How to find eigenvectors of a nxn matrix

Question

Let $A$ be the $n\times n$ matrix with a 1 in every entry. How to find the eigenvectors and eigenvalues of $A$? We cannot find them by the ordinary method using characteristic polynomials, right? Then what shall we do?

Answer 1

HINT

Note that such matrix is a projection (with dilation) onto the one dimensional subspace spanned by $\vec v=(1,1,...,1)$.

Answer 2

The matrix has rank $1$, so its null space has dimension $n-1$. If $n>1$, this means $0$ is an eigenvalue with geometric multiplicity $n-1$, so its algebraic multiplicity is $\ge n-1$.

On the other hand, $n$ is an eigenvalue of the matrix (why?), so its algebraic multiplicity is $\ge 1$.

Therefore $0$ has algebraic multiplicity $n-1$ and $n$ has algebraic multiplicity $1$, because the sum of the algebraic multiplicities is $n$; no other eigenvalue can exist.

The characteristic polynomial is thus $\det(A-\lambda I)=(0-\lambda)^{n-1}(n-\lambda)$. (This also holds for $n=1$, of course, because $(0-\lambda)^0=1$.)