Bv\| \leq \|B\|\|v\|$ so can't...
I was going in a very different direction. Thank you.
@lavish jewel Yep. The issue is dealt with 😄
lavish jewel
👍
fleet orbit
this basically sums up the whole topic of linear algebra 🥲
Timeline of how I look after encountering an issue: 
small glen
Can I please get an explanation of this notation? Are double || another way to indicate a square root?
lavish jewel
the double bars denote vector norm
the square root of the sum of squared components of a vector
on the first line, they factored out a k^2 from everything, so that you have sqrt(k^2(u_1^2 + ...))
then you can apply the sqrt to k^2 and to the sum of u_i^2
sqrt(k^2) = abs(k)
or in other words, the equality written above is identical do the one written below by simply using the definition of the vector 2-norm
small glen
Thank you!
north anvil
Given two vectors, what's the easiest way to find their cross product?
north anvil
I'd have to practice that or just memorize the formula - tyty
Also for a cross product, given two vectors A and B, A x B = ABsin(θ), how does this transform into a vector quantity with components if ABsin(θ) is a real number?
lavish jewel
it doesn't
nocturne jewel
Also for a cross product, given two vectors A and B, A x B = ABsin(θ), how does ...
$a\cross b =\norm{a}\norm{b}\sin{\theta}\hat{n}$ for the normal unit vector n
stoic pythonBOT
Mosh
lavish jewel
that's the magnitude of the vector you get if you do it the other way
north anvil
oh okay
nocturne jewel
however finding n is extraneous work since you can just... find the cross vector
north anvil
So ABsinθ = ||A x B||
nocturne jewel
yes
lavish jewel
yep
wintry steppe
$\mathbf{a} \times \mathbf{b} = ``\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}"$
nocturne jewel
why quotes on =?
north anvil
it's not necessarily equal to that expression
wintry steppe
It's just formal notation, right?
north anvil
you perform certain things on the matrix to get the cross product i think
nocturne jewel
No, they're equal
wintry steppe
They're equal but it's just a mnemonic to remember the cross product
stoic pythonBOT
Prudence
wintry steppe
There we go
lavish jewel
yeah, it's something that works, but is more like an abuse of notation
north anvil
i component is det(matrix with a2, a3, b2, b3)
j component is det(matrix with a1, a3, b1, b3)
k component is det(matrix with a1, a2, b1, b2)
i am pretty sure
lavish jewel
yep
north anvil
What if it's the cross product of 2d vectors?
Is the component of k = to 0 for all 2d vectors, i guess?
lavish jewel
the cross product is only traditionally defined for vectors in R^3
you can do it with vectors whose 3rd component is 0, but then the result will only have a component in that coordinate
i.e. 0,0,z
lofty scroll
My mind/understanding just got nuked. The dimension of a vector space is the size of its basis right?
lavish jewel
sounds right
lofty scroll
The rank of a transformation is the dimension of its range.
lavish jewel
mhm
lavish jewel
no
lofty scroll
Since the basis of the set showed is $(1,x^2,x^3)$ ?
stoic pythonBOT
tuturuuu
lofty scroll
hmmmm, ok I'm having some unlearning to do
lavish jewel
it's only rank 2 because x^2 and x^3 have the same coeff
if they were different, then it would be rank 3 as you say
but the text even goes out of its way to explain that
"must have the same coefficient of x^2 as of x^3."
lofty scroll
Yeah, I thought previously that its all about the basis
Rightttt, the 3 set is not a basis for that range space!
lavish jewel
idk if this will help or make it worse for you, but you could just turn that matrix into a vector and find the transformation matrix from R^4 to R^4 by checking what happens to the canonical basis when you transform it
lofty scroll
I'm sorry, what does canonical basis mean?
lofty scroll
I guess, I'm not quite understanding it, can you walk me through as to how the linear map I sent earlier can be transformed into a matrix?
lavish jewel
let's say the input vector is (a,b,c,d)
and let's build the polynomial in the output in ascending powers, because why not
the constant term is a + b + 2d
this means the first row of the matrix is [1 1 0 2]
the second row is all 0s, because the linear term is multiplied by 0
the third term is the coefficient for x^2, which is c. this means the third row is [0 0 1 0]
the last row is also [0 0 1 0]
the full matrix is:
1 1 0 2
0 0 0 0
0 0 1 0
0 0 1 0
you can immediately see this has only 2 linearly independent columns
i.e. the matrix is rank 2
lofty scroll
Righttt, I guess transforming it into a matrix clears it up, thank you!
