number^matrix plz help me on this, I cant understand this, plz help me

number^matrix example a=[1 2;3 4] is a 2x2 matrix 3^a= [87.8864,127.1198;190.6797,278.5661] how come these numbers? [87.8864,127.1198;190.6797,278.5661]

plz help me on this algorithm of calculation these numbers...

Answer 1

[Teja Muppirala]

If the exponent was a scalar, say a=2, then :

a = 2;
3^2 = 9 = exp(2 * log(3))

If you define z = 2*log(3), then

3^2 = 9 = exp(z) = z^0 + z^1 + z^2/2 + z^3/6 + z^4/24 + z^5/120 + ...

The matrix exponential can be defined in the same way

a = [1 2; 3 4]
3^a = exp([1 2; 3 4] * log(3))

If you define z = a*log(3), then

3^a = z^0 + z^1 + z^2/2 + z^3/6 + z^4/24 + z^5/120 + ...

where a and z are now 2x2 matrices.

This infinite sum will converge to the answer. The actual method MATLAB uses is somewhat different (it doesn't use the Taylor series, it uses a Pade approximation) but the general result is the same.

[Jan]

@Teja: Where did you find this information?

[Teja Muppirala]

I remember learning this in a control systems class back in graduate school. The matrix exponential comes up as part of the general solution to a state space system.

dx/dt = A*x + B*u

y = C*x + D*u

--> y(t) = (some long expression with e^(At) in it somewhere)

Answer 2

[Jan]

The documentation explains it: "Z = x^Y is x to the Y power [...]. Computed using eigenvalues and eigenvectors."

Well. Hm. Everything clear now?

[EDITED] See this link: http://freemat.sourceforge.net/help/operators_matrixpower.html. But here some strange limitations are embedded: "(problems arise with non-symmetric matrices b, so let us assume that b is symmetric)" - without any further explanations.

Answer 3

[John D'Errico]

Start with the (updated) classic reference.

They did a better job of explaining it all than I will do in a short space anyway.

Answer 4

[aliya]

no my dear, I know that. my purpose is 3^a....

Answer 5

[aliya]

but how? how Computed using eigenvalues and eigenvectors? what is this algorithm?

[Jan]

+1 for this fair question. Matlab's documentation reminds me to a German commercial: "Contains the best of a 1/4 liter of milk" - when pure fat is meant.

20 minutes with Google did not increase my knowledge about a^B, but I've learnt something about "quantum healing".

Answer 6

[Honglei Chen]

Hi ailya,

As Teja mentioned above, the key thing to understand is a^x can be expanded using Taylor series. The second piece that the matrix power can be carried out using eigen-decomposition. For example, assume that

A = EDE^(-1)

where E is the eigenvector and D is the diagonal matrix of eigenvalues, then the power of A can be written as

A^2 = EDE^(-1)*EDE^(-1) = ED^2E^(-1)

Similar derivation can show that

A^n = ED^nE^(-1)

So putting both pieces together, you get

3^A = e^(A*ln3) = 1+A*ln3+A^2*(ln3)^2+... = 1+ln3*EDE^(-1) + (ln3)^2*ED^2E^(-1)...

HTH,

Honglei

Answer 7

[aliya]

thanks to all of you. and sorry Honglei Chen, I couldn't accept two answer, but your answer was nice too. thanks teja Muppirala thanks to all of you again.

Answer 8

[aliya]

but I try this 3^a = exp([1 2; 3 4] * log(3)) and the answer was 3.0000 9.0000 27.0000 81.0000 and was not equal to 3^a answer! why?

[Jan]

EXP works elementwise. What happens with EXPM? Just a guess - I cannot test this current because the Matlab computer is shut down already.