derivative of x to a matrix power.
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- เผยแพร่เมื่อ 28 ก.ย. 2024
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The last case can be written as AX^(A-I) like other cases
Yeah, I was looking at it like "what would you naturally expect here?" and calculated Ax^(A-I) and it was right. Surprised Michael didn't pick up on that.
I was waiting the entire video to see if the pattern holds, glad it does.
Can we conclude based on these cases that derivative of any matrix A is always Ax^(A-I) ?
It would be amazing if you'd continue with the differential forms series, you didn't get to explain the generalised stokes theorem and it would be awesome, that series helped me a lot, guys like this comment to make it visible to Dr. Penn!
Your editor seems to be leaving in a lot more cases where you said the wrong thing and then jump cutting to you correcting yourself rather than cutting out you saying the wrong thing
19:26 Follow-up question
19:41 Good Place To Stop
Very cool!
How do you prove that log(x^A)=A*log(x)?? I don't think that's a-priori clear for matrices?
I think he's just defining x^A when A is a matrix. It's clear how to define e^A thanks to the power series, and you can exploit this definition since, for real numbers, x^a = e^(a*log(x)). By analogy, you can define x^A to be e^(A*log(x)).
ln(x)*lambda = ln(x^lambda). How about x=1/2+3*I*Pi, lambda= 1/4+5*I*Pi.
Wolfram Alpha says that the real parts of those answers are the same, but the imaginary parts are different, so these are not equal, but that is because 1 or both of x and lambda are complex in your example, but I assume that Micheal was assuming that both x and lambda are only real numbers.
it feels like a magic
Its half-about
Matematik güzel şey
In a sense it is really - there is a start point and a hoped for, nay intended end point sooo ... how to make it fit?
In a sense it is a bit like epsilon-delta-N-n stuff.
We know the start, sense the end point then fudge about in the middle to justify both?
And maybe (?) (!) that is the job of abstract algebra. To show why something works and why it should work most of the time, what pitfalls there are and identify real or imaginary humongous monsters.
For example what effect does chaos theory and minor perturbations have on matrix calculations?
@@Alan-zf2tt Well, consider the fact that all scalar values are 1x1 Matrices and start from there since a 1x1 matrix is still a square matrix. Food for thought!
I would do d/dx x^A = d/dx exp(A log x) and then using the well known formula for derivative of a matrix exponential, d/dt exp(A t) = A exp(A t) you get d/dx x^A = (1/x) A exp(A log x) = (1/x) A x^A = A x^{A-1}
Is that last matrix for the non diagonalizable case not just Ax^(A-I) as well? I did the matrix multiplication in my head so it might be wrong but it seemed to check out.
Good point
Interesting. If true then that would correspond nicely to the usual power rule, except that the exponent is a matrix rather than a number.
@@Jono98806 Yup. Another commenter has commented the bigger idea. If you show that a) The derivative exists, is well defined, and is continuous for an arbitrary matrix (this is not difficult from the definition of the derivative) and b) On a dense set of matrices (the diagonalizable ones), it has the formula d/dx(x^A) = Ax^(A-I), then the formula works everywhere.
The derivative of x^A is straight forward, if A is diagonizable (as you showed). Since d/dx x^A is continous and almost all matrices are diagonizable (over C), this proves d/dx x^A = A x^{A-I} for all matrices A (in arbitrary dimension)
Turns out that not all matrices are diagonizable even over C since the eigenspaces may be degenerate, which means 1 dimensional space for a 2 fold eigenvalue
@@Idk-hp3oo "almost all" means the subset of diagonalisable matrices is dense in the set of matrices. Therefore by the argument of the continuity of d/dx, the conclusion follows
This is where my mind went first.
@@panadrame3928is continuity enough here or would you need an identy principle of holomorphic functions?
how do you know d/dx x^A is continuous wrt A? It seems true but non-trivial to prove.
loving all these linear algebra vids recently!
Very nice, linear algebra + calculus = math porn.
Or in other words, calc 3.
There's no need to split into cases as long as you differentiate term wise in the definition of the exponential (in the following, d represents derivative with respect to x):
d(x^A)=d sum_{n=0}^infty A^n [ln(x)]^n / n!
d(x^A)=sum_{n=0}^infty A^n n [ln(x)]^(n-1) / [n! x]
d(x^A)=(A/x) sum_{n=0}^infty A^n [ln(x)]^n / n!
d(x^A)=(A/x)*x^A
d(x^A)=Ax^(A-I)
This is how I did it too
only if we can justify the change of order between differentiation and limit.
@@ranshen1486 but you can justify it because the sum converges absolutely
Have you been assuming that the eigenvalues do not equal to 0 for the diagonalizable cases? Those don't seem to follow the same power rules.
What do you mean? Using the power rule for the exponent 0 we get d/dx x^0 = 0x^(-1) = 0 which is correct.
for diagonal matrices you can just use that f(A) is the same as the diagonal matrix A where f is applied to each entry of the diagonal.
you can also use the Cayley-Hamilton-Theorem again in the form i mentioned in your last video to calculate matrix functions very easily.
This guy sucks. He has over crossing lines and fragmented derivation paths going from one "step" to the next in his "jump about " explanation of what he's doing. AND he stands in front of the board while ha writes out and "explains" what he's writing down.
What is the significance of a matrix power? What does it mean geometrically?
Great video
Very interesting sir
i love it when he casually says "so that's pretty cool" about stuff i'd loudly claim to be unbelievably beautiful
I think the way I would write the last matrix is
(λ/x)*(x^(A-λI))
What is practical use for this. How does the need for this come about?
I think you're on the wrong channel, dude
Bro you are much more boring than a book in which theories are complicated.Your explanations are one sided and goes over head.