wintry steppe
I like linear algebra
but row operations? like trying to find the inverse of a function, or finding reduced echelon form for an augmented matrix or whatever.. I assume most of the work is unnecessary when we have the beautiful and fantastic invention that is computers
it also seems like darn guessing game, I mean I know there's a certain pattern to it but once you do one thing wrong everything just collapses
I'm just learning linear algebra so there are probably more things to be learned and more things to be forgotten, but really I'd just like some advice for learning it
especially intuitively cuz unlike calculus and a couple of other subject it seems the more memorization prone . . . for now anyways
hollow finch
for row reducing i just do each column one at a time
once you get the first column to have only an entry in the first row then youre never going to be adding a multiple of it to any other row.
im guessing thats what you meant by "once you do one thing wrong everything just collapses"
like if you have $\begin{bmatrix}1&1&1\0&1&1\end{bmatrix}$
stoic pythonBOT
nix (@ me for the love of euler)
hollow finch
a rookie mistake is to add -1 of the first row to the second to eliminate the 1s in the second row but you should instead add -1 of the second row to the first to isolate the pivot in the 22 entry
idk if thats what you mean though. the process/algorithm for row reduction is relatively straightforward
wintry steppe
seems like the more practice with the more used to it I will be ... which can be said for everything but yea
thanks 
hollow finch
np
wintry steppe
it's just long and tedious but I guess that's what happens you like maths
well sometimes it can be
hollow finch
i think its good to be comfortable with row reduction on your own, but yeah i would generally recommend using a computer if you can. if you only take a few things away from linear algebra, i wouldnt prioritize row reduction over some of the other more important things you cover
wraith patio
To prove that a transformation is an isomorphism is it sufficient to just find the inverse transformation?
like they give me two isomorphisms $T:V\to W$ and $U:W\to Z$ and ask to prove that $UT:V\to Z$ is an isomorphism
stoic pythonBOT
taxminion
wraith patio
i want to just say that bc they are isomorphisms there exist inverse transformations of $T$ and $U$ so the inverse of $UT$ would just be $(UT)^{-1}=T^{-1}U^{-1}$
stoic pythonBOT
taxminion
wraith patio
is that enough to prove that UT is an isomorphism?
north hedge
how do u know the inverse of UT exists and also that it is linear
(its 'obvious,' but the thing they are asking u to prove is also obvious so i dont think its what they want)
i guess they want you to work from definitions (prove composition of linear maps is linear + composition of bijective maps is bijective)
stoic pythonBOT
Tim O'Brien
forest quiver
Does it mean "solution set is empty" because there are inf. solutions?
I guess I am confused what "general solution means," is it the values for each variable?
Oh crap I was looking at the wrong solutions section
never mind me
tidal wharf
your solution is correct haha
forest quiver
yeah I got it
drowsy flower
If a matrix is not symmetric, is it not orthogonally diagonalizable?
and if it is also not Hermitian, is it not unitarily diagonalizable?
torn hornet
Yes. Try writing A=QDQ^T where D is diagonal and Q is orthogonal
What is the transpose of A
drowsy flower
um would it be A^T = (QDQ^T)^T = QD^TQ^T ?
torn hornet
Hmm and what’s transpose of the diagonal D
torn hornet
Yep. For unitary I think you need real eigenvalues and I don’t know if being Unitaraly diagonalizable implies real eigenvalues
drowsy flower
yeah so I also wanted to ask about that. Lets say I have n by n matrix A with ONLY real entries. Now if I want to check if it is unitarily diagonalizable, then I would probably have to compute the eigenvalues right and check if it is ALL real right?
torn hornet
Yeah, but if all entries are real and it is Unitarily diagonalizble it’s easy to show it’s hermitian by the same logic
drowsy flower
Ah yeah I see
If A is hermitian matrix then all the eigenvalues are real and eigenvectors are orthogonal to each other
torn hornet
But I think complex matrices in general won’t work, like the identity times i say
drowsy flower
and if A is hermitian, then it must be unitarily diagonalizable.
tidal wharf
If you only know that the matrix is unitarily diagonalizable (with possibly complex eigenvalues), then it is normal (AA* = A*A)
drowsy flower
Maybe but in my scenario, I dont have complex matrix, so I think it should work fine right
torn hornet
Yeah sure
drowsy flower
If you only know that the matrix is unitarily diagonalizable (with possibly comp...
