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.
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 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 which, if D is the identity matrix, which is certainly possible since the three eigenvalues are "1", then .
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"
if the eigenspace is of dimension 1 or
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
Of course you would still have the problem that, since these are NOT the identity matrix, and will not cancel so the will still depend on the eigenvectors.