# Spectrum of a Subspace of Matrices

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.

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 $$n-1$$).

• Excellent! Thank you for a clear and thorough answer. 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 $$t=0$$).

• Thank you for the example, I guess sometimes $2\times 2$ isn't big enough :P Sep 19, 2019 at 3:06
• 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. Sep 19, 2019 at 8:29
• @AlexMeiburg Nice one, guess I really had no excuse here then Sep 19, 2019 at 8:34