I dont know if it is unitarily diagonalizable
I am just given a matrix and asked to determine if it is unitarily diagonalizable or not. But I think I get how to do it now
torn hornet
Oh real eigenvalues are not enough to show it’s unitarily diagonalizable
tidal wharf
there's a theorem that says it is unitarily diagonalizable if and only if it is normal
where normal means it commutes with its conjugate transpose
tidal wharf
e.g. unitary matrices are unitarily diagonalizable since they are normal, but they are not hermitian
drowsy flower
right I see
fleet orbit
Someone able to help with this problem? I know that I have to find the standard matrix A of the transformation, but I'm not sure how and I don't understand how expressing v as a linear combination of the other vectors will help me
Edit: I guessed [3, 2] and it was right, but I do still want to know how to approach this
native rampart
Let's say v=a_1w_1+a_2 w_2+a_3 w_3
Then T(v)=a_1 T(w_1) + a_2 T(w_2) +a_3 T(w_3)
You don't need to find the standard matrix of transformation here
Just plug in the values of a_1,a_2,a_3,T(w_1),T(w_2) and T(w_3) and you get your answer
fleet orbit
I see by solving the linear combination for a_1, a_2, and a_3 I can use that to substitute into the second equation to find T(v), I didn't know I could approach it like that!
I was trying to do it this way below like an algebra problem, is it possible to find A like this or for that matter find A at all without information on its dimensions?
wraith patio
how do u know the inverse of UT exists and also that it is linear
thats my point. im asking if i can prove it exists by finding it explicitly. i know that U and T have inverses, and can show that T^-1 U^-1 functions as an inverse of UT
native rampart
I see by solving the linear combination for a_1, a_2, and a_3 I can use that to ...
I don't think there is any nice way of finding A(matrix of T wrt standard basis)
You could write (1,0,0) as a linear combination of w_1,w_2 and w_3 and repeat the same thing as above
Similarly for (0,1,0) and (0,0,1)
teal grotto
Someone able to help with this problem? I know that I have to find the standard ...
put the vectors w1, w2, w3 as columns into a matrix M and find its inverse. then M^-1 (v) will give you the coefficients a,b,c to v = aw1 + bw2 + cw3 in terms of the ordered basis {w1, w2, w3}
then you should be able to solve completely
forest quiver
This was originally a matlab question for me
I was able to solve it with coding, but I wanted to also solve it with an augmented matrix
and this a_0 term is throwing me off a little bit
I plugged the values into the p(t) there (since there are 5 inputs, we use a 5th degree polynomial)
Should I omit the first column? That represents the a_0 values, which are 1 for each because there is no variable to change it
yeah erasing the first column worked out
But I wonder why
Oh wait
a_0 = 0
because p(0)=0
omg nevermind
drowsy flower
Hi, another linear algebra question...... I am working with complex numbers and my space is hermitian inner product spaces equipped with standard inner product.
stoic pythonBOT
meguuuuu
drowsy flower
The problem is that both of my attempts for showing Col(A)^perp is subset of Null(A*) and Null(A*) is subset of Col(A)^perp is way too similar, which is why it feels like I am doing something wrong. Here is my attempt
stoic pythonBOT
meguuuuu
drowsy flower
They both might be completely wrong as well.. Can someone tell me if this is valid or not?
forgot to mention but there is a property in our book that says <Az, w> = <z, A*w> which helped me a ton
teal grotto
col(A) and the orthogonal complement of col(A) intersect trivially. if y is in the orthogonal complement of A, and like you have said, there is an x so that Ax = y, then you would imply that y is also in col(A), which would mean y = 0. i dont think thats what you want to do here
drowsy flower
ah shoot does that mean both of my attempts are wrong?
teal grotto
yes i think so. looks like you've made the same error in both attemps
drowsy flower
wait
for the first attempt, if I just say y belongs to Col(A), doesn't it work?
not y belongs to Col(A)-perp
and since their inner product evaluates to 0 when y belongs to Col(A), then it must be case that Null(A*) belongs to Col(A)-perp
I still however don't know for other direction
teal grotto
so im not quite sure im following what you're saying. if you just say y is in col(A), then you have not shown that the orth. comp. of col(A) is a subset of null(A*)
drowsy flower
because if <a, b> = 0 and a belongs to col(A) then it must be case that b belongs to orth comp of col(A)
yeah
but for other direction
no clue
teal grotto
have you tried to do a dimension count?
drowsy flower
hmm not really
how does it work?
