Functions of matrices

>This is nice thank you. I also would like to show it converges to said \sqrt{I+A}

Yes, that might be the harder part. I would suggest writing out N+1 terms of the series for sqrt(1+x) to get a polynomial of degree N, call it p(x). Calculate p(x)^2-(x+1), get something like x^N*(some small thing). if ||A||

By the way, I'm not sure sqrt(I+A) is well-defined, i.e. there may be many square roots.

Do you have special class of matrices? Symmetric would be easy.

is f applied element-wise?

Ok I get the point I will try this. Also where does idea come from that for any powerseries I only have to switch in the matrix. And what about derivatives. Why am I not taking derivatives of the matrix?

A is symmetric. But I am trying to get the general idea of matrix power series also.

not necessarily. Only if the definition requires it to be (which it does only for diagonal matrices)

the function doesn't even have to do anything with the problem. I mean all this is saying is that if you have a convergent power series, then it is also convergent when considered as a matrix power series.

>not necessarily. Only if the definition requires it to be (which it does only for diagonal matrices)

this is a confusing question. if f is function of more than one variable, your ith partial derivatives will have different dimensionality from the (i+1)th partial derivatives.

i think this guy got it and now you're just fucking with us.

OP here

Sorry if I have been unclear.
1. I want to understand and proof (if this works) that I can do a taylorseries of a matrix function.
2. Also I don't know why I am not taking derivatives. ( 2 might be answered by 1)
3. I want to solve the particular problem (which I assume to be straight forward then)

en.wikipedia.org/wiki/Matrix_function

I understand why it is convergent, as I have said in

so basically you're asking if there is a version of taylor's theorem for matrix-valued functions of a single real variable? that's a very different question that what's in the first post..

> matrix-valued functions of a single real variable
[math] C^{n x n} \rightarrow C^{n x n} [/math] i believe. at least that's how i saw it being defined. I.E. Mapping a matrix to a matrix. In the specific problem [math] f(A) = \sqrt{I + A}

[math] f(A) = \sqrt{I + A}[/math]

this doesn't even make sense. a matrix can have several or none square roots, so the square root is not a well defined function.

would it be well defined if [math] (I + A) = B*B^T[/math] (looking for the B)

It's easy for the particular case of a symmetric matrix, because it is diagonalizable. Write out finitely many terms of the series, factor out the diagonalizing matrix and it's inverse on the left and right.. you essentially are applying the series term by term to the diagonal elements of a diagonal matrix. This will also give an easy way to show that the result B satisfies B^2 = I + A.

In general it is harder. You have a series for some analytic function f. Plug in A in the series to get f(A), and you can show that this is well-defined because the series converges for a certain class of A's.

But whether or not f(A) has the properties you expect, based on properties of f(x) for real or complex x is another question. There is also the question of uniqueness. If D is real NxN, diagonal, with positive elements on the diagonal, then there are at least 2^N real diagonal matrices B that satisfy B^2 = D. You can choose the + or the - square root for each diagonal element.

>1. I want to understand and proof (if this works) that I can do a taylorseries of a matrix function.

The OP pic explains exactly how to do that.

>2. Also I don't know why I am not taking derivatives. ( 2 might be answered by 1)

[math]f(0),f'(0),f''(0),...[/math] are just scalars. You get them the same way as any other Taylor series.

>3. I want to solve the particular problem (which I assume to be straight forward then)

Finding the Taylor series of ? Just take the real-valued Taylor series and replace your input variable with a matrix.

>But whether or not f(A) has the properties you expect ... is another question
how do I answer this ? the covergence for ||A|| < r part is solved and understood.

>The OP pic explains exactly how to do that.
I can see why it converges. But I don't understand how this explains that it has the same properties as the f(x) case, i.e. "proves this shit works".

>You get them the same way as any other Taylor series.
There are multidim. taylor series where I take derivatives, why am I not taking derivative w.r.t. matrix A (or am I) and how would they even turn out to be scalars?

the real question is: does the function [math]f(A) = \sqrt{A}[/math], defined by the taylor expansion of the real-valued function, have the property [math]f(A)f(A) = A[/math] ?

Yes and I would be interested in why it does. Then I could understand how this concept was developed.

if f is a function of several variables, then there will be more (i+1)th partial derivatives than ith partial derivatives so the dimensionality doesn't agree for the sum or the products.

if f is originally a function of a single variable, then how does it apply to a matrix if not elementwise?

the wiki article is confusing.

it would make more sense if they said

g(A) = f(0) + f'(0)A + f''(0)A^2/2! + ....

which i think is what they mean. the language they use is confusing.

i have watched this lecture, which describes some of the stuff in the wiki article: youtube.com/watch?v=UXWMYr0LQAk

Still not sure how to approach this. I suppose using jordan decomposition it could work.
I don't understand why we don't use the function on the rotation matrices.

i also think that this poster has the correct interpretation of the statement.

f(x) and the matrix function are defined as a taylor series, so of course the taylor approximation would converge to the actual value of the function.

this poster i meant:
>

why is it that would converge to the same value ?

from the wiki
>If the real series converges for |x| < r,

if your interpretation is correct, why would they need to say this when they defined f by its taylor series?

in the article they say that f(x) = f(0) + f'(0)x + f''(0)x^2/x! +...

so the function f is equal to its taylor series

>f(x) and the matrix function are defined as a taylor series, so of course the taylor approximation would converge to the actual value of the function.

maybe I am missing something, but this is not an explanation.

or equal to its taylor expansion, i should say

but for which input space ? i assume i could not input a vector for example. they claim that it works with a matrix and that it converges for ||A||

So, like, on a physical observable level what is the Taylor serious used for?

this one is used for estimating the squareroot of a matrix by taylor approximation. Computing the general formula upfront saves reduces the O(n^3) complexity to O(n^2)

>estimating the squareroot of a matrix

Well it's a start. Thank you.