For $a\in \mathbb{R}$ let $A\in \mathbb{R}^{4\times 4}$ the matrix \begin{equation*}A=\begin{pmatrix}2 & 2 & 3 & 4 \\ 1 & 3 & 3 & 4 \\ 1 & 2 & 4 & 4 \\ a & 2a & 3a & 4a+1\end{pmatrix}\end{equation*}
Let $\Phi=\Phi_A:\mathbb{R}^4\rightarrow \mathbb{R}^4$ the endomorphism of $\mathbb{R}^4$, that is given by $v\mapsto \Phi (v):=A\cdot v$.

I want to show that $1$ is an eigenvalue of $\Phi$ and to determine the dimension of the corresponding eigenspace $\text{Eig}(\Phi, 1)$.

To show that $1$ is an eigenvalue do we have to show that $\det (\Phi -1\cdot I)=0$ ?

If $A$ is a $5 \times 5$ invertible matrix and that $A^{-4} = 3A^T$, compute for $\det \, A$.

What I did was since $A$ is invertible, then $\det A^{-1} = \frac{1}{\det A}$ and since $\det (AB) = \det A \det B$ this should give

$$
\begin{align}
\det \left(A^{-4} \right) & = \det \left(3 A^T \right) \\
\left(\det A^{-1}\right)^4 & = 3^5 \det A^T \\
\left(\frac{1}{\det A} \right)^4 & = 3^5 \det A^T \\
1 & = 3^5 \left(\det A \right)^5 \\
1 & = \det A
\end{align}
$$

wherein $\det A = \det A^T$ in the second to the last line. I'm not entirely sure if what I did was correct, particularly in the…