I would probably set basis for col(A) then right?
ah I feel tired, I will come back to this tomorrow. Thanks for helping me out again coycoy
teal grotto
@drowsy flower yea. im pretty tired too. maybe somebody will get back to you when you wake up. just a quick question, dont you want A to be a square matrix? thats the only way you can really use <Ax,y> = <x,A*y> im pretty sure
teal grotto
nah, just say that A is in Mat_{n x n}(C)
nothing else should have been effected by that
teal grotto
oof. that is unfortunate
lavish jewel
you don't need it to be square for that to work
at least for the product to be defined, i mean
teal grotto
you don't need it to be square for that to work
really? thought the inner product needed to be defined as a bilinear form V x V —> F
lavish jewel
well
<Ax,v> is, say, V x V -> F
<x, A*v> would then be U x U -> F
if A is V -> U, that is
they are two different inner products, but both are defined and equivalent
teal grotto
u mean A : U -> V right?
lavish jewel
yeah oops
teal grotto
okay. those inner products are relative to U and V then. cool. didn’t know that
lavish jewel
that inner product there is the same as saying v*Ax
and the two definitions are the result of associativity
(v*A)x and v * (Ax)
it's the same operation in the end
feral jetty
Hi, I was looking at the barycentric form for lagrange polynomials, and was wondering if it is possible to use it to compute l_j(X) / l(x) ?
screenshot took from https://en.wikipedia.org/wiki/Lagrange_polynomial
l_j(X) / l(X) from the formula would be $1 /l'(x_j)(x-x_j)$
native rampart
Sure why not
stoic pythonBOT
hello123456789
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
feral jetty
Sure why not
Would it not be uncomputable at l_j(x_j) / l(x_j) ?
Maybe I am not understanding the formulae
feral jetty
Would it not be uncomputable at l_j(x_j) / l(x_j) ?
because there is x - x_j in the denominator... sorry if question is not clear, let me know and I can redo it for you
lofty scroll
The column space is basically the range of the linear map. Is there a similar connection with the row space?
lavish jewel
the range of the transpose of the map
dire thunder
@pastel moth please if u are being serious clarify ur question
simply saying if you have function f of x domain specifies where from you can take x's and range is set of the possible values of f
yaw and roll
your question is very vaguely statet
what are 'yaw', 'roll', 'pitch' and 'whole' and 'component' in this context
the problem is that in current statement nothing makes sense at all
mild crypt
What does the determinant mean intuitively for fields that aren't the reals or the complex numbers?
leaden tide
The determinant really isn't tied to fields
I'd say that in general, the determinant is a way of quantifying whether a given function (resp. set of vectors) is an isomorphism (resp. a basis) using a scalar
(the two are very closely related since any isomorphism sends any basis onto another basis)
mild crypt
oh interesting that is not what i understood from my prof's lecture
can you elaborate a bit?
leaden tide
Our teacher decided to approach it from a different angle, talking about matrix determinants last
Basically, we consider a certain type of function that takes multiple inputs
It has to be :
- alternating : if two of its inputs are the same, it is equal to 0
- antisymmetrical : if you flip two of its inputs, its sign flips as well
- n-linear : if you fix every input but one, and let that input vary, then the function is linear
Let's start with two dimensions
Take f(x,y) such a function
then f(x,x) = 0 and f(x,y)= - f(y,x)
(note : the first two properties are equivalent so long as you aren't working in 𝔽₂)
mild crypt
alright im with you so far
leaden tide
Let's take some basis B=(e1,e2) , such that x and y have coordinates x1, x2, y1, y2 respectively
(note : there is a dependence on the choice of basis here)
If you replace x and y with their expression, use n-linearity (or multilinearity) and cancel out terms using aforementioned properties, you get f(x,y) = (x₁y₂-x₂y₁)f(e₁,e₂)
leaden tide
It has to be :
- alternating : if two of its inputs are the same, it is equal t...
What we just showed is that, if you set a basis, all functions satisfying this are proportional to each other
And in particular proportional to x₁y₂-x₂y₁ (I wonder where I've seen that before)
We then declare the determinant in basis B to be the only function such that f(e₁,e₂)=1
(works, since the function is entirely determined by its input on B)
In higher (finite) dimensions, the same things apply
All such functions are proportional to each other, and uniquely determined by their input on the chosen basis ; we set the determinant to be the nicest one
Now, let's take the time to check that the determinant does what we want it to : check if something is linearly independent/a basis
if you plug in 0 anywhere the result is 0 since it's multilinear
if you plug in the same thing twice then it's 0 because it's alternating
and if you can express one vector as a linear combination fo the others, you can use multilinearity to expand that into a sum of terms that all vanish because it's alternating
mild crypt
interesting
mild crypt
If you replace x and y with their expression, use n-linearity (or multilinearity...
