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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

Redeldio
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  • Duplicate is not the same linear combination of identity and transpose as in this question, but there is no point in having separate questions for every possible linear combination of them. My answer there shows how to handle general such linear combination in general. – Marc van Leeuwen May 09 '18 at 04:47

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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. $n=4$

QuantumEyedea
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  • Thank You for the answer. I note that, if we take the determinant of both members of the equation $\Lambda^T=\frac{\lambda-1}{3}\Lambda$, we get that $\left(\left(\frac{\lambda-1}{3}\right)^4-1\right)\operatorname{det}\Lambda=0$, which is satisfied for $\lambda=-2, 4$ but also for $\lambda=1\pm3i$. I do not understand how to exclude there two last eigenvalues.. – Redeldio May 09 '18 at 09:58
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    Let $\Lambda^{T} = \frac{\lambda - 1}{3} \Lambda$ be Equation (A) and let $\left( \left[ \frac{\lambda - 1}{3} \right]^2 - 1 \right)\det(\Lambda) =0$ be Equation (B). It is (A) that must hold: you get (B) only by taking the determinant of (A), so (B) is just an equation that must necessarily hold as a result of (A). You'll find that if you plug $\lambda = 1 \pm 3 i$ into (A) that there exist no solutions $\Lambda$ so in this sense the Eigenspaces $\mathbb{E}_{1 \pm 3 i}$ are empty. – QuantumEyedea May 10 '18 at 20:23