i.e., p(lambda) = (1 - lamba)^3. Find A^25.
It was kind of my instinct to choose [1, 0, 0; 0, 1, 0; 0, 0, 1] as my answer, but I'm not sure the proper way of going about it.
Hey biggerleaffan.
You need to know the eigen-vectors in addition to the eigen-values to know the original matrix.
The eigen-values give the information about how the matrix scales a vector with respect to some eigen-value but it doesn't give the direction.
If the matrix A were diagonalizable then we could write $\displaystyle A= B^{-1}DB$ where D is the diagonal matrix having the eigenvalues of A along its diagonal and B is the matrix havig the eigenvectors of A as columns. That is why we cannot determine A itself from its eigenvalues alone- we need to know the corresponding eigenvectors also.
However, whatever B is, in that case we can write $\displaystyle A^{25}= (B^{-1}DB)^{25}= (B^{-1}AB)(B^{-1}AB)...(B^{-1}AB)= B^{-1}D^{25}B$ which, if D is the identity matrix, which is certainly possible since the three eigenvalues are "1", then $\displaystyle A^{25}= B^{-1}IB= B^{-1}B= I$.
However this is, as I said, "If the matrix A were diagonalizable". Since 1 is a triple eigenvalue it is possible that there do NOT exist three independent eigenvectors. The "eigenspace" may be a subspace of dimension 1 or a subspace of dimension 2 in which we cannot diagonalize A. The best we could do would be the "Jordan Normal Form"
$\displaystyle \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$
if the eigenspace is of dimension 1 or
$\displaystyle \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
if the eigenspace is of dimension 2.
And finding powers of those is much harder.
Added: Well, not that much harder! A little calculation shows that
$\displaystyle \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n= \begin{bmatrix}1 & n & 1 \\ 0 & 1 & n \\ 0 & 0 & 1\end{bmatrix}$
and
$\displaystyle \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}^n= \begin{bmatrix}1 & n & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
Of course you would still have the problem that, since these are NOT the identity matrix, $\displaystyle B$ and $\displaystyle B^{-1}$ will not cancel so the $\displaystyle A^n$ will still depend on the eigenvectors.