i am a bit confused by this step here, what exactly is the expression you are cancelling things with
i am liking this interpretation so far though
leaden tide
Well, time for some expansions
latex though
$f(x,y)=f(x_1e_1+x_2e_2, y_1e_1+y_2e_2) $ \
$= x_1f(e_1, y_1e_1+y_2e_2) + x_2f(e_2,y_1e_1+y_2e_2)$ by linearity of the first argument \
$= x_1y_1f(e_1, e_1) + x_1y_2f(e_1,e_2)+x_2y_1f(e_2, e_1)+x_2y_2f(e_2, e_2)$ by linearity of the second argument \
$ = x_1y_2f(e_1,e_2)+x_2y_1f(e_2, e_1) $ because $f$ is alternating \
$ = (x_1y_2-x_2y_1)f(e_1,e_2)$ because $f$ is antisymmetrical
stoic pythonBOT
Syst3ms
leaden tide
(what terms cancel and where the negative signs appear has a lot to do with the symmetric group ; i'll spare you the formula that involves it, but it exists and it is convenient, theoretically at least)
follow so far?
mild crypt
ooooh i see now
that is cool
i like that
so then the determinant gives us a solution for f(x, y) = d * f(e1, e2) ?
for the set of functions that are obeying these properties
leaden tide
when f is the determinant, then det_B(x,y)=x₁y₂-x₂y₁
the factor disappears, that's how it was defined
leaden tide
and the functions satisfying such properties are all proportional to each other
saying which one is the determinant is then really just a matter of convenience
and 1 is convenient
mild crypt
That makes a lot of sense
leaden tide
I'll move on to linear maps, which will segue nicely into matrices
mild crypt
Yeah lets do that because my previous intuition was basically 3b1b so i wasnt sure how it could be generalized to a vector space of a nonreal field before
leaden tide
Actually, some minor details before we do that
I said that the determinant changes between bases, but it's a bit more explicit than this
Specifically, $\det_{B'} = \det_B'(B)\det_B$
stoic pythonBOT
Syst3ms
leaden tide
That is not how i expected it to look, but i suppose this is fine
Anyway, you can change bases
And in the 2D/3D cases, you can interpret that as just changing units of volume
mild crypt
but volume isnt defined in every field tho right? idk anything about measure theory
leaden tide
Indeed, it's just a representation that helps visualize what we're doing
And which is most easily seen in the 2D/3D case
We're saying that "the volume in B' is just the volume in B adjusted by how the volume of B looks like in B'"
mild crypt
i see, that makes sense
leaden tide
Now, I don't quite remember how the proof goes
But I think I recall that it's usually simpler than I think it is
oh wait
yeah, I got it
Hmm, I got the formula backwards
$\det_{B'} = \det_{B'}(B)\det_B$
stoic pythonBOT
Syst3ms
leaden tide
Yeah, this makes more sense now
Since all determinants are proportional, $det_{B'} = \alpha\det_B$. Plug in B and you get $\alpha = \det_{B'}(B)$
stoic pythonBOT
Syst3ms
mild crypt
here $\det{B'}$ just means the determinant with respect to basis $B'$?
stoic pythonBOT
united9jackson
stoic pythonBOT
united9jackson
leaden tide
here $\det{B'}$ just means the determinant with respect to basis $B'$?
correct
Anyhow, linear maps
mild crypt
Apologies, my internet is not having it
no worries, i really appreciate you taking the time to explain this
leaden tide
For any linear map $u$, $\det_B(u(v_1),\ldots,u(v_n)) = \alpha\det_B(v_1,\ldots,v_n)$ where $\alpha$ only depends on $u$, not the choice of basis
stoic pythonBOT
Syst3ms
leaden tide
We define $\alpha = \det(u)$
stoic pythonBOT
Syst3ms
leaden tide
This is actually quite easy to prove, just consider that the function on the left (with v_1,...,v_n as inputs) is alternating, skew-symmetric (got the wrong term before) and multilinear : hence, it is proportional to the determinant in basis B
and as for why it doesn't depend on the choice of basis, multiply by det_B'(B) on both sides and you get the same identity with basis B'
Now, here come a lot of the properties you're used to
First of all, $\det(f\circ g) = \det(f)\det(g)$
stoic pythonBOT
Syst3ms
leaden tide
proof : use the identity that defines det(f) and det(g) in succession and you get the identity for det(f . g)
It should also come as no surprise that det(id)=1
And from this you deduce that det(f⁻¹)=det(f)⁻¹
And as for invertibility/linear independence, the determinant of a linear map also behaves as desired
If u isn't an isomorphism, then det(u)=0
If u isn't an isomorphism (in dimension n), then the image of any set of n vectors by it will be linearly dependent, which gives a determinant of 0 as mentioned previously
Great, this works for linear maps
Now for matrices, it's really just a shift in perspective
if you've watched 3b1b, he's probably drilled into your head that matrices are a representation for linear transformations
And this is very true : a linear map is uniquely defined by how it transforms each basis vector of a given basis
Then, the matrix in basis B = (e_1,...,e_n) of a given linear map f is defined as such : take f(e₁) and express in in vector form with respect to basis B ; this is the first column of your matrix. Do the same for every basis vector, you have yourself a representation of f.
