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12:02 AM
Derivatives, credit risk, fixed income derivatives
The latter two are half semester courses
 
sweet. let me know how these go. i'm interested in those as well. how do you like them
 
Okay its getting better, we're using O notation on $\Delta t$,
 
hmm.. so? (as part of some derivative definition?)
 
Limits, not derivatives. Nothing is differentiable here
 
there's the ito derivative
 
12:17 AM
We're not doing that right now
Just statements about orders
 
hm. they sound like super interesting classes though.
4 classes total?
 
Yes but 2 of them are half semester classes
So 3 at any one time
The credit risk class I like the most so far
 
Problem: If $n > 2$, then $f(x) := x^n + ax + b$ has at most 3 roots...I could use a hint on solving this problem.
 
12:35 AM
@user193319: I think this is easy.
Have you considered Rolle?
 
Hmm...Let me see if that theorem is available to me.
 
(3 real roots, I presume)
 
@Semiclassic: That's a gratuitous remark.
 
Hmmm...Unfortunately Rolle's theorem doesn't appear in the section from which this problem comes.
 
What does?
 
12:40 AM
In fact, it doesn't look like it appears in my book at all.
 
It's just a special case of the Mean Value Theorem.
 
I have the MVT, so I suppose I could prove Rolle's theorem as a lemma to this problem.
 
No, no. Mean Value Theorem is just fine.
 
Okay, let me think about how to use it.
 
@GFauxPas are these offered year round, or just in the fall? also, do they count as two separate classes, or just one?
 
12:46 AM
hi @GFaux and @JoeShmo
 
howdy, ted
saw gromov in the flesh today
on the 8th floor of the courant institute
 
I saw you said that.
He's even scarier than I am.
 
how so
it was like witnessing a rare pokemon
you haven't played nintendo much back in the day have you..
 
never
 
trust me when i say the reference was brilliant
 
12:49 AM
I guess I know more superstars than you do :P
 
oh i don't doubt that for a moment
 
@TedShifrin The most I am able to conclude is that $f'$ is $0$ at least 4 values. But I don't see any contradiction in this.
 
no, no, @user193319. How'd you get that?
 
@JoeShmo eh, you could have made it shiny :P
 
@TedShifrin By assuming that $f$ has at least 4 roots, say $x_1 < x_2 < x_3 < x_4$. Then since $f(x_i)=0=f(x_{i+1})$, there exists $c_i \in (x_i,x_{i+1})$ such that $f'(c_i) = \frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}=0$.
 
12:52 AM
Right, @user193319. So how many roots can $f'$ have?
 
...three?
 
Why's that?
 
Not sure...In fact, it doesn't sound right.
 
@Semiclassical haha
 
best way to one-up an obscure reference: with an even more obscure reference
 
12:57 AM
@user193319 you know what $f'(c_i)$ looks like
 
@user193319: Have you calculated $f'(x)$?
 
I just did, actually: $f'(x) = nx^{n-1}+a$, so $0 = f'(c_i) = nc_i^{n-1} + a$.
 
Don't bother writing $c_i$. How many roots does $f'$ have?
The very first thing you should have done — 10 minutes ago — was compute $f'$.
 
I should think it would depend on $n$.
 
Well, sure. So tell us the exact answer.
 
1:00 AM
drumrolls
 
Hint: It also depends on $a$.
 
There's a few possibilities, but not many possibilities. So lay them out.
 
If $n$ is odd, then there are at most two; if even, then only 1....
 
OK, so what do you conclude?
 
That we have a contradiction, since we've shown that $f'$ has at least 4 roots.
 
1:04 AM
No you haven't
 
Didn't I do this above using the MVT?
 
If $f$ has 4 roots, then how many would $f'$ have?
 
do you get any special powers as room owner?
 
yes. refer to his scary comment from earlier
 
LOL, that was a non-power.
I get to put people in time-out and remove messages to trash.
I think that's about it. And — again — I wasn't even consulted about this appointment.
I guess robjohn and anon abdicated. I miss them both dearly.
 
1:08 AM
@TedShifrin with great power, comes great responsibility.
 
There isn't great power, guy.
But I'm not all that irresponsible, thank you.
(Check out the logic!) :D
 
great power => great responsibility
<=>
! great responsibility => ! great power
 
Yeah, so?
 
the statement you made is independent of either of those
 
@MikeM: Here's an interesting post.
Probably so, @JoeShmo. They were two independent statements.
 
