Xjyuk

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 $lambda_i$,
$$
U_1 = leftpmatrixlambda_1&t\0&lambda_2: t in Bbb Cright, quad
U_2 = leftpmatrixlambda_1&0\0&t: t in Bbb Cright.
$$
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.

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

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 $lambda in mathbbC$, 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 $1-lambdaneq 0$.) 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 $ntimes n$ matrices, the spectrum is either finite (of size at most $n$), or it is $mathbbCsetminus K$, where $K$ is a finite set of exceptions (of size at most $n-1$).