So, provided you choose a basis, a matrix is just a glorified way of identifying a linear map (or is a linear map just a glorified way of identifying a matrix? doesn't matter, they're the same thing)
And no, it doesn't matter that we may not be working in ℝⁿ all the time, since choosing a basis gives you an isomorphism between that and any space of dimension n
Usually, the determinant of a matrix is defined using some horrendous formula which I won't scare you with, but it has the following nice properties
If A represents the linear map f in any basis, then det(A)=det(f)
(it doesn't depend on any basis for the same reason that the determinant of a linear map doesn't depend on any basis)
Moreover, det(A) the determinant of its columns with respect to the canonical basis {(1,0,...,0), (0,1,...,0), ..., (0,...,0,1)}
And everything that I said about linear maps also holds for matrices : they're the same anyway!
I think that just about wraps up how the topic was explained to us
You haven't said much in a while, everything alright?
mild crypt
Yea just taking it all in lol
leaden tide
Yeah, it is a lot
mild crypt
I'm starting to get a better understanding, probably going to take this and do some proofs/mess around after
leaden tide
But yeah, I kind of regret the way the matrix determinant is introduced : it's this really arbitrary quantity and it just has nice properties for whatever reason
It obfuscates a lot of really clever things about matrix representation of linear maps
Also, the proof that det(AB)=det(A)det(B) gets a whole lot easier with this approach
mild crypt
But yeah, I kind of regret the way the matrix determinant is introduced : it's t...
true i was super lost trying to figure it out and some of the proofs in the text from my school were literally using the ugly sum formula
leaden tide
instead of repulsive sum manipulation, you get matrix -> linear map -> linear map rules -> matrix
(in case i didn't mention it, multiplying two matrices is equivalent to composing the linear maps they represent)
mild crypt
needless to say i kinda ditched the text the class is using and picked up axler
leaden tide
(which also gives a nice reason why matrix multiplication isn't commutative)
leaden tide
why, of course it does
as Henri Poincaré once said, math is the art of giving the same name to different things
This determinant approach also gives a very nice way of showing that if AB=I then A and B are invertible and inverses of one another
Without having to check BA=I
Instead of doing intermediate difficulty proofs with properties of matrix multiplication, you get this (miraculous) theorem by commutativity of multiplication 
actually that's the wrong argument, but if det(AB)=1, then surely neither det(A) nor det(B) are 0, so they're invertible, and the inverse is unique
leaden tide
it all connectsss
notice that for every fact true about matrix representation of linear maps, matrix multiplication works really nicely
needless to say that isn't a coincidence
mild crypt
so matrix multiplication just comes from a way to represent linear maps algebraically?
leaden tide
Matrices are really deep, it's just hard to explain all of this depth in a suitable order
I don't know which came first to be honest
mild crypt
yeah i am glad though that im still getting some of it from the course im in (and you obviously) because in high school for me it was just "this is a matrix it holds numbers deal with it"
leaden tide
But the overall formula for matrix multiplication is a generalization of matrix-vector multiplication
Which itself is a generalization of row-vector multiplication
So yeah
"Why is matrix multiplication not commutative?"
Short answer : "try an example"
Long answer : "function composition isn't commutative"
mild crypt
alright well
I really appreciate this im gonna go back and read through this while going through my course notes
leaden tide
sure, and don't hesitate with the questions
mild crypt
love having a good understanding/foundation/intuition ('basis' if you will 😉 ) of math
leaden tide
nice
wintry steppe
hello
yeah so i checked google
for a simpler defintion
and it said
row replacemen t= row + multiple of another row
slide share
yes
it says the same in the yotube video i was watching
so what is row replacement
wait so can u send me a link for a full course in linear algebra
so row replacemnt isnt a thing
oh ok
wait
cause i have started learning linear algebra today can u tell me which youtuber explains the full course well
like from the basics ?
ok thx
marble lance
It should read:
Given $g \in W^*$, $T^g \in V^$ is given by $T^g(v) = g(Tv)$ for all $v \in v^$.
