Find eigenvalues and eigenspaces of a linear map in $\mathbb{C}_{4\times4}$

Question

Let $f:\mathbb{C}_{4\times4}\longrightarrow\mathbb{C}_{4\times4}$ be the linear map defined by $f(A)=A+3A^T$. Determine eigenvalues, the corresponding eigenspaces and their dimension.

I cannot start this exercise. Some helps?

Thank You

Answer 1

Suppose $\lambda \in \mathbb{C}$ is an eigenvalue of $\ f:\mathbb{C}_{4 \times 4} \to \mathbb{C}_{4 \times 4}$, with corresponding eigenvector $\Lambda \in \mathbb{C}_{4 \times 4}$. Then we have $f(\Lambda) = \lambda \Lambda$. By the definition of $f$ this means: $$ \Lambda + 3 \Lambda^{T} = \lambda \Lambda $$

Re-arranging the above equation, we get $\Lambda^{T} = \frac{\lambda - 1}{3} \Lambda$.

In this suggestive form, we notice that $\lambda=4$ yields $\Lambda^{T} = \Lambda$, which is satisfied when $\Lambda$ is a symmetric matrix. We also notice that $\lambda = -2$ yields $\Lambda^{T} = - \Lambda$, which is satisfied when $\Lambda$ is a skew-symmetric matrix.

So summarizing there are two eigenvalues, $\lambda \in \left\{ 4, -2 \right\}$. We have corresponding eigenspaces:

$\mathbb{E}_{4} = \left\{ \Lambda \in \mathbb{C}_{4\times 4} | \Lambda^{T} =\Lambda\right\}$ where $\dim\left( \mathbb{E}_{4} \right) = \frac{n(n+1)}{2}$

$\mathbb{E}_{-2} = \left\{ \Lambda \in \mathbb{C}_{4\times 4} | \Lambda^{T} =- \Lambda\right\}$ where $\dim\left( \mathbb{E}_{4} \right) = \frac{n(n-1)}{2}$

See this post about the dimensions of the space of symmetric and skew-symmetric matrices. Notice also that $\frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2$, which means that $\mathbb{E}_{4} \oplus \mathbb{E}_{-2} = \mathbb{C}_{4\times 4}$ and so we're done.

P.S. $n=4$