1:11 AM
I don't have the heart to look at anyone else's math tonight.
 
Fair enough. I apologize.
Well, one good thing came from my idiocy with non-orientable Gauss Bonnet. I'm now going to go up to Irvine and see Dick Palais and Chuu-Lian Terng for Chinese dinner :P
 
Hey everyone!
 
Heya Demonark. Just in time for me to leave for quick dinner and then bridge at my house.
 
wait ted before you go
 
You back to your usual indolent self by now? :D
yes?
 
1:13 AM
hot take. is information theory a gimmick?
 
I know nothing.
 
hrmp
 
Yup, finally recovered from the GRE I think
 
So your family recognizes you again, Demonark?
 
its potentially offered next semester. gotta know if its any different than dynamic systems / stochastic calc
 
1:13 AM
Though I guess I've still gotta be ready to get back in action just in case I have to do October
 
/is it a gimmick
 
My parents said that in the days before the test I was looking really tense and drained, and now I'm far more relaxed
 
It is a modern subject, @JoeShmo. I certainly wouldn't say that makes it a gimmick. But I truly am ignorant.
I believe that, Demonark.
 
Bob
can I ask a math question?
 
If you'd taken my bootcamp courses, you'd have been drained earlier along :P
Of course, @Bob. Just do.
 
Bob
1:15 AM
if I have a system of two non-linear equations what do I need to do to show that here is only one solution?
 
And tbh I think I'm probably making a bit more progress on math than when I used it to procrastinate tbh. Going through some AT now (gotta at least learn fundamental group and covering spaces properly, having done problems, before class this fall)
 
Oh, ugh. That's hard, @Bob.
Demonark, as I said earlier, you have definitely evolved. :)
Two variables, I assume, @Bob?
 
Bob
Two variables
it seems to me that if all the partial derivatives are positive then the solution is unique.
 
Well, positive partial derivatives doesn't quite settle it.
 
Bob
what else would I need?
 
1:18 AM
You'd need the two solutions to be uphill from one another (left to right), not downhill.
With just that criterion, I can give you one with two solutions.
 
Bob
what does it mean for the two solutions to be uphill from one
when I say a unique solution, I mean something like (1,1)
 
$(x_1,y_1)$ and $(x_2,y_2)$ with $x_2>x_1$ and $y_2>y_1$.
I know what unique solution means.
There's something else going on besides just positive partials.
You need the derivative matrix invertible.
Sadly, I need to get going. I have company showing up soon.
Bye all
 
Bob
does that mean it has to be invertible on the entire interval?
 
See you!
 
Bob
Thank you for your time @TedShifrin
 
2:00 AM
Happy Tuesday everybody! Enjoy your math-tastic day :)
 
bye ted
 
2:37 AM
probably a dumb question but is log(c*n) = O(log(n))?
I want to say yes, but I don't often see the constant inside the log function
 
log(c*n)=log(n)+log(c)
so it's just log shifted by a constant
 
right, of course
but lets say I have
log(c*n + d) where c and d are constants
I want to justify what to do with that d
 
3:00 AM
Maybe something like... if $d$ is negative, then for $n > 0$, $\log(cn + d) < \log(cn)$, and if $d$ is positive, then for $n > 1$, $\log(cn + d) < \log(cn + d + d(n - 1)) = \log n + \log(c + d)$.
 
3:29 AM
right, cool, thanks
:D
 
3:51 AM
Hi everybody, I created a new room called Social Behavior Mathematics (Game theory). If anyone is interested please join the room and say hi. Cheers.
 
0
Q: Limit of Product of Functions Where One Function Converges to Infinity

user193319 Let $A \subseteq \Bbb{R}$, let $a$ be a limit point of $A$, and $f,g : A \to \Bbb{R}$ functions. If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = L \in (0,\infty)$, then $\lim_{x \to a} f(x)g(x) = \infty$. I first tried to solve this problem "from the definitions", but I couldn't...

I could really use help on this problem...It's midnight where I am this is due early this morning...
 
Pig
4:10 AM
what have you tried?
 
Did you follow the link? I outlined what I tried.
 
Pig
sorry, to be more specific, what did you try along the lines of doing it by definition (when you say you don't see how to do this?)
 