stoic pythonBOT
Lunasong the Supergay
mild current
I see @marble lance , TYVM
marble lance
np
vagrant hazel
Hi looking for some help with this
quartz compass
you can use linearity to construct them
vagrant hazel
I think e1 = [1,0,0] e2 = [0,1,0] e3 =[0,0,1]. the basis given here is [100][110][111]
quartz compass
[1,1,0] - [1,0,0] = [0,1,0]
operate with T and you can write down what the thing on the left is to get the thing on the right
see what I mean?
vagrant hazel
no sorry Im not fully with you. the computation you did which column was this for?
quartz compass
so we want to know T[0,1,0] right
but we don't have [0,1,0]
so we can look at the vectors that T is acting on
[1,1,0] - [1,0,0] = [0,1,0]
T[1,1,0] - T[1,0,0] = T[0,1,0]
[1,1,1] - [2,1,0] = T[0,1,0]
vagrant hazel
ok I think i see where you're goin
quartz compass
yeah, same kind of trick works for e_3
vagrant hazel
sweet thanks:D
quartz compass
yup you're welcome
vagrant hazel
What is the standard matrix A := Mat (T ) of T ? would that be the result from above as A = [[2,1,0],[-1,0,1],[-1,0,-2]]
nocturne jewel
What is the standard matrix A := Mat (T ) of T ? would that be the result from...
well what did you get for part a?
vagrant hazel
T(e1) = [2,1,0] T(e2) =[-1,0,1] T(e3)=[-1,0,-2]]
nocturne jewel
then yes that'd be the matrix representation of T
vagrant hazel
thanks:D sorry for the stupid Q's just fairly unsure with LA at the min
the minute I think I have something whoosh its wrong :/
radiant yarrow
Do I just define real valued functions like that? Or do I show that they're differentiable?
lavish jewel
define them how?
lavish jewel
i would do something like take two functions that satisfy the properties and then exploit the linearity of differentiation
lavish jewel
let g: R->R and f:R->R be differentiable.
quartz compass
continuity doesn't really matter, although true
radiant yarrow
Okay
lavish jewel
consider f(x) + g(x), x \in R
take the derivative of this sum, i.e. d(f(x) + g(x))/dx
this is equal to d(f(x))/dx + d(g(x))/dx
blah blah
radiant yarrow
Okayy
lavish jewel
just use the def of vector space and the knowledge that derivatives behave nicely like that
quartz compass
yeah, just list out the vector space axioms and start proving them one by one and see which ones you get stuck on
lavish jewel
this one is really very easy, just repeat stuff step by step
radiant yarrow
Just showing how you can have the properties for the functions after addition and scalar multiplication?
Like I fear that I might be being a little hand wavy
lavish jewel
first, do you know the properties of a vector space?
radiant yarrow
Yes
commutativity
associativity
scalar multiplication
distributive property
existence of zero vector
additive and multiplicative identity
Existence of negative vectors to each vector which sum to zero vector
lavish jewel
aight
we show all of these exist for the "vectors" you were given
the "vectors" are differentiable functions from R to R
radiant yarrow
Okay I need to show that the sums and scalar multiples are differentiable?
radiant yarrow
Is writing in limit form the right way?
lavish jewel
i don't think that's needed at all here
radiant yarrow
Oh okay
lavish jewel
it should be super simple
like
take f(x) and g(x). f(x) + g(x) is done with the usual sum, which is commutative, so f(x) + g(x) = g(x) + f(x)
lavish jewel
similarly for associativity
now take f(x) + g(x) + h(x)
again, we are using the usual sum, which is also associative, and so (f(x) + g(x)) + h(x) = ...
leaden tide
edd, what are you trying to show here
lavish jewel
hmm?
radiant yarrow
Don't I need to do like f'(x)+g'(x)=g'(x)+f'(x)?
radiant yarrow
unless... the 8 axioms?
yeah I'm new
lavish jewel
the 8 axioms indeed, since they said vector space and not subspace
leaden tide
I mean, that doesn't mean you have to use the 8 axioms
The easiest to prove something is a vector space still is to prove that it's a subspace
we had like 4 exercises on our sheet that actually used the 8 axioms and i precisely did 0 of them
lavish jewel
anyway, yeah, you can add in for those 2 that (f(x) + g(x))' = f'(x) + g'(x)
(differentiation is linear)
lavish jewel
but it should be a pretty basic property of differentiation
you might be able to just straight up use it
that the derivative of a sum is the sum of the derivatives
that should be enough
well, when the corresponding derivatives exist
radiant yarrow
Got it
lavish jewel
which they do here, because we are working precisely with differentiable functions
radiant yarrow
Yeah
lavish jewel
if you say it like that, that should be enough
radiant yarrow
Let V = {0 } consist of a single vector 0 and define 0 + 0 = 0 and
c0 = 0 for each scalar c in F. Prove that V is a vector space over F.
