I conjecture the following:

Let $U \subset \Bbb C^{n \times n}$ be an affine subspace, and let $S_U$ denote the "spectrum of $U$", that is $$ S_U = \{\lambda \in \Bbb C : \det(A - \lambda I) = 0 \text{ for some } A \in U\}. $$ Then either all elements of $U$ have an identical spectrum, or $S_U = \Bbb C$.

Is this correct? Some simple examples of each case: for any fixed $\lambda_i$, $$ U_1 = \left\{\pmatrix{\lambda_1&t\\0&\lambda_2}: t \in \Bbb C\right\}, \quad U_2 = \left\{\pmatrix{\lambda_1&0\\0&t}: t \in \Bbb C\right\}. $$ Clearly, $S_{U_1} = \{\lambda_1,\lambda_2\}$ and $S_{U_2} = \Bbb C$. Are there any other possibilities? I suspect that there is a quick algebraic-geometry approach here that I am missing.

An aside: I would also be interested in the case of real affine subspaces of $\Bbb C^{n \times n}$, if anyone has insights on what the possibilities are there. Note that the symmetric real matrices are an example of a subspace where we attain $S_U^{(\Bbb R)} = \Bbb R \subsetneq \Bbb C$. We can clearly attain any "line" in $\Bbb C$, but I wonder if there are other possibilities.


2 Answers 2


To expand on Christian Remling's answer a bit: in his example, setting $det(A(t) - \lambda) = 0$ becomes $(1-\lambda)t+\lambda^3=0$, and solving for $t$ gives $t = x^3/(x-1)$ -- so for any $\lambda \in \mathbb{C}$, we have $x \in S$ by choosing this $t$, with the exception of $\lambda=1$.

In general for an affine subspace of dimension $d$, we can define the space by $A(t_1,t_2,\dots t_d)$, and all entries in $A$ are linear in the $t_i$. Then $\det(A(t) - \lambda)$ is an $n$th degree polynomial in all the $t_i$ and $\lambda$. It is possible that the determinant has no dependence on any $t_i$, in which case we have a finite spectrum. Otherwise, we have a dependence on the $t_i$. Then $\det(A(t)-\lambda)=0$ has a solution in $t_1$ whenever at least one of the non-constant coefficients is nonzero. (In the above example, as long as the linear coefficient $1-\lambda\neq 0$.) Since the leading coefficient (that is not a constant 0 polynomial) is an $n-1$th degree polynomial in $\lambda$, it is zero at a most $n-1$ different places, and this is at most $n-1$ places that the spectrum can fail to be continuous. The spectrum only fails to be continuous when all of the coefficients in the $t_1$ polynomial are zero, so this is an upper bound. So, theorem:

For an affine subspace of $n\times n$ matrices, the spectrum is either finite (of size at most $n$), or it is $\mathbb{C}\setminus K$, where $K$ is a finite set of exceptions (of size at most $n-1$).

  • $\begingroup$ Excellent! Thank you for a clear and thorough answer. $\endgroup$ Sep 19, 2019 at 3:01

This is not true. Consider $$ A(t) = \begin{pmatrix} 0 & t & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} . $$ Then $\det (A(t)+1) = 1$, so $-1\notin S$, but clearly the spectrum of $A(t)$ is not constant (for example, $0$ is in the spectrum if and only if $t=0$).

  • $\begingroup$ Thank you for the example, I guess sometimes $2\times 2$ isn't big enough :P $\endgroup$ Sep 19, 2019 at 3:06
  • $\begingroup$ Actually @Omnomnomnom, $2\times 2$ can be enough. Consider [[t 1] [1 0]], which spectrum $\mathbb{C}\setminus\{0\}$. :) For any t, having an eigenvalue of 0 would imply a determinant of 0, but the determinant of that matrix is always -1. $\endgroup$ Sep 19, 2019 at 8:29
  • $\begingroup$ @AlexMeiburg Nice one, guess I really had no excuse here then $\endgroup$ Sep 19, 2019 at 8:34

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