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 $\lambda = 2.$
$$ \begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ \end{bmatrix} $$
The matrix is a unitary conjugate (via permutations) of $$\begin{bmatrix}1&1&0&0\\1&1&0&0\\0&0&1&1\\0&0&1&1\end{bmatrix} $$(concretely, use the permutation corresponding to the exchange of the second and third elements of the canonical basis). So the characteristic polynomial is the square of the characteristic polynomial of the $2\times2$ matrix with all ones, which has eigenvalues $0$ and $2$. We get, then, $$\lambda ^2 (2-\lambda)^2. $$
This matrix has a simple block form $$ \begin{pmatrix}I&I\\I&I\end{pmatrix}$$ where $I$ is the $2\times 2$ identity, so you can eyeball the eigenvalues of the $2\times 2$ all-ones matrix (which are $2$ and $0$) and then realize that they will both contribute twice since each eigenvector of this matrix ($(1,1)$ and $(1,-1))$ will correspond to a two-dimensional invariant subspace for the full $4\times 4$ matrix ((a,b,a,b) and $(a,b,-a,-b))$
By inspection we see that $A (1,1,1,1)^T = 2 (1,1,1,1)^T $.
Similarly, $A (1,0,-1,0)^T = 0 $, $A (0,1,0,-1)^T = 0 $ and $A(1,-1,1,-1)^T = 2 ((1,-1,1,-1)^T$.
All four vectors are linearly independent, hence constitute a basis of eigenvectors.
Denote $e_1 = \begin{bmatrix}1\\0\end{bmatrix}$ and $e_2 = \begin{bmatrix}0\\1\end{bmatrix}$.
Let the matrix in question be $A$.
Observe that $A = \begin{bmatrix}I&I\\I&I\end{bmatrix}$, where $I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$.
Then, $A$ is similar to $B = \begin{bmatrix}1&1\\1&1\end{bmatrix}$, whose eigenvalues are easily found by the determinant method:
$$\begin{array}{rcl} \det(B-\lambda I) &=& 0 \\ (1-\lambda)^2 - 1 &=& 0 \\ \lambda^2 - 2\lambda &=& 0 \\ \lambda(\lambda - 2) &=& 0 \\ \end{array}$$
Or from inspection: $B\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix}$ and $B\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$.
From that, we easily find two eigenvalues: $A \begin{bmatrix}\vec v\\\vec v\end{bmatrix} = \begin{bmatrix}2\vec v\\2\vec v\end{bmatrix}$ and $A \begin{bmatrix}\vec v\\-\vec v\end{bmatrix} = \begin{bmatrix}\vec 0\\\vec 0\end{bmatrix}$.
Each eigenvalue is with multiplicity $2$, as $\vec v$ is a vector of $2$ dimensions.
The first two columns are obviously linearly independent, while the last two columns are duplicates of the first, so the nullity of this matrix is 2, which means that it has $0$ as an eigenvalue of multiplicity two. The row sums all equal $2$, so that’s another eigenvalue with associated eigenvector $(1,1,1,1)^T$ (right-multiplying a matrix by a vector of all 1’s sums its rows). The last eigenvalue can always be found “for free:” the trace of a matrix is equal to the sum of its eigenvalues. The trace of this matrix is equal to $4$, therefore the fourth eigenvalue is $4-0-0-2=2$.
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,
$$(1-\lambda)((1-\lambda)^3+\lambda-1)-(1-\lambda)^2+1=l^4-4l^3+4l^2.$$