Well, by definition we have that for every $\epsilon > 0$, there exist $\delta>0$ such that $|x-a| < \delta$ implies $f(x) > \epsilon$ and $K-\epsilon < g(x) < K + \epsilon$. Hence $f(x)g(x) > \epsilon(K-\epsilon)$. My idea was to choose $\epsilon > 0$ so that $\epsilon (K-\epsilon) > \epsilon$...
But that's true if and only $K-\epsilon > 1$, which isn't necessarily the case.
 
Pig
so $\epsilon$ is generally supopsed to be small
but you know that $f(x) \to \infty$
 
Yes.
 
Pig
4:15 AM
in particular, $f(x)$ should be bigger and bigger as you get closer to $a$
So that means there shouldn't be too much reason to choose the same $\epsilon$ there
in particular, one way to think about it is
 
You mean choose an $\epsilon$ for $f$ and choose a "different" $\epsilon$ for $g$?
 
Pig
yep
In particular, take $M > 0$, you want to find some $\delta > 0$ such that $|x-a| < \delta$ implies $f(x)g(x) > M$
 
And in such a way that $f(x)g(x) > \epsilon$ is the end result?
 
Pig
yes
(continuing what i was saying)
You can imagine that when $x$ is close to $a$,
1. $f(x)$ can be as big as you want
2. $g(x)$ is roughly $L$
which means you want to be in the situation where e.g.
2. $g(x) > L/2$ (which should be possible as long as $x$ is close enough to $a$)
1. and then $f(x) > M/(L/2)$
 
Why $L/2$? How do you get $g(x) > L/2$? Shouldn't $K$ figure in their somewhere?
 
Pig
4:23 AM
Your link has $L$, not $K$
 
Oh, you're right. Sorry!
 
Pig
The reason i had $L/2$ is just that I want to produce $f(x)g(x) > M$ by saying something like $f(x) > ?$ and $g(x) > ?$
Saying $f(x) > ?$ is very easy, since we know that $f(x) \to \infty$
 
@Pig You should copy and paste what you wrote into an answer.
 
Pig
Saying $g(x) > ?$ is harder. But you know that $g(x) \to L$, so what that means is - for anything smaller than $L$, e.g. $L/2$, or $L/3$, or $0.99999L$ - once you are close enough to $a$, then $g(x)$ would be bigger than that thing ($L/2$ or $L/3$ or $0.99999L$)
$L/2$ is just a convenient choice unless I want to troll and do $0.314159265L$ instead
and meh I'm too lazy to write a full fledged answer based on this. If you are working on an assignment or something like that though, it maybe a good exercise to write that out once yourself
 
There is no $\epsilon$ for $f$ (going back a bunch of statements) ... :)
 
4:34 AM
Hi @TedShifrin !
 
hi @Abcd
 
Pig
hi @Ted
 
hi Piggy
2
 
Pig
do you still think about geometry these days? :P
 
Who starred that, for goodness sake?
 
Pig
4:37 AM
oh lol
 
I still answer questions on MSE, yeah, and occasionally discuss geometry in here.
 
 
Pig
(not me by the way)
and cool!
 
Do you know about eigenvalues, @Abcd?
 
@TedShifrin What has he done in the step "Eliminating p, q from 1 and 2"? How did he get that determinant?
@TedShifrin No
 
4:38 AM
Write (1) and (2) as a system of HOMOGENEOUS equations ($0$ on the rhs). It has a nontrivial solution. Therefore the matrix of coefficients has determinant 0.
 
Rehi Ted!
 
rehi Demonark
 
@TedShifrin So it means the solution of the two equations would be coinciding lines?
 
The homogeneous equations? Yes.
One equation is a multiple of the other.
I guess therefore it's true for the original system, too.
It is.
 
What is eigenvalue? What is its use here?
 
4:41 AM
The fact that $AB = B$ (and $B\ne 0$) says that $1$ is an eigenvalue.
 
But what is eigenvalue?
 
It's a scalar $\lambda$ so that $Av=\lambda v$ for some nonzero vector $v$.
This means that the determinant of $A-\lambda I$ must be $0$. So you solve for $\lambda$ to find the eigenvalues.
@Piggy: Last time I checked I had way more points for diff geo answers than anyone else on MSE. Some of them are slightly interesting :)
 
Pig
nice :D
@Ted, how was your mathematical path like? In particular, how was Chern as an advisor?
 
