# Eigenvalues of a complex $4 \times 4$ Matrix

Let $A\in \mathbb{C}^{4×4}$. We are given the equation

$$A^{4}+12A^{2}=6A^{3}+8A$$

and

$$\textrm{rank}(A)=2\cdot\textrm{rank}(A-2I_{4})=4$$

I already found out that $2$ is an Eigenvalue, but how do I determine the other ones ? I'm mainly looking for hints.

• Is it $\textrm{rank}(A)=2\times\textrm{rank}(A-2I_{4})=4$ or $\textrm{rank}(A)=2$ and $\textrm{rank}(A-2I_{4})=4$? – Bill O'Haran May 15 '18 at 12:11
• Your last equality cannot be , since it is clear zero is one of the eigenvalues and thus $\;A\;$ cannot have full rank (in this case, full rank$\,=4\,$ )...or directly one can see $\;A\;$ cannot be invertible as it is a zero divisor... – DonAntonio May 15 '18 at 12:13
• @ChristianSinger Then the problem is exactly wrong. – DonAntonio May 15 '18 at 12:19
• I see your point. Do you think I should delete this question or keep it because the info was nonetheless helpful to me in some way (and maybe for others aswell) – Christian Singer May 15 '18 at 12:25
• @ChristianSinger Keep it, by all means! As you say, even as the question seems to be wrong, both comments and answers help fpr better understanding certain features. – DonAntonio May 15 '18 at 12:32

The polynomial $x^4-6x^3+12x^2-8x = x(x-2)^3$ annihilates the matrix $A$ so $\sigma(A) \subseteq \{0,2\}$.

If $\sigma(A) = \{0\}$ then $\operatorname{rank}(A-2I) = 4$ so $\operatorname{rank}(A) = 2\operatorname{rank}(A-2I) = 8 > 4$ which is a contradiction.

Assume $\sigma(A) = \{0,2\}$.

If $\operatorname{rank}(A) = 0$ then $A = 0$ so $\sigma(A) = \{0\}$, a contradiction.

If $\operatorname{rank}(A) = 2$ then $\operatorname{rank}(A-2I) = 1$ so $$\operatorname{null}(A) + \operatorname{null}(A - 2I) = 2 + 3 = 5 > 4$$

If $\operatorname{rank}(A) = 4$, then $0 \notin \sigma(A)$, a contradiction.
Therefore, the only option is $\sigma(A) = \{2\}$, which implies $\operatorname{rank}(A) = 4$ and $\operatorname{rank}(A-2I) = 2$. An example of such a matrix is
$$A = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix}$$
• @ChristianSinger The spectrum i.e. the set of all eigenvalues of $A$. – mechanodroid May 15 '18 at 12:37
• @ChristianSinger It just means that $A^4-6A^3+12A^2-8A = 0$. If a polynomial $p$ annihilates a matrix (i.e. $p(A) = 0$), then the minimal polynomial of $A$ divides $p$, so $$\sigma(A) = \{\text{ zeros of } m_A\} \subseteq \{\text{ zeros of } p\}$$ – mechanodroid May 15 '18 at 12:46
• @ChristianSinger You can also see it like this: assume $p(A) = 0$ and take an eigenvalue $\lambda$ of $A$. There exists $x \ne 0$ such that $Ax = \lambda x$. Then $$0 = p(A)x = p(\lambda)x \implies p(\lambda) = 0$$ so $\lambda$ is necessarily a zero of $p$. – mechanodroid May 15 '18 at 12:51
• The given example Jordan form for $A$ has minimal polynomial $(x-2)^3$. Another dissimilar example would be $A=\left(\begin{smallmatrix}2&1&0&0\\0&2&0&0\\0&0&2&1\\0&0&0&2\end{smallmatrix}\right)$ with minimal polynomial $(x-2)^2$. – ccorn May 16 '18 at 10:08