(V is called the zero vector space.)
For this do I use like c*0 = 0 for all c in F, therefore V = F?
nocturne jewel
You're showing that {0} is a vector space over any field F with those definitions of + and *
that means go through the axioms and show they all hold
lavish jewel
the thing is this one is not necessarily the usual sum nor scalar product
nocturne jewel
just note it's gonna look and feel weird cause it's all 0 and scalars
lavish jewel
it's gonna feel weird, yes
but read it over a few times
they don't actually say what this 0 element is, nor what F is
nocturne jewel
ex: $0+0=0+0=0$ so + is commutative
lavish jewel
it's more general than you'D think
stoic pythonBOT
Mosh
lavish jewel
not quite, that's only for one or so of the axioms
radiant yarrow
Okay
radiant yarrow
they don't actually say what this 0 element is, nor what F is
So I can't equate it to 0?
my bad they've mentioned V is a zero vector space
lavish jewel
they don't say what the field is
radiant yarrow
I didn't copy the text
radiant yarrow
My bad they've said "the" zero vector space
lavish jewel
for example, that definition works for any N dimensional vector of all zeros over the reals and complex numbers
radiant yarrow
(V is called the zero vector space.)
nocturne jewel
Yeah $V={0}$
stoic pythonBOT
Mosh
lavish jewel
the whole purpose of these exercises is to notice that vectors are not just list of numbers
but ANYTHING that has a handful of properties
nocturne jewel
pretty much anything that has a meaningful notion of addition and scaling
lavish jewel
this 0 vector is a very abstract meaning
aside from the ones i gave above, there's the 0 function among the differentiable functions from R to R, too
etc.
it's the zero vector space over some generic field F as long as it has those properties
doesn't matter much what F is
or what exactly the vector is
just its properties
radiant yarrow
Then how do I prove it's a vector space over F if I don't know what F is
lavish jewel
just use the definitions
that's the point
it's very abstract and rather boring, but also super useful
radiant yarrow
okayy
lavish jewel
(0 + 0) + 0
then, by the given definition of 0+0=0, we get 0 + 0
using the definition again, 0+0 = 0
and similarly associating the last 2 zeros instead
you say "i know that", but you don't really
lavish jewel
they could've said "let v = {0} satisfy the property that 0 + c = 1"
that it is written as "0" means nothing
don't look at the symbols
read the definitions
forget about numbers
lavish jewel
forest quiver
lavish jewel
the whole idea is for you to learn that linear alg is not just about lists of numbers, it's about any abstract thing that "behaves" like vectors do
radiant yarrow
V={0} and defined as 0+0=0
lavish jewel
whatever those things may be
they told you 0+0=0, yes
but they never said + is the usual sum
radiant yarrow
I only have to use this to define it's a vector space over field F
lavish jewel
it could be that 0 + 5 = 10
radiant yarrow
nocturne jewel
To define it as a vector space, it has to pass all 8 axioms
lavish jewel
you should only use what you know for certain, and that is only what you were told
sully you, not me
you're missing the point
- is not just addition of numbers
it's a generic binary relation
it can be defined in several ways in different contexts
forest quiver
<@!247157233228447746>ask your question
I figured it out
radiant yarrow
ok
Let V denote the set of ordered pairs of real numbers. If (a1, a2) and
(b1, b2) are elements of V and c ∈ R, define
(a1, a2) + (b1, b2) = (a1 + b1, a2b2) and c(a1, a2) = (ca1, a2).
Is V a vector space over R with these operations? Justify your answer.
How do I show commutativity? do I interchange positions of a1, a2 or are the elements everything inside the brackets
Do I have to show $(a_1 , a_2) + (b_1 , b_2) = (b_1 , b_2) + (a_1 , a_2)$?
stoic pythonBOT
Researcher in Pre-algebra
lavish jewel
Do I have to show $(a_1 , a_2) + (b_1 , b_2) = (b_1 , b_2) + (a_1 , a_2)$?
this
radiant yarrow
Okay
lavish jewel
this is an example of what i said, of the sum not being the usual sum
here "sum" even has multiplication in there
radiant yarrow
Okay
how do I show associativity here
it seems weird
Oh wait
how do I show existence of zero
lavish jewel
you say that, but do you get 0 if you do that?
lavish jewel
more like -a1, 0, yeah?
radiant yarrow
yeahh