Very inspiring.
 
Pig
4:57 AM
What made him good? Was he inspiring mathematically or elsewhere too?
 
We became good friends, and his wife liked me cuz I was into food and invited them to nice dinners :)
 
Pig
haha nice
 
 
1 hour later…
6:21 AM
If A is skew symmetric then $A^{2n}$ is symmetric and $A^{2n+1}$ is skew symmetric how?
 
@Abcd just compute it
 
The tables below show the values of predicates P(x, y), Q(x, y), and S(x, y) for every possible combination of values of the variables x and y. The row number indicates the value for x and the column number indicates the value for y. The domain for x and y is {1, 2, 3}.

https://i.imgur.com/tJeR8WA.jpg

Indicate whether the quantified statements is true or false
∃x ∀y P(y, x)
I cant quite figure out how to deal with the swapped x,y --> y,x
I understand perfectly when the order of x,y is flipped like this ∀x∃y y2 = x, but they've described a table and how to read x,y on it, so I dont know whether I invert the manner in which i would be reading it or what
 
6:57 AM
"Let $f(x)=x/3$ and $g(x)=3x$, $f,g:\mathbb R \to \mathbb R$. Determine whether they are topologically conjugate".

My thinking: h=1/x clearly satisfies $h\circ f = g \circ h$ but is not a homeomorphism on the real line since it is not continuous at x=0. But this does not show that $h$ is the only function that can be chosen. How do I determine this more generally?
 
 
1 hour later…
8:05 AM
@JasperLoy Hi, Jasper. All is explainable after reading this meta link math.meta.stackexchange.com/questions/29007/…
Happily, after that, not sure exactly when, things returned back to normal (however, not sure for how long).
@JasperLoy Hope you're fine. Talk some other time since now I'm overloaded with a new mathematical project consuming all my time.
 
 
3 hours later…
11:05 AM
Can someone please tell me how author got $\det{(A^*)}= \det{(A)}$ at the end of his first line?
 
11:29 AM
I have to evaluate $\int_0 ^{1/2} dx/((1+x^2)\sqrt(1-x^2)) ... I substitute x=1/t to and then it simplified into a simple integral. Is there any objection or incorrect mathematics in this substitution?
 
@Abcd didn't I already tell you to burn your book
 
@mercio It has good problems (not theory) thats why I cant burn it
 
det(conjugate(A)) = conjugate(det(A))
 
And there's literally no good high school maths textbook in India!
 
yes but you are on the internet right now
the internet is full of ressources
so they are wrong about that line. However they actually never used it because they didn't go as far as they should have and stopped at det(conjugate(transpose(A))) * det(A) = I
from there the correct deduction is conjugate(det(transpose(A))) * det(A) = 1
and then conjuagte(det(A)) * det(A) = 1
and then square of modulus of det(A) = 1
and then modulus of det(A) = 1
 
11:46 AM
@mercio I think you have missed a det before I
 
yeah
I fixed the typo in the next lines o..o'
it's okay if you look at the screen from far enough cuz I looks like 1
also I forgot to mention that det(transpose(A)) = det(A)
which is equally as important as det(conjugate(A)) = conjugate(det(A))
and your book messed up at combining the two
 
@mercio how did you get that?
 
conjugation commutes with sum and products
the determinant is just a huge sum of huge products
of the coefficients
for example in 2 by 2 matrices
det(A) = ad-bc
conjugate(det(A)) = conjugate(ad-bc) = conjugate(a)conjugate(d) - conjugate(b)conjugate(c) = det(conjugate(A))
 
got it! Thanks!
 
the Khan Academy is a great resource
2
 
11:54 AM
@ForeverInactive I received your chat messages. If you want you can visit my new youtube channel (in my profile) and there you will see my new email address. If you want you can contact me and we can talk privately there. Take care!
@Abcd I am not sure which high school textbooks are good actually. But for some high school mathematics, you may want to see Basic Mathematics by Serge Lang.
@Abcd If you want you can buy the books on cambridge.org that are endorsed by Cambridge for the GCE O level and A level exams. Those are usually more well written than other books I have come across.
 
