Raising Matrices to power of n. (complex)

[Epsillon]

So let's say you have a matrice that has a det of 0

So X= [1 1; 1 1]

So using some verification and algebra u arive at

X^n= 2^(n-1)[1 1; 1 1]


So I am trying to define what n can be.

I already established n cannot be a - since one would get infinity for an answer since it is a singular matrix.

Another limiation is 0. Since that just gives you an identity which the formula cannot express.

But I am confused with fractions.

Technically they should work but I do not know how to explain it.
Same with irrational numbers.

so can n=1/2?




and if that is not hard enough what about imaginary numbers?


For some reason I know they work with non singular matrices .

 

[Epsillon]

anyone?

 

[alexgs]

Your formula does work for n=1/2. (If you square 2^(1/2-1) [1,1;1,1] you get back your original matrix.) In fact, the formula works for n=1/b for any positive integer b. Since you know it works for positive integers it also works for n=a/b for positive integers a/b. So it works for positive rational numbers.

For irrational and imaginary n you have to decide what it means to raise a matrix to this type of power. Hopefully someone here will have an idea.

Mathematica actually let's you raise your matrix to a positive irrational power and it gives back your formula. (It gave an error message when I tried to do imaginary powers). I don't know how to interpret this result, though.

 

[Ben Niehoff]

For imaginary (or in general, complex) powers, the definition for ordinary numbers (not matrices) is

$$\alpha^{\beta} = e^{\beta \ln \alpha}$$

For a matrix A, this is only defined if ln(A) exists. This is not always the case (in fact, I think det(A)=0 is precisely one case for which ln(A) is not defined...the determinant is something like the "magnitude" of a matrix).

Note: One can take arbitrary functions of matrices f(A), so long as f(A) can be defined as a convergent Taylor series. The matrix logarithm can be given by

$$\ln (1+A) = A - \frac12 A^2 + \frac13 A^3 - \frac14 A^4 + ...$$

which does not converge for your matrix.

 

[Epsillon]

I used MATLAB and it gave values for fractions. So I am confused there.

Is there a proof for fractions?

I want to algebriacly proove whyy fractions work. Perhaps through induction?

 

[HallsofIvy]

A proof of what for fractions? You haven't stated what it is you want to prove.


The matrix
$$A= \left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]$$
has two distinct eigenvalues and so is diagonalizable. That is, there exist a matrix P such that $P^{-1}AP= D$ where D is a diagonal matrix having the eigenvalues 0 and 1 on the main diagonal. Then $A= PDP^{-1}$ and
$$A^n= (PDP^{-1})(PDP^{-1})(PDP^{-1})\cdot\cdot\cdot(PDP^{-1})= PD^nP^{-1}$$
where Dⁿ is easy- it is the diagonal matrix with 0 and 2ⁿ on the main diagonal.

Specifically, an eigenvector corresponding to the eigenvalue 0 is <1, -1> and an eigenvector corresponding to the eigenvalue 1 is <1, 1> so we can take
$$P= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]$$
and then
$$P^{-1}= \left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right]$$

That is
$$\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right]$$

and so
$$\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]^n= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2^n\end{array}\right]\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right]$$


You can show that the formula works for 1/n as well by reversing it:
$$\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2^{1/n}\end{array}\right]^n\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right]$$

 

[Epsillon]

). I basicly want to prove that when you have a singular matrix X. And you raise it to the power of 0<n<1 one can yield results.

Thanks for the exmplanation I get how to do it now. But I need to research more on eigenvalues and eigenvectors since in the course I am in currently we do not learn about those key properties.

Couple more things:
Without using eigen values can you simply prove that by saying
X$$^{1/n}$$=M$$^{n}$$
Where M is the result?

and I gota questions can you also use taylor series to explain why irrational numbers work?

Or perhaps like a pi expansion where you put fractions for n for your required precesion?


An why exactly doesn't ln of X work when logs work.

 

[HallsofIvy]

If X^1/n= M then Mⁿ= X, not "X^1/n= Mⁿ".

I don't know what you mean by "why exactly doesn't ln of X work when logs work. "

ln (the natural logarithm) works whenever any log "works". log_a(X)= ln(X)/ln(a).