derivative of x to a matrix power.

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[mesomatics]

The last case can be written as AX^(A-I) like other cases

[birefringent]

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}

[kiranduggirala]

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.

[kono]

loving all these linear algebra vids recently!

[paosusuu]

i love it when he casually says "so that's pretty cool" about stuff i'd loudly claim to be unbelievably beautiful

[NutziHD]

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)

[demenion]

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.

[Circuito]

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!

[Keithfert]

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)

[romajimamulo]

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

[s]

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.

[goodplacetostop]

19:26 Follow-up question
19:41 Good Place To Stop

[tharunsankar]

What is the significance of a matrix power? What does it mean geometrically?

[digxx]

How do you prove that log(x^A)=A*log(x)?? I don't think that's a-priori clear for matrices?

[hgnb]

Very nice, linear algebra + calculus = math porn.

[Suprdud]

I think the way I would write the last matrix is

(λ/x)*(x^(A-λI))

[christianwolinski]

ln(x)*lambda = ln(x^lambda). How about x=1/2+3*I*Pi, lambda= 1/4+5*I*Pi.

[bradfordtaylor]

What is practical use for this. How does the need for this come about?

[vectorshift]

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.

[stanleysimon]

Bro you are much more boring than a book in which theories are complicated.Your explanations are one sided and goes over head.