Singapore math is very popular over at Mathematics Educators.SE
 
I don't know anything about that. What do you mean by popular?
 
a lot of high school teachers in the US use it
 
I do not care about popularity or rankings. Most rankings are superficial to me. I just judge a book, course, syllabus, or teaching method based on its own merit.
Most math teachers I have come across know very little about math, and teach their students the wrong things.
The same might be said of those who write books for school students.
 
12:09 PM
there isn't much incentive for high quality math teachers in high school
 
Yes, teachers are often overworked and underpaid.
 
hello, i need help with the integral e^(x^0.5).
Or, i see the solution in front of me, but i don't understand the substitution made there (u := x^0.5). It seems like i dont understand the concept of integration by substitution right... i thought the derivative of that what i substitute has to appear as a factor in my integrand, but this is not the case here?..
 
 
1 hour later…
1:28 PM
Can someone please help me with the proof of $A (adj (A))= \det(A)I_n$ ?
 
1:41 PM
nvm I am done.
But can someone please explain this thing: Sum of the products of the elements of any row or column with the cofactors of the corresponding elements of any other row or column is 0?
 
2:00 PM
For what kind of measure spaces $(X,\mathcal A,\mu)$ can the $L^p(X)$ spaces be finite dimensional? $p=\infty$ in particular
 
> Moreover, the Liouville numbers form a dense subset of the set of real numbers.
So does that mean the Liouville numbers are dense in the reals, or only a subset of the reals?
because the plot here suggested they are dense at where the blue lines are, but sparse elsewhere
Base 10 here for comparison
 
2:40 PM
I have to evaluate $\int_0 ^{1/2} dx/((1+x^2)\sqrt(1-x^2)) ... I substitute x=1/t to and then it simplified into a simple integral. Is there any objection or incorrect mathematics in this substitution?
 
@tatan mathematica gives the result as simply $\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}$, so the suggestion that said substitution simplifies matters seems reasonable
 
How to prove the following
Let ord(a) = n in group G, n in Naturals. prove that $ord(a^k)|n$ for any k in integers
 
Consider the subgroup generated by $a$, and remember that order of an element divides the order of the group
 
3:02 PM
Why is $\operatorname{adj(A^T)}= (\operatorname {adj} A)^T$ ?
 
if A*adj(A)=adj(A)*A=det(A)I, then upon transposing both sides you get $A^\top (\text{adj } A)^\top = (\det A)I_n = (\det A^\top)I_n$
and since the adjugate of A^top should satisfy $A^\top (\text{adj }A^\top)=\det (A^\top)I_n$, we see that they'd better agree
(not sure that's a sufficient proof tho. all it shows is that the two matrices satisfy the same equation; one has to further argue that only one such matrix exists.)
 
Still having a little bit of trouble seeing this. How to prove the following
Let ord(a) = n in group G, n in Naturals. prove that $ord(a^k)|n$ for any k in integers
So the subgroup generated by $a$ will be $e,a,a^2,...,a^{n−1}$
How do we show that $ord(a^k)|n$ ?
 
I guess one place to start: What's the largest possible value of ord(a^k) ?
 
It would be n right?
 
Right. At that point you've got (a^k)^n = (a^n)^k = e^k = e
 
3:17 PM
Yup
 
@Semiclassical What have you done? Can you please explain? Its not obvious.
 
my proof brain isn't working at the moment, but if p=ord(a^k) doesn't divide n, you'd have (a^k)^p =a^{kp} = e
and k can't divide n---it's less than n, after all, and must be a nonnegative integer
Not sure where to go from there
 
29 mins ago, by Sharath Zotis
Let ord(a) = n in group G, n in Naturals. prove that $ord(a^k)|n$ for any k in integers
@SharathZotis prove a more general theorem: if $b^m = e \in G$, then $\operatorname{ord}(b) \mid m$
an even more general theorem: the stabilizer of a group action is a subgroup
 
@Abcd start from $A(\text{adj }A)=\det(A)I_n$. Since adj(A) is proportional to the inverse of A, the two commute and you can equally well say that $(\text{adj }A)A=\det(A)I_n$.
 
(look at the action of $\Bbb Z$ on $G$ by sending $(n,b)$ to $b^n$, and deduce the first theorem I quoted, using the fact that every subgroup of $\Bbb Z$ is cyclic)
 
3:25 PM
@Leaky pedagogical tip: gauge the level of the target audience, don't try and show off to them
2
 
@Semiclassical Yes I know that then?
 
@ÍgjøgnumMeg well I'll admit that the last theorem may not be very useful
but the first theorem definitely is
 
what is cl(Liouville numbers)?
 
You can then transpose both sides to get $A^\top(\text{adj A})^\top = \det(A)I_n$
 
3:26 PM
@Semiclassical ok, then?
 
Writing $B=A^\top$, you have $B(\text{adj }A)^\top = \det(B^\top)I_n=\det(B)I_n$
oof
 
@LeakyNun Doesn't that theorem come from the fact order is defined as the least positive integer $m$ to satisfy $b^m = e$. And thus no matter what ord(b) | m ?
 
But the adjugate of $B$ should satisfy $B(\text{adj }B)=\det(B)I_n$
 
no.
 
@Semiclassical then?
 
3:28 PM
@SharathZotis what's the problem if ord(b)=7 but b^9 = e?
 
I see
 
so if there's only one matrix C such that $BC=\det(B)I_n$, then we conclude that $C=(\text{adj }A)^\top = \text{adj }B=\text{adj}(A^\top)$
 
Wait a second
We get:
$A^T \text{adj} (A^T) = A^T (\text{adj} A)^T$
 
hey guyz
 
Right.
 
3:31 PM
But we cant cancel the A^Ts
because cancellation laws do not apply to matrix multiplication
 
In general, no. But if the adjugate is going to make sense, what can you say about det(A) ?
 
det(A)= det(A^T) (thats another property btw)
 
more basic than that.
Can det(A) = 0, if we're going to have A*adj(A) = det(A)I ?
 
@Abcd do you know the other definition of determinant?
 
@LeakyNun What is that, I just know the cofactor definition ad-bc
 
3:33 PM
one of them is Leibniz, one of them is Lagrange, I can never tell which is which
 
Actually, hmm. This approach may only make sense if det(A) !=0
but the adjugate matrix is well-defined even if det(A) = 0
 
i wanted ur opinion ive asked this before. I failed my classic differential geometry class the previous semester and now i have 2 option. Take the course again or take measure theory class. What would be more beneficial. (my interests are more towards algebra). Taking a new course is always more fun but .. i dont know..
 
For which values of $n (where n ≥2) is A_n abelian$
why?
 
Anyways. If det(A) is nonzero, then so is det(A^T) and therefore A^T is invertible
 
@Semiclassical Its given that the theorem is only for invertible matrices
 
3:35 PM
in which case you can cancel it from both sides to get adj(A^T) = (adj A)^T
 
So if A is invertible, then adj(A^T)=(adj A)^T
 
Surely you must be joking Joseph Liouville, there is no way to explicitly plot all of these?
 
@SharathZotis have you tried some examples?
 
@Semiclassical Please tell me when can we cancel and when cant we in matrix multiplication
 
3:36 PM
But I think that adj(A^T)=(adj A)^T is supposed to hold even if A isn't invertible :/
 
AB = CA doesnt mean C = B
 
If I understood correctly, that union and intersection is basically saying that the Liuoville numbers are those that are arbitrarily close to the center of the band of irrationals that surrounds each rational
 
Suppose $AB=AC$. If A is invertible, then $A^{-1}$ exists and we can multiply both sides on the left by $A^{-1}$ to get $B=A^{-1}AB=A^{-1}AC=C$
 
What if AB = CA?
 
Then you can't do anything in general.
 
3:38 PM
what is squat
 
english idiom
 
But see
AB = CA
 
"you can't do squat" = " you can't do anything"
 
then also we can multipliy by A^-1
 
Sure. You'll have $B=A^{-1}CA$
 
3:39 PM
oh ya.
Okay thanks!!
 
in such cases, the matrices are said to be similar
which can be handy, but it's not as simple as just $B=C$
Similarly, if AC=BC with C invertible then A=B
So the rule is basically that, if a matrix is invertible, you can cancel it from the same side
 
yes
 
The problem is that this approach relies on A being invertible. But the adjugate of A exists even when A is singular
so this probably isn’t the best proof
 
@Abcd what is your definition of adj?
 
@LeakyNun matrix formed by the tranpose of the matrix formed by cofactors of all elements
 
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