Linear Algebra and Eigenvalues

Bluesman01
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Suppose A is a diagonlizable nxn matrix where 1 and -1 are the only eigenvalues (algebraic multiplicity is not given). Compute A^2.

The only thing I could think to do with this question is set A=PD(P^-1) (definition of a diagonalizable matrix) and then A^2=(PD(P^-1))(PD(P^-1))=P(D^2)(P^-1)

This is how I left it on the test but I am sure this isn't right. How can you solve this without having the original matrix A or the algebraic multiplicity of the eigenvalues?
 
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Bluesman01 said:
Suppose A is a diagonlizable nxn matrix where 1 and -1 are the only eigenvalues (algebraic multiplicity is not given). Compute A^2.

The only thing I could think to do with this question is set A=PD(P^-1) (definition of a diagonalizable matrix) and then A^2=(PD(P^-1))(PD(P^-1))=P(D^2)(P^-1)
Since A is diagonalizable, and its only eigenvalues are 1 and -1, then what form must D take? Even more to the point, what does D2 have to be?
Bluesman01 said:
This is how I left it on the test but I am sure this isn't right. How can you solve this without having the original matrix A or the algebraic multiplicity of the eigenvalues?
 
But if the multiplicity of either eigenvalue is more than one, D changes form. Both P and D do. That's where I got stuck.
 
Bluesman01 said:
But if the multiplicity of either eigenvalue is more than one, D changes form. Both P and D do. That's where I got stuck.

Of course D changes form. D^2 doesn't. What is it? Once you figure out what it is, P won't matter.
 
Ah I see it now. Thanks to all who replied.
 
That would be all two of us: Dick and myself...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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