Eigenvalue by inspection?

Question

Can I guess the eigenvalues of a $3\times3$ matrix having all entries $1$? for e.g., consider the matrix

$ \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right)$

Answer 1

Yes, it has eigenvalues $3$ and $0$.

One eigenvector is $x=\begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix}$ - it is actually the transpose of one of the rows of the matrix. The associated eigenvalue is $3$.

Any vector $\ne 0$ orthogonal to $x$ is an eigenvector to the eigenvalue $0$.

Answer 2

You see that the kernel is two-dimensional, so you have that 0 is an eigenvalue with multiplicity two. Moreover if you act on the vector $(1,1,1)^t$ you obtain $(3,3,3)^t$, so 3 is the other eigenvalue.