15
$\begingroup$

Let $A$ and $B$ be complex $4\times 4$ matrices. Assume both are Hermitian, and that they are linearