#linear-algebra
zinc lava
dot product gives you a number. π
what class is this for?
are these vectors in the plane?
zinc lava
ah okay. So v = (a1, a2, a3, ... an) and -v = (-a1, -a2, -a3, ..., -an)
What is the dot product?
gaunt field
-a1^2 - a2^2 - a3^2... ?
zinc lava
yep and if a vector is a unit vector what is the length?
er oops you need a -(an)^2 at the end of your sum.
gaunt field
hmm im confused
zinc lava
What is | |v| | ?
gaunt field
1 right?
zinc lava
well in terms of a1 and a2 and a3 ... an?
zinc lava
just length of v = square root of ((a1)^2 + (a2)^2 + (a3)^2 + ... (an)^2) but you probably knew that!
zinc lava
yep ... so we know, after a bit of annoying algebra ... that 1 = (a1)^2 + (a2)^2 + (a3)^2 + ... (an)^2 π
gaunt field
yeah
zinc lava
but our answer was -(a1)^2 - (a2)^2 - (a3)^2 - ... (an)^2 ... so can we simplify this?
gaunt field
ok let me write it down
gaunt field
oh so its just -1
zinc lava
kk
gaunt field
okay well for part b, am I allowed to expand it into like v^2 - vw + wv - w^2 or you can't do that?
well i guess not
gaunt field
i mean its no right
zinc lava
v = (a1, a2, a3, ...an) u = (b1, b2, b3, ... bn) w = (c1, c2, c3, ... cn)
We're looking for v(u + w) let's say.
is that vu + vw?
v(u + w) = (... at(bt + ct)...) = ( ... atbt + atct ... )
vu + vw = (... atbt + atct)...)
moral: distributive property works. (pretend I did both the left and right side)
gaunt field
okay
zinc lava
so yeah ... (v + w) (v - w) = v (v -w) + w (v - w) = v v - vw + w v - w w
gaunt field
So v * v is just 1? cuz I already found that - v * v = -1
zinc lava
yes
gaunt field
so its 4?
zinc lava
vv -vw + wv - ww = 1 + (curious thing) -(related curious thing) - 1 π
zinc lava
-vw = -a1(b1) - a2(b2) ... - an(bn)
gaunt field
oops yea
zinc lava
yep
gaunt field
well its kinda hard... length of v and w and both 1 tho
zinc lava
ah in this case, you don't have much room to simplify. Think about the other expression first.
vv check. ww check, vw check ...
zinc lava
wv π
gaunt field
okay, it seems like it would be wv = a1b1 + a2b2 etc.
oh so if you add it to -wv then it cancels
zinc lava
quick rehash ... (a + b)c = ac + bc and (a + b)(c + d) = ac + ad + bc + bd and (-a)(a) = -(aa) and -ab = (-a)(b)
don't ever prove them again cause that's a waste of time :). (unless you're expected to)
zinc lava
do you mean the coordinate proofs?
anyway I'm on my way out :). Pretty sure you can handle part C.
gaunt field
okay thanks
zinc lava
yar. you are welcome.
gaunt field
I need help with b) https://gyazo.com/24831a46121f5af0ea60d99e39c1c251
wintry steppe
@gaunt field This is a straightforward application of the basic laws of the inner product.
gaunt field
What do you mean?
wintry steppe
Yes
gaunt field
Yeah I tried doing the dot product but im not really sure what to do
wintry steppe
What dot product did you look at?
gaunt field
u and v+w, U and v-w, u and v, u and w
i mean i don't really know what to do with the equations
wintry steppe
What are the properties of the dot product?
There's one property that is more important than anything else you'll learn about it.
gaunt field
i mean i know that they are equal to zero cuz of the orthongonality
wintry steppe
That's a definition, but that's also important to know here π
wintry steppe
What do you know about the dot product?
gaunt field
if u = (a1, a2, a3) and v = (b1, b2, b3) and they are orthogonal then a1b1 + a2b2 + a3b3 = 0
thats all the comes to mind π¦
wintry steppe
Okay.
Well, there's one property that should be the first and only thing you ever think of with regard to it π
And that is -- the dot product is bilinear
gaunt field
What does that mean?
wintry steppe
It's an operator with two arguments, right?
gaunt field
yeah
wintry steppe
u . v
well, if you fix v as constant
then it's linear in u
so (u + u') . v = (u . v) + (u' . v)
and (au) . v = a(u . v)
It's also symmetric: so u . v = v . u
gaunt field
okay
wintry steppe
Those two properties (that it is bilinear and that it is symmetric) give you everything you need to know about it.
wintry steppe
Just a vector
gaunt field
okay
wintry steppe
For any vectors, named u u' and v, then that equation holds
I called it u' but I could have said w
or u2
or whatever I might
gaunt field
yeah
well i keep manipulating the variables and moving stuff around but idk what im looking for
so u.(v+w) = 0 and u.(v-w) = 0 then u.(v+w) = u.(v-w)
wintry steppe
Can you write down what your assumptions are? What your goals are?
ok, those are your assumptions right there π
gaunt field
assuming its orthogonal
wintry steppe
What's your goal?
wintry steppe
If what is true or not?
gaunt field
oh thats actually the assumption right
im trying to figure out if u is orthogonal to v and w
wintry steppe
ok
gaunt field
so i somehow have to come up with something like u . v = 0 and u . w = 0 right
wintry steppe
yep
gaunt field
oh well I found that u w = 0 after I did this
so u.(v+w) = 0 and u.(v-w) = 0 then u.(v+w) = u.(v-w)
@gaunt field
wintry steppe
So you are starting with the two equations:
u . (v + w) = 0
u . (v - w) = 0
And you want to end with
u . v = 0
u . w = 0
gaunt field
yeah
I think I found that u . w = 0
cuz u(v+w) = u(v-w) then if you distribute both sides
uv cancels out on each side
then 2uw =0 so uw = 0
wintry steppe
I don't think that works
gaunt field
rip
gaunt field
i mean subtract both sides with uv
wintry steppe
To be sure, this isn't multiplication we're doing. You can't divide. That's not one of the rules.
gaunt field
hm ok
wintry steppe
This is where you need to use that bilinear property I was talking about.
When you see u . (v + w), you should be dying inside to expand it out
gaunt field
okay yeah i did that
wintry steppe
what was the result?
gaunt field
uv + uw
gaunt field
and i got uv - uw on the other side of the eq
wintry steppe
Yes.
wintry steppe
That's actually
a good question
I said you can do it with addition
Do you think it works for subtraction?
gaunt field
mmm probably cuz even with addition, w could be w = -w'
wintry steppe
Cool. Then yeah, bilinearity works over addition and subtraction
gaunt field
nice
wintry steppe
So that move you made is justified
gaunt field
okay well i have uv + uw = uv - uw
but u said before I can't subtract both sides with uv
wintry steppe
oh
no you can subtract both sides
You just hadn't said it in the chat exactly what you were doing, so I thought you were skipping a step.
wintry steppe
Yeah. Once you expand it out, you get an expression which is just adding up a bunch of uv and uw 's
gaunt field
so u is orthongoal to w
wintry steppe
sweet
gaunt field
but theres no v π¦
wintry steppe
Well, you got a lot more equations you've derived
I'm sure you can figure out how to use them in a similar way to prove uv = 0
wintry steppe
π
gaunt field
dont see it
gaunt field
i was trying to plug in 0 for u.w
wintry steppe
Restating your progress:
So you are starting with the two equations:
u . (v + w) = 0
u . (v - w) = 0
You derived:
uv + uw = 0
uv - uw = 0
uw = 0
And you want to end with
u . v = 0
u . w = 0
wintry steppe
π
wintry steppe
np
gaunt field
can anyone help with c)? https://gyazo.com/24831a46121f5af0ea60d99e39c1c251
so i know length of u and v is 1..
torpid portal
and u-v is the hypotenuse of the triangle
so you can use pythagoras theorem
to get that |u-v|=sqrt(2)
since obviously |u| and |v| are both 1
gaunt field
damn i didnt think of that :/
about the right triangle
is |u-v| = |u| - |v|?
nvm ofc not
lul
gaunt field
yeah yeah
i drew that
so sqrt 2 = u - v
but its not exactly sqrt 2 = |u -v |...
oh wait
nvm
torpid portal
well i mean ||u-v|| is the length of the hypotenuse right?
gaunt field
cuz they are already unit vectors
yah
i don't know what to do, should i be trying to read more information or just keep trying to do problems until i am able to do it myself? cuz i dont think i should be taking so long to figure this stuff out
okay well for d) https://gyazo.com/24831a46121f5af0ea60d99e39c1c251 i think it's false because if you have a vector (0,5) and a vector (5,0), then both have magnitude 5, so on the left side of the equation it would be 50, whereas on the right side it would be 100 so it's not true, but how would you prove it without just plugging in values
old flame
Hello, I'm stuck on this question, are there any hints
Suppose that $U$ and $V$ are finite-dimensional vector spaces and that $S \in L(V,W)$, $T \in L(U,V)$. Prove that $\dim nullST \leq \dim nullS+ \dim nullT$.
stoic pythonBOT
Otoro:
spiral star
@old flame try the rank nullity theorem together with subspace properties
i.e. how are dimensions of subspaces relate
old flame
@spiral star actually heres my attempt, but I'm stuck
Since both $U$ and $V$ are finite dimensional, by the rank-nullity theorem, $\dim U= \dim null T + \dim range T$ and $\dim V= \dim null S +\dim range S$. Since $null ST \in V \Rightarrow \dim null ST \leq \dim V$ and $\dim null T \leq \dim U$, subtracting both yields $\dim null ST - \dim null T \leq \dim V - \dim U$, substituting the relations in, $\dim null ST - \dim null T \leq \dim null S + \dim range S - \dim null T - \dim range T$
stoic pythonBOT
Otoro:
spiral star
@old flame you could try to restrict S to range(T). that makes working with the composition easier
gaunt field
ok yeah I figured it out @dusky epoch
spiral star
in the first part where you show that subspace implies b = 0
you are done after the first step
because you already conclude b = 0
dreamy iron
So Iβm thinking, what if the other Axioms yields inclusive results for b. Or worse, they lead to the wrong value for b.
spiral star
your assumption is that all vector space axioms hold and that what you have is a vector space
dreamy iron
Yes
spiral star
so there is nothing to check
if the vector space axioms imply b = 0 from addition, you dont have to check anything else
dreamy iron
All three imply b = 0
spiral star
yes, of course
but you only need to check one
the other ones are unnecessary
pick one
dreamy iron
Can I just choose which ever I want to leave on there?
spiral star
you just did 3 times the work
dreamy iron
Oh, kk
spiral star
there are multiple ways to derive b = 0 but you only need one of course
dreamy iron
I was worried that maybe one of the Axioms could lead to a b that was the wrong value, truly.
I see now that didnβt happen.
spiral star
if one of the axioms leads to b = 0 and another one leads to b = 1 then you would have a contradiction to the assumptions that you had a vector space in the first place
dreamy iron
Oh
Iβm just going to leave all three in there and warn the reader to pick just one.
spiral star
i would choose the shortest one and drop the rest. good proofs are concise and easy to understand
dreamy iron
TYVM flow
spiral star
the more you write, the harder it will be on the reader, and the easier it is to make mistakes in notation
dreamy iron
Yeah
spiral star
$\int (f+g) = \int f + \int g = b + b = b \implies b = 0$
stoic pythonBOT
Flow:
spiral star
that's all you need
dreamy iron
the b+b = b part needs justification i think.
spiral star
no
dreamy iron
I have to force the sum to be in the vector space
spiral star
the sum is in the vector space by assumption
native rampart
dusky epoch
@dreamy iron your assumption is that f+g is in V_b, whence the "= b"
spiral star
you dont have to justify what is already true by assumption
dreamy iron
Hence int f+g = b ??
dreamy iron
okay. Thatβs throwing me for a loop. Haha
spiral star
you have a function space here, the vectors are functions
the addition of vectors takes two functions and produces a new function by pointwise addition
dreamy iron
the subspace is a set of functions.....WITH THE ADDED requirement that their integral be zero.
spiral star
yea
its a subset of some function space
and your assumption is that if you equip that subset with the restricted operations of the original function space, then it already forms a vector space
dreamy iron
Iβm just so new to this, haha.
spiral star
you dont have to justify that
that's your premise
you only have to justify that b = 0
and b = 0 follows from vector addition for example
dreamy iron
i need to marinade in this.
spiral star
take two vectors from your space, call them f and g
since its a vector space, it must be closed under addition
that is, (f+g) is also a vector in the space
f+g is a function given by (f+g)(x) = f(x) + g(x)
dreamy iron
KK, yeah. I follow that
spiral star
since f+g is in the vector space, its integral must be $\int (f+g) = b$
stoic pythonBOT
Flow:
dusky epoch
i mean tbh
it's easiest to use the zero vector axiom for this since it just gives you b=0 directly
spiral star
yea i guess
dreamy iron
I have that as part C.
spiral star
that's why i said you can choose any axiom you like
dreamy iron
i like that one cuz itβs clean.
But the added explanation is MOST WELCOME, so if you wanna keep talking, Iβm all ears
spiral star
anyway, to finish the addition thing: you know that $b = \int (f+g) = \int f + \int g = b + b $ and thus $b = 0$
stoic pythonBOT
Flow:
dreamy iron
Okay. That argument, i followed.
spiral star
since your space is a subspace of that function space, it must necessarily contain the zero vector, which is the zero function. and then $\int 0 = b = 0$
stoic pythonBOT
Flow:
spiral star
this would be the shortest way i guess
old flame
@spiral star may I ask how do you restrict ? is it like defining a new map from range T to range S ?
dreamy iron
mniip says my notes are verbose, im just trying to make them βninny fool proofβ so in the event I need to refer to these in the future, i can get right into the same frame of mind.
spiral star
@old flame you just make the domain smaller
dreamy iron
.... and by extension, understand what I was doing.
spiral star
@old flame
instead of S: V -> W you only consider im T -> W
im T is a subset of V
$S|_{\im T}(v) = S(v)$
stoic pythonBOT
Flow:
old flame
so its still S, but just a smaller domain ?
spiral star
it's the "same" function but defined for a smaller domain
old flame
all good thanks
spiral star
@dreamy iron i can understand that. i've also written very verbose proofs at first :p
but a good quality proof is correct and easy to process
your proof is basically complete after a few steps but then you go on to show the same thing again xd
so for "subspace ==> b = 0" i would choose what ann suggested
and for "b = 0 ==> subspace" you just show that its closed under addition and scalar mul.
which is one line for each
and then your whole proof fits on half a page
dusky epoch
have you already shown that the kernel of a linear map is a subspace of the map's domain?
cause if so then b=0 => subspace can be proved by invoking that lemma
spiral star
using linearity of the integral? π
dusky epoch
yes
spiral star
that's pretty cool lol
π
@dreamy iron with ann's suggestion your proofs would condense to this lmao:
dusky epoch
i'd add a reference to the "kernel is a subspace" theorem
spiral star
yea i guess
it's sort of a central theorem that you need everywhere tho
just like the linearity argument
T(0) = 0
spiral star
i guess a lot of small facts come together here
i dont think it gets much shorter than this
that's pretty cool
dreamy iron
This is Axler. His pace is different. Iβve not yet encountered kernels, nor have I proved that integrals are linear.
But Iβll book mark this conversation to reference for later.
spiral star
oh so this is about subspaces before you learn linear maps?
dreamy iron
Yeah
spiral star
the proof of linearity for integrals isnt really part of linear algebra, it comes from the definition of the integral, so that's an analysis question
you already used linearity of the integral in your proofs
dreamy iron
Linear transformations are chapter 3.
spiral star
i guess without linear maps you can't argue in that short way yet. but with the subspace properties it's still really short
dreamy iron
you already used linearity of the integral in your proofs
@spiral star it was some hack-eyed line saying something to the effect βrecall from basic calculus that such and such about integrals is true.β
spiral star
i wouldnt say it's basic lol. there goes quite some analysis into showing those properties
but you can assume that the reader of your proofs understands this
dreamy iron
The baby calculus classes all stem students take just say βmemorize these and calculate.β
Iβm sure theyβre not actually basic.
I didnβt have enough skill points in analysis so I procrastinated this question.
spiral star
that's alright. answering those questions are not part of linear algebra
axler assumes you know this stuff anyway
dreamy iron
he really does!
spiral star
yea don't worry. you wont have to prove analytical theorems in linear algebra lmao
but whenever you show in analysis that something is linear, then it becomes usable in linear algebra
integration and differentiation for example
(u+v)' = u' + v' and so on
but you will see that soon i guess when you get to those chapters in the book
dreamy iron
Axler sprinkles them into the exercises.
The problem you helped me with is four, about integration.
Problem three, which Iβve skipped, is about differentiation....
Iβm gonna attack that one next.
spiral star
π
old flame
@spiral star I am once stuck again, sigh
Proof : Since $U,V$ are finite dimensional, by the rank-nullity theorem, $\dim V= \dim null S +\dim range S$ and $\dim U=\dim null T + \dim range T$. Since $null ST$ is a subspace of $V$ this implies $\dim null ST \leq \dim V$, $\dim null ST \leq \dim null S + \dim range S$. Consider the restricted domain of $S$, where $S : range T \rightarrow W$, since $ST$ is only defined when $T$ maps to the domain of $S$, this implies $T$ is surjective, so $\dim V = \dim range T$. Subtracting both equation $\dim U - \dim range S = \dim null S + \dim null T$.
stoic pythonBOT
Otoro:
spiral star
T stays the same, yea
the idea is that the composition doesnt change if you restruct S to range(T)
you basically strip away the unimportant part of S
when you compose S and T, then you will only ever map things from range(T) with S
so you might as well restrict S to range(T) in the first place
ST(x) = S(Tx)
and Tx is in range(T)
you can focus on the part of T that maps U -> range(T) and the part of S that maps range(T) -> W
range(T) is a subspace of V that you can reason about easily
because the rank nullity theorem applies
old flame
I guess I think I'm getting confused between ST and S itself
old flame
ah okay thanks, will get back to you soon π
spiral star
but you see what i mean, right?
restrict T's codomain to range(T) and restrict the domain of S to range(T)
old flame
yeah but restricted S is different than S, since S also maps other vectors in V where those are not from Tu
spiral star
yea S and the restriction are not the same
but that doesnt matter
$S|_{\im T} \circ \widehat{T} = S \circ T$
stoic pythonBOT
Flow:
old flame
wait how is T and T hat different ?, both maps $u \in U$ to range T
stoic pythonBOT
Otoro:
spiral star
$T = \iota \circ \widehat T$ where $\iota\colon \im T \to V$ with $\iota(x) = x$
stoic pythonBOT
Flow:
spiral star
basically, T is $\widehat T$ but we made the codomain larger
stoic pythonBOT
Flow:
spiral star
you can extend the codomain of any function by inclusion into a superset
but the key idea is, that you want range(T) in the middle of the diagram instead of V
old flame
you mean for the first one ?
spiral star
you want to go from this
to this
because if you apply the rank nullity theorem to the second one, it gives you the answer straight away
the proof writes itself almost
old flame
alright............
so I guess, I need to define new maps T hat and restricted S yeah ?
spiral star
and since the the two compositions are equal, you can use the lower diagram to reason about the upper one
well yea, but their definitions are trivial
T hat is just T
and S restricted to range(T) is just S but with smaller domain
here is a hint: can you figure out what $\ker (S|{\im T})$ and $\im(S|{\im T})$ are?
stoic pythonBOT
Flow:
old flame
$\ker (S|{\im T})$ is $null T$ ? and $\im(S|{\im T})$ is a subspace of $range S$ ?
spiral star
no to the first one
and yes to the second one but you can say more
$\ker (S|_{\im T})$ cant be $\ker T$ because they are not even in the same vector space
stoic pythonBOT
Flow:
old flame
ah big mistake sorry
spiral star
im just gonna tell you i guess. $\ker (S|_{\im T}) = \ker S \cap \im T$
stoic pythonBOT
Flow:
old flame
its a subspace of range T
spiral star
yea
it should be intuitive. its the part of null(S) that is also in range(T)
because when you restricted S, you threw away everything of V that is not in range(T)
old flame
yeah ?
spiral star
and the restriction of S still maps exactly like S
old flame
just less vectors though
spiral star
so if S(x) = 0 then the same is true for the restriction
but you only allow those x that are also in range(T)
the proof for this is super simple, you can just show that those sets include each other
alright
but what about the image of the restriction
$\im (S|_{\im T}) = \im ST$
stoic pythonBOT
Flow:
spiral star
that one is trivial
$\im T = T(V)$ and $\im ST = (ST)(V) = S(T(V)) = S(\im T) = \im(S|_{\im T})$
stoic pythonBOT
Flow:
spiral star
that's just a fact by definition
spiral star
i think with $\ker (S|{\im T}) = \ker S \cap \im T$ and $\im (S|{\im T}) = \im ST$ you can see how the dimension formula applies now right?
stoic pythonBOT
Flow:
spiral star
dimension formula = rank nullity theorem
apply it to
in particular, $(S|_{\im T}) \circ \widehat{T} = ST$
stoic pythonBOT
Flow:
spiral star
so it starts like this: $\dim (\ker ST) = \dim U - \dim (\im ST)$
stoic pythonBOT
Flow:
spiral star
and now use $\im ST = \im(S|_{\im T})$
stoic pythonBOT
Flow:
spiral star
and carry it through
old flame
alright thanks for the help, show u when im done
spiral star
π
winter lake
dire thunder
privyet @wintry steppe
wintry steppe
privyet vimes #βhow-to-get-help
dire thunder
no u
wintry steppe
old flame
@spiral star something like this ?
Proof : Suppose $U,V$ are finite dimensional, and $S \in L(V,W), T \in L(U,V)$.
\par Consider $\widehat{T}$ such that $T = \iota \circ \widehat{T}$, where $\iota : range T \rightarrow V$ and $S_{\im T} : range T \rightarrow w$. By rank-nullity theorem, $\dim null ST = \dim U - \dim range ST$, since $range ST = range S_{\im T}$, $\dim null ST = \dim U - \dim range S_{\im T}$.
\par Once again by rank-nullity theorem, $dim U= \dim null \widehat{T} + \dim range \widehat{T}$, once more by rank-nullity theorem $$\dim range \widehat{T} = \dim null S_{\im T} + \dim range S_{\im T}$$, substituting this into the previous equation, $$\dim null ST = \dim null \widehat{T} + \dim null S_{\im T} + \dim range S_{\im T} - \dim range S_{\im T} = \dim null \widehat{T} + \dim null S_{\im T}$$.
\par As $null S_{\im T}$ is a subspace of $null S$ and $\dim null \widehat{T} = \dim null T$. This results in $$\dim null ST \leq \dim null T + \dim null S$$.
spiral star
oof
old flame
why does it look so bad LOL
wintry steppe
might help the formatting to put the equations and/or inequalities in double dollar signs
would massively improve readability
old flame
alright lemme do some edits
spiral star
i think sometimes an equation might say more than a long text :p
old flame
so u think I should skip some description ?
spiral star
yea
also, you dont have to mention when you substitute something in, the reader can guess that easily from a sequence of equations
old flame
oh okay haha
Suppose $U,V$ are finite dimensional, and $S \in L(V,W), T \in L(U,V)$.
Consider $\widehat{T}$ such that $T = \iota \circ \widehat{T}$, where $\iota : range T \rightarrow V$ and $S_{\im T} : range T \rightarrow w$. By rank-nullity theorem, $\dim null ST = \dim U - \dim range ST$, since $range ST = range S_{\im T}$, $\dim null ST = \dim U - \dim range S_{\im T}$.
By rank-nullity theorem, $\dim U= \dim null \widehat{T} + \dim range \widehat{T}$, once more by rank-nullity theorem $\dim range \widehat{T} = \dim null S_{\im T} + \dim range S_{\im T}$. $$\dim null ST = \dim null \widehat{T} + \dim null S_{\im T} + \dim range S_{\im T} - \dim range S_{\im T}$$ $= \dim null \widehat{T} + \dim null S_{\im T}$.
As $null S_{\im T} \subset null S$ and $\dim null \widehat{T} = \dim null T$. Thus $$\dim null ST \leq \dim null T + \dim null S$$
stoic pythonBOT
Otoro:
Compile Error! Click the 
reaction for details. (You may edit your message)
spiral star
i guess algebra becomes a bit heavy on the notation sometimes :p
old flame
unfortunately yes
spiral star
maybe a bit too much for the bot if you dont add your own macros
spiral star
you can add your own latex macros to the preamble that texit uses
so you can get a command for range(T) and null(T) for example
and other stuff
old flame
oh how to do that ?
spiral star
that would take me too long to explain now if you havent used tex much before
old flame
I have latex though
spiral star
anyway, i have a bit of trouble parsing your post
so im just gonna post my solution and you can compare it lol
old flame
the difference was, I broke down $dim U$ instead
stoic pythonBOT
Otoro:
old flame
and broke down $range T$ hat too
stoic pythonBOT
Otoro:
old flame
and stuff cancelled, then since null restricted S is a subspace of null S, and null T hat = null T , inequality follows
spiral star
well $\widehat Tu = Tu$ so then $\ker \widehat T = \ker T$ and $\im \widehat T = \im T$
stoic pythonBOT
Flow:
stoic pythonBOT
Otoro:
Compile Error! Click the reaction for details. (You may edit your message)
spiral star
i dont see how breaking down anything else would be helpful 
but in the end its just applying the rank nullity theorem twice
but with the restricted function you get the right dimensions
old flame
same same, I used it twice, but once to break U and once to break $range \widehat{T}$
stoic pythonBOT
Otoro:
old flame
I' just thinking whether mines worked, since I broke down different objects
spiral star
what did you do with U
maybe you should rewrite your proof. if you have latex locally, write it there and just post a screenshot
spiral star
yea that works too
but the notation is still horrible xd
dont mix im / ker notation with null / range in your proof
i guess you havent seen function restrictions before?
when you have a function $f\colon A \to B$ and a subset $X \subseteq A$ then you can define the restriction of $f$ to $X$ by ${f|_X\colon X \to B, \ x \mapsto f(x)}$
stoic pythonBOT
Flow:
spiral star
if the function $f$ is linear, and $X$ is the carrier of a subspace, then the restriction $f|_X$ is also linear. that's very easy to see
stoic pythonBOT
Flow:
spiral star
and yea, either use $\operatorname{range}(V)$ and $\operatorname{null}(V)$ or $\im(V)$ and $\ker(V)$ but not mixed
stoic pythonBOT
Flow:
spiral star
write a macro for those things
maybe something like
\DeclareMathOperator{\vnull}{null}
\DeclareMathOperator{\vrange}{range}
or be more creative with the names, whatever you like
then in tex you can write \vrange T and so on
basically, it's common to just declare a bunch of macros for the things you need.
that's easy when you work locally anyway
hollow oar
yea i do a lot of homework in latex so i made a .sty file where i declare a ton of new commands and import all my packages
its very nice
spiral star
yep it's nice to have a collection of macros that are useful for lots of stuff π
and then some locally defined ones for convenience
old flame
ohhhhhhhh, I see
I think I have seen restriction but very rarely so I may have forgetten
sounds good, I guess I need working on my latex layout LOL
one question, $T=\iota \circ \widehat{T}$, what is the reason for this ? a bit confused
stoic pythonBOT
Otoro:
old flame
cause don't both map to range T ?
spiral star
it's not relevant for the proof. that was just to explain how you get T back by including the image of T hat into V
maybe a picture explains it better
you can factor any function f
into a function that just maps to the image of f
and then you take an inclusion into a superset of the image
old flame
but isn't B the image of f ?
spiral star
only when f is surjective
old flame
ohhhhhhhhhhhh nevermind lol
spiral star
the image is always a subset of the codomain
old flame
sometimes I do hear people use the terms image and codomain exchangebly
spiral star
thats wrong :p
spiral star
the important part of the function is contained in the image
like, if you take the domain and the image of the function, you get its graph
of course you can make the codomain as big as you like, as long as it contains the image
but you can also "cut off" the part of the codomain that isnt part of the image
i used iota as an inclusion map. it is just the identity function from the image to the bigger set
old flame
so in this case, $\widehat{T}$ and $T$ are the same, but they may not be, so ur definition of $\iota$ helps with this ambiguity right ?
stoic pythonBOT
Otoro:
spiral star
T hat and T have the same graph, but they are not the same function
a function is a tuple (domain, graph, codomain)
they have the same domain and graph, but a different codomain
so they are not equal
but they have the same domain, the same graph and the same image
they are for all intents and purposes interchangeable
often we only care about the graph of a function and kind of ignore the codomain
old flame
they have different codomain because one is $range T$ and one is $V$ right ?
stoic pythonBOT
Otoro:
spiral star
range(T) is a subset of V but not necessarily equal
old flame
yeah so thats the case am I right ?
old flame
alright, good to know
spiral star
all information about the function is encoded in the domain and the graph
the codomain doesnt carry any information really
it can be chosen as any superset of the image
it doesnt change any behavior of the function
that's why it's just a technical thing
spiral star
often you will actually make the codomain smaller so it matches the image
for example, injective functions are called 1-to-1
but the 1-to-1 relation is between the domain and the image, not the domain and the codomain
but you can reduce the codomain to the image, and then any injective function becomes bijective
this is something you do quite often to get bijective functions
the codomain only matters if you want to talk about surjectivity
otherwise it just gives you a hint for "the image is a subset of this"
if you define a function on real numbers by f(x) = xΒ² then its convenient to say f: R -> R even though the image is non-negative
stoic pythonBOT
Otoro:
spiral star
yea or $\bR_{\geq 0}$
stoic pythonBOT
Flow:
old flame
but its just for a simpler notation ?
spiral star
yea, we could choose any superset of the image as codomain
we could also make it f: R -> C
doesnt change anything
old flame
true
barren blaze
So I am back with a "how do I efficiently solve these type of questions?" question. I need to determine if U is a linear subspace of R^n. What should I be looking at first? The question has 6 of these to solve and there must be a fast way of doing this right?
wintry steppe
not really, either its visible that something is not a linear subspace and then you can quite fast show why, but if it is a linear subspace you need to check all the properties
spiral star
checking properties is usually super quick anyway
and if it doesnt work its rather obvious
barren blaze
I guess I am looking for an easy way out because I don't intuitively understand some of the definitions for x
spiral star
well the one you posted should be easy
also hint: if a subset involves a norm that isnt defined by a scalar product, then it's probably not a subspace
for your example, just pick a standard basis vector, like e_1 = (1, 0, ..., 0) and then multiply it by 2
e_1 is in U but 2 e_1 is not, so it cant be a subspace
barren blaze
Thanks for the explanation, I think I am getting there. I will try other examples and see
half storm
I've never seen a linear map between two vector spaces over different fields. It seems that it's always implicit that the map is between two vector spaces that are over a common field. But sometimes my book makes explicit when stating certain theorems that the field most be common between the two vector spaces. Do there exists linear maps - or analogous to them - between vector spaces V and W that are vector spaces over different fields?
spiral star
you need the same field to define linearity
scalar multiplication isnt necessarily preserved or can even be defined
you cant say f(cv) = cf(v) when c is not even in the field for cod(f)
half storm
Yea I was thinking that it wouldn't even make sense.
winged iris
you can have a linear map between different vector spaces
half storm
but sometimes the book will define it expliclty and sometimes it won't.
spiral star
that's normal
sometimes you mention the field
some theorems hold for all fields, some only for R or C
sometimes you need to mention the field so you give it a name
if your theorem or proofs needs to take scalars from the field you need to give it a name
winged iris
you define the map in terms of the bases of the codomain
rose coral
If you are curious, there are ways of having "linear maps" between vector spaces over different fields. If you have a way of relating two fields (with what is called a field homomorphism), you can define maps that incorporate this relation in the linearity (I think these are called semi-linear maps). For example if you have two fields F and K, and vector spaces V over F and W over K, a map h:F to K that relates the field, you can call a map T:V to W semi-linear (probably with respect to h) if it respects the addition and has T(cv) = h(c)T(v). But really don't worry about this unless you are curious.
spiral star
yea you need the field hom as well
but it isnt equivalent to the definition of linearity
funny thing is, i know semi linear maps as something different
but yea, field homs must be used and it isnt equivalent to linearity anymore
just something similar
half storm
I had a feeling that's exactly what you needed to do.
some sort of map that relates the two fields.
so a field homomoprhism 
rose coral
With fields it probably isn't so interesting as field homomorphisms are very restrictive. Essentially you can only have inclusions of fields like the inclusion of Q into R or R into C.
I'm sure you could study some wacky stuff if you use rings instead, but I think that's way outside of linear algebra and I really I'm just making up stuff
spiral star
random question: taking conjugates in complex numbers is a field automorphism right?
spiral star
okay, then what i know as semi linear maps lines up with that def
i just didnt think about it lol
rose coral
Oh cool
spiral star
i learned it only with complex conjugate, but its basically a field hom
rose coral
Yeah that makes sense, so like a complex inner product is linear in one entry and semi linear in the other
spiral star
yea
i think that is where i learned semi linear
but it wasnt obvious to me that it applies to any field hom
but it makes sense to me to generalize like that π
not sure if it's very useful tho
going into a different field
rose coral
I think it only really makes sense when you talk about semi-linear maps with different field homs (like the identity and conjugation for the inner product)
Otherwise if you consistently use the same field hom h:F to K you can just think of F being a subfield of K, and its like you are considering the codomain as a vector space over F instead of K
spiral star
makes sense
barren blaze
I am trying to find the best approximation vector(might translate differently to english) of a vector on a linear subspace. The question gives me a linear subspace in form of U=span{v_1,v_2} and the vector v_3. If I am understanding right, I need to find the orthonormalbasis for the subspace which I did. But I am not sure what to do next? Am I doing this all wrong?
half ice
@barren blaze
Are v1, v2, v3 in RΒ³?
barren blaze
@barren blaze
Are v1, v2, v3 in RΒ³?
@half ice yes they are
half ice
Okay so you've got the orthonormal basis, you can continue in that direction. Get a new vector u, that is orthogonal to v1 and v2. Then, there is a unique way to write:
v3 = av1 + bv2 + cu
Since u is orthogonal to the plane,
av1 + bv2
is the best approximation.
The cross product can find u, if you're willing to go that route
barren blaze
Would the new orthogonal vector u be an element of U? Asking this because apparently I need to find a u which is an element of the subspace.
half ice
No
av1 + bv2 is
My bad, I should have used w or something.
This vector isn't going to be in the answer, but it will be used to find the answer
barren blaze
I asked it because the question goes: "find the best approximation of v3 through/with the element u* E U"
half ice
Okay so you've got the orthonormal basis, you can continue in that direction. Get a new vector w, that is orthogonal to v1 and v2. Then, there is a unique way to write:
v3 = av1 + bv2 + cw
Since w is orthogonal to the plane,
av1 + bv2
is the best approximation, and is in U
w itself won't be in U
barren blaze
so perhaps I can write u* = av1 + bv2?
half ice
Yeah!
barren blaze
Thanks a lot!
torn tangle
Hi guys, are you still doing the question or can I ask one really quick?
barren blaze
go for it!
half ice
Do you know what R is?
torn tangle
oh
half ice
Of course, we call this RΒ² normally, but that's how the product works on R
torn tangle
so we're just meshing the elements
half ice
Now RΓ RΒ² takes two elements, the first from R and the second from RΒ²
torn tangle
so for every element in R^1, there is three in R^3?
half ice
And bundles them. Of course, this just ends up being 3 elements from R
torn tangle
so for a point say
(1,2,3,4)
We could do take (1,2,3) from Z^3 and (4) form Z^1, or (1,2) from Z^2 and (2,4) from Z^2
half ice
Yeah, sure haha
wintry steppe
so for every element in R^1, there is three in R^3?
you have to be careful when saying stuff like this since |R| = |RΒ³|
half ice
Γ is normally something we consider between sets
If A and B are sets, then AΓB is a set of "bundled elements" where the first is from A and the second from B
torn tangle
Yup
And then as you go up higher dimensions you have even more different combinations from different sets right?
half ice
There's a bunch of ways to make Rβ΄, sure. But there is only one Rβ΄
torn tangle
could Z^6 = Z^2 x Z^2 x Z^2
torn tangle
nice
Is it ever the case we can produce a set from "bundling" sets of different types
like uh
half ice
Yeah, good point I should have thought to use that
torn tangle
(3,sqrt(2))
torn tangle
yup
half ice
ZΓR is a set. (3, β2) is one of its elements
torn tangle
And then also
ZxQ'
This probably sounds silly but uh
Is division a thing in this context? Removing coordinates?
half ice
You can make a function for that
Ο1 is a function RΒ² β R that takes the first value
Ο1(1,2) = 1
torn tangle
ah
half ice
I suppose "first two values" could be done haha
torn tangle
It's not possible for there to be a function where it removes all coordinates is it? like Ο1(1,2) =
half ice
Every function needs an output and a set it outputs to
torn tangle
true
half ice
But I'm sure something like that could be made to work
torn tangle
I understand it now, thanks heaps !
rose coral
Usually you consider R^0 = {0}
And so if you map into R^0 everything gets sent to 0. I kinda makes sense cause R^2 is a plane, R is a line, and R^0 is a point.
Also makes sense in terms of vector spaces
honest notch
I apologize for the poor quality, is this a suitable explanation
I'm not sure if I've understood the question correctly
honest notch
I see, cheers for picking up on that
half storm
To say that the nullspace of a $nullity(A^t) = dim(N(A^t)) = 0$ means that $A^t$ is invertible. Meaning that $A^t x = 0$ is 1-1 and has only 1 solution; $x = 0$.
stoic pythonBOT
JohntheDon:
half storm
You want to prove that $dim(N(A^{t}))} = 0$. So you have to show that $N(A^{t}) = { 0 }$.
stoic pythonBOT
JohntheDon:
Compile Error! Click the reaction for details. (You may edit your message)
half storm
So suppose a $x \in N(A^t)$ So $A^tx = 0.$
You need to show that $x = 0$ you might be able to figure it out from there.
stoic pythonBOT
JohntheDon:
raven parrot
So, during the summer, I was taught ALOT of linear algebra from a relative, then he talked about the CayleyβHamilton theorem.
Which fascinated me.
honest notch
thanks for your help, ill see how i go from here
stoic pythonBOT
JohntheDon:
limber sierra
ah yes, $\det(\lambda) = \det(\lambda I_n - A) \implies \det(A) = \det(AI_n - A) = \det(A - A) = 0$, my favourite proof
stoic pythonBOT
Namington:
limber sierra
totally valid dont question it at all
ignore the fact that applying this to literally any other multilinear form leads to a contradiction
honest notch
@half storm is the working out above not sufficient to prove this
honest notch
Ok, thanks for the clarification. Makes a lot more sense now
stoic pythonBOT
JohntheDon:
limber sierra
....thats what i said
half storm
but A may not be square
@limber sierra Are we forced conclude that A is a square matrix from the out set of the problem?
limber sierra
huh?
half storm
O.k.
limber sierra
i was just trying to point out a flaw in their proof
they assumed A was square when they wrote RREF(A) as an identity matrix
half storm
I just thought I saw you say that A may not be square.
half storm
I'm confused.
limber sierra
from the start of the proof, we dont know whether A is square or not
so asserting that rref(A) = identity matrix
is wrong
at least at the beginning
since we dont know whether A is square
and if it isnt, clearly its RREF cant be square.
half storm
but then you eventually have to show that the $dim(N(A^t)) = 0$
stoic pythonBOT
JohntheDon:
half storm
which would imply that A^2 is injective yea.
nvm
so you cna't do it their way.
don't worry about it.
limber sierra
okay, that doesnt mean you can just assert that A is square at the beginning without proof
half storm
I figured out what was wrong.
cool
I guess you can say that m is necessarily less than or equal to n.
@honest notch yea so that doesn't work. I think you have to do it the way I was thinking of initially
Well you probably don't have to
there's probably another way to prove it.
honest notch
@half storm youre saying the way i currently have it doesnt work?
half storm
Yea
honest notch
I see
half storm
You've assumed that the matrix is square in the beginning.
half storm
No, the theorem will hold regardless whether or not the matrix is square.
This theorem holds for any matrices where the number of rows is less than or equal to the number of columns i.e. $m \leq n$.
stoic pythonBOT
JohntheDon:
stoic pythonBOT
Sage:
honest notch
is this correct?
half storm
Sorry, I'm not sure what the question is.
honest notch
all good, ignore that then
just not sure how to change my proof for what youve just gone over
half storm
My proof would look something like this:
By our assumption, we know that $yA = 0$ has a unique solution. Meaning that there is a solution and there is only one solution. Consider the zero vector $0 \in \mathbb{R}^m$. It's easy to show by the definition of matrix multiplication that $0A = 0$. It follows that y = 0 is a solution and the only solution.
stoic pythonBOT
JohntheDon:
half storm
Now suppose that $x \in N(A^T)$. So $A^T x = 0 \implies {(A^Tx)}^T = 0^T \implies x^T A = 0^T $. But by the uniquness 0, then $ x = 0$. So $N(A^T) = {0} \implies nullity(A^T) = 0$.Now by rank-nulllity theorem $dim\mathbb{R}^m = nullity(A^T) + rank(A^T) \implies m = rank(A^T)$ The rank of a matrix and it's transpose are equal. So $rank(A^T) = rank(A) = m$. \qed
stoic pythonBOT
JohntheDon:
half storm
@honest notch does that proof make sense?
honest notch
${(A^Tx)}^T = 0^T \implies x^T A = 0^T$ How did you get to this
stoic pythonBOT
Sage:
half storm
It's a property of taking the transpose of the product of matrices. Given the product of two matrices A and B ${AB}^T = B^TA^T$. And that ${A^T} ^T = A$
stoic pythonBOT
JohntheDon:
honest notch
Ah i see
ok i understand this
are you able to clarify by the uniquness 0, then x = 0
half storm
By assumption, it was claimed that y is a unqiue vector in R^3 such that yA = 0. We showed earlier that this mandated that y = 0.
Since 0A= 0. And uniqueness means that one and only one such factor exist that such that yA = 0. So we are forced to conclude that 0 is the one and only such vector.
After some algebra, we ended up with the statement that $x^T A = 0^T$. So $x^T = 0 \in \mathbb{R}^m \implies {x^T}^T = 0 \in \mathbb{R}^n \implies x = 0 \in \mathbb{R}^n$
stoic pythonBOT
JohntheDon:
half storm
π
mortal stream
is this correct or is this an error?
half storm
definitely an error lol.
mortal stream
if Ax = 0 has one solution, the nullity is 0 right
half storm
Yup
mortal stream
thanks!
rocky wharf
Ofc, thank you.
old flame
Hello guys, this time I really need some hints on how to even approach this problem. Thank you
Prove that if V is finite dimensional with dim V > 1, then the set
of noninvertible operators on V is not a subspace of L(V ).
One note there is a theorem where if V is finite dimensional, then given $T \in L(V)$, T is invertible, so I suppose this is useful in some way ?
stoic pythonBOT
Otoro:
bleak gorge
That is a false statement I believe.
Try a few examples and see if you can find a counterexample I guess
to elaborate you should be able to pick a V and then define two linear operators, say T, U in L(V) and then show that their sum is not in L(V) (i.e. T, U are noninvertible but T+U is invertible)
rose coral
I think that's what @old flame said, but I don't know what they meant in their last sentence
spiral star
@old flame if you are struggling with how to define the operators in a general setting here is a hint: since V is finite dimensional, any T in L(V) is fully described by the images of a basis of V.
old flame
@bleak gorge I'm sorry but which statement you meant was wrong ?
spiral star
the last statement you made. T in L(V) is not necessarily invertible
old flame
Yeah I think I misread the theorem, it's supposed to be T is invertible if T is injective or surjective
spiral star
yea
old flame
So I guess the starting point now is to define some operators and compute the addition ? Like what Jesse said ?
spiral star
yes
old flame
Okok tyty
spiral star
construct two non invertible operators whose sum is invertible
old flame
So I assume the operators are non invertible at first ?
spiral star
your subspace consists of non invertible operators
you have to construct two specific ones
sometimes the sum of two non invertible operators will still be non invertible
but sometimes the sum becomes invertible
you need to give an example where the sum of two non invertible operators becomes invertible
just construct them
old flame
I see so it's to prove that space is not closed under addition
spiral star
yes!
old flame
So from what u said, construct a subspace of V filled with non invertible operators, then prove not closed under addition, so it's not a subspace of V in the first place
spiral star
well, technically it will never be a subspace. you just take the subset
the set is already given
spiral star
take the set of non invertible operators, and show that it cant be the carrier set of a subspace of L(V)
because it wouldnt be closed under addition
zinc copper
In comparing example 1.10 and 1.12, how could 1.12 compare to 1.10? The naturals are countably infinite while the reals are not, so how could we unwrap the reals into an infinite sequence such that it could fit into a vector?
There's another line that I missed in the shot that says "The difference between this and Example 1.10 is the domain of the functions", which is why I ask
spiral star
you should try to move away from viewing vectors as tuples
the concept of vector space is not tied to that at all, even though its a the classic intro example
any set can be made into a vector space if you can define the necessary operations on it
zinc copper
alright
it just seemed as though they were suggesting a tuple interpretation was possible
I realise it is a vector space
I checked all the axioms
spiral star
if you take any set and you can define a vector addition and scalar multiplication on it that follows vector space axioms then it becomes a vector space
zinc copper
I was just wondering if a tuple interpretation was possible
spiral star
in particular, any set of functions with a field as codomain can be made into a vector space
zinc copper
right
spiral star
tuple interpretation breaks when the domain isnt countable
spiral star
π
i think its better to view vector spaces as something where you can add vectors and multiply with scalars
inspecting the internals of a vector isnt part of the theory, that's specific to the sets you are working with
zinc copper
true
spiral star
but linear algebra isnt concerned with that
except if you want to count the relation to F^n for finite dimensions
quartz compass
just to give an example, the set of continuous functions on some domain can be seen as a vector space
try writing out tuples for that just to see
zinc copper
yeah that was what I was originally asking about haha
old flame
Yup
spiral star
V - range(S) isnt a vector space
but the idea is not bad
just not formulated well xd
but you can take a non invertible S in L(V). then range(S) is not V but a strict subspace. then take T in L(V) with range(T) as the complement of range(S)
old flame
But isn't the complement of range S the same as my exclusion ? What's the difference hmmm
spiral star
by saying, the direct sum range(S) + range(T) = V
set complement does not apply
for example, you would remove the zero vector from the set
because the zero vector is in all subspaces, what you are left with cannot be one
old flame
Ohhhhhhhhhhhhhh
So it's in the right direction, but the definitions needs to be a bit more intricate right ?
spiral star
yes, it's the right direction
usually you decompose vector spaces into direct sums
$V = U \oplus W$
stoic pythonBOT
Flow:
old flame
But didn't u say complements don't apply ? How can u define range T then
old flame
May you explain the difference please
spiral star
two subspaces $U,W \leq V$ are complements if $U \oplus W = V$
stoic pythonBOT
Flow:
spiral star
meaning $U + W = V$ and $U \cap W = \qty{0}$
stoic pythonBOT
Flow:
spiral star
and $U + W = \qty{u+w \mid u \in U, w \in W}$
stoic pythonBOT
Flow:
spiral star
if $U \oplus W = V$ then a basis $(u_1, \dots, u_m)$ of $U$ and a basis $(w_1, \dots, w_n)$ of $W$ will form a basis of $V$ as $(u_1, \dots, u_m, w_1, \dots, w_n)$
stoic pythonBOT
Flow:
spiral star
this is how split vector spaces
their dimensions will also add up, $\dim U + \dim W = \dim V$
stoic pythonBOT
Flow:
spiral star
this is all stuff you have to prove of course
set complements do not work with vector spaces
old flame
Encountered it already
Yeah
Oh I didn't knew vector space complements are the ones in direct sums
Haven't heard of that definition yet
So if you're defining a complement in vector space does the notation $^{c}$ still apply ?
stoic pythonBOT
Otoro:
spiral star
havent seen it used
old flame
So are there notations for th complement ?
spiral star
only the one i have showed you
you can basically say, "extend the basis to a full basis of V"
then you have the complement
old flame
Complement as in the extension yeah ?
spiral star
if you have U <= V and you extend a basis of U to a basis of V then the new basis vectors you added span the complement of U
old flame
Oh, so in the direct sum, the complement naturally shows itself then
spiral star
yea, whenever two subspaces form the original space as a direct sum, they are complementary
old flame
Alright then, let me rewrite my proof, I'm just very glad I got the correct direction
spiral star
you can also get around all of this if you just define the maps by a basis
since your approach is already correct im not gonna give away any answers by showing my solution i guess
spiral star
End(V) is the set of endomorphisms on V, i.e. L(V)
and Aut(V) is the set of automorphisms on V, i.e. invertible operators in L(V)
old flame
Oh are those like more advanced terminologies lol
spiral star
well not that advanced
it's just terminology that applies to all of algebra, not just linear algebra
old flame
Means the same thing ? Throughout algebra ?
spiral star
yes.
homomorphisms are structure preserving maps
endomorphisms are homomorphisms with the same domain and codomain
automorphisms are bijective endomorphisms
in linear algebra a homomorphism is a linear map
it preserves the structure of a vector space
but the same terminology applies to all of algebra: group theory, ring theory, etc.
that's also why i use ker and im instead of null() and range()
old flame
That is very interesting