How do I show that $1$ is not an eigenvalue for $A$, by showing that there are no eigenvectors for $\lambda = 1$.

Question

Consider the matrix $$A=\begin{pmatrix}-1 & 3& 3& 3\\ 3& 1& -1& 5\\ 3& -1& 7& -1\\ 3&5& -1&1\end{pmatrix}.$$

How do I show that $1$ is not an eigenvalue for $A$, by showing that there are no eigenvectors for $\lambda = 1$?

This what I've done so far:

$$\left ( A - \lambda I \right )\vec{x} = \vec{0},$$

$$A-I =\begin{pmatrix}-2 & 3& 3& 3\\ 3& 0& -1& 5\\ 3& -1& 6& -1\\ 3&5& -1&0\end{pmatrix}.$$

I have do the Gaussian elimination to find the solution.

This is what my augmented matrix looks like in row-reduction form:

$$\begin{pmatrix}-1 & 3& 2& 8\\ 3& 0& -1& 5\\ 0& -7& -5& -0\\ 0&1& 0&-1\end{pmatrix}.$$

What do I do from here?

Answer 1

Showing that $1$ is not an eigenvalue of $A$ is the same as showing that the matrix $A - I$ is invertible.

You can do this by checking that its determinant is nonzero.

Alternatively, you can perform row reduction starting with $A - I$. Once you arrive at an upper triangular matrix with no zeros on the diagonal, you can be certain that $A - I$ is not invertible.

The matrix you wrote is not nearly in row-echelon form, because of the nonzero entries in the $(2,1)$-, $(3,2)$- and $(4,2)$-positions.

Answer 2

A nice trick to show that $\lambda=1$ is not an eigenvalue of $A$ is to consider that over $\mathbb{F}_2$: $$\det(A-I) =\det\begin{pmatrix}0 & 1& 1& 1\\ 1& 0& 1& 1\\ 1& 1& 0& 1\\ 1&1& 1&0\end{pmatrix}=1,$$ hence the determinant of $A-I$ is odd, so it is for sure different from zero.

As a curiosity, this is related with the fact the the number of derangements (permutations without fixed points) in $S_{2n}$ is always odd.