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Let $A$ be a $4 \times 4$ matrix and let $\dim \mathcal N (A) = 2$. What can you tell about the eigenvalues of $A$?

Rodrigo de Azevedo
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ProJaqf
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  • Welcome to Mathematics Stack Exchange. Two of them are $0$ – J. W. Tanner May 31 '20 at 15:59
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    As $A$ isn't invertible surely $0$ is a double eigenvalue of $A$, see https://math.stackexchange.com/questions/503585/show-that-a-matrix-a-is-singular-if-and-only-if-0-is-an-eigenvalue/503644#503644. – Michael Hoppe May 31 '20 at 16:01

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As $\dim(N(A))=2$, $0$ is an eigenvalue with multiplicity 2 in $A$'s characteristic polinomial $\chi_A$ then $\chi_A=x^2p(x)$.

Now, if $A\in \mathbb{R}^{4\times 4}$ $\chi_A$'s coefficients are real so if $p(x)=(x-\lambda)(x-\mu)$ then either $\lambda,\mu\in \mathbb{R}$ or if $\lambda\in \mathbb{C}\setminus \mathbb{R} \implies \mu =\overline{\lambda}$. In the former case you have two eigenvalues which may be equal or not. In the latter $0$ is the only real eigenvalue of the matrix.

if $A\in \mathbb{C}^{4\times 4}$ then $\chi_A=x^2(x-\lambda)(x-\mu)$ with any $\lambda,\mu \in \mathbb{C}$. Then you have two more eigenvalues which may be equal.

N. Pullbacki
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  • You can’t conclude that the algebraic multiplicity of $0$ is equal to $2$, only that it is at least $2$. Consider, for example, $\tiny{\pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}}$ and $\tiny{\pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}}$, both of which satisfy the conditions of the problem and have characteristic polynomial $\lambda^4$. – amd Jun 01 '20 at 00:27
  • Thanks, I think I fixed it. – N. Pullbacki Jun 01 '20 at 00:30