Let $A \in M_n (\mathbb{R})$ be a matrix whose all entries are $0$ except those on the diagonals above and below the principal diagonal, which are all $1$. That is, $a_{ij} = 1$ if $|i-j|=1$ and $0$ otherwise.
Show that the eigenvalues of the matrix are symmetric around origin
I have to show the eigenvalues of this matrix are symmetric with respect to origin.
If I understand correctly, I'm required to show if $\lambda$ is an eigenvalue of $A$ then so is $-\lambda$. My idea is to somehow show $ \text{Ker}(A+\lambda I)$ is non trivial when I know $ \text{Ker} (A-\lambda I)$ is non trivial, but I'm not being able to deduce this.
Best Answer
Let $\lambda$ be an eigenvalue of $A$ corresponding to the eigenvector $v$ of $A$, then $Av=\lambda v$ where $$v=\begin{bmatrix} x_1\\ x_2\\x_3\\ x_4 \\\dots \\ x_{n-1}\\ x_n\end{bmatrix}.$$
Now take $w=\begin{bmatrix} x_1\\ -x_2\\x_3\\ -x_4 \\\dots \\ x_{n-1}\\ -x_n\end{bmatrix}$. Then $Aw=-\lambda w$ and thus the proof follows.