TITLE: Finding eigenvalues shortcut
QUESTION [5 upvotes]: I am being asked to find the eigenvalues for this matrix. It mentions that some tricks can be used instead of having to use $det(A-\lambda I)$.
I understand how to do it that way, but what is a shortcut I can use for this matrix? Thanks.
Solution is $(\lambda -2)^2 \lambda ^ 2$
$\lambda = 0$ and $\lambda = 2.$
$$
        \begin{bmatrix}
        1 & 0 & 1 & 0\\
        0 & 1 & 0 & 1\\
        1 & 0 & 1 & 0\\
        0 & 1 & 0 & 1\\
        \end{bmatrix}
$$

REPLY [2 votes]: Developing $|A-\lambda I|$, on the first row you get
$$
        (1-\lambda)\begin{vmatrix}
        1-\lambda & 0 & 1\\
        0 & 1-\lambda & 0\\
        1 & 0 & 1-\lambda\\
        \end{vmatrix}+
        \begin{vmatrix}
        0 & 1-\lambda & 1\\
        1 & 0 & 0\\
        0 & 1 & 1-\lambda\\
        \end{vmatrix}.
$$
Then developing again,
$$(1-\lambda)((1-\lambda)^3+\lambda-1)-(1-\lambda)^2+1=l^4-4l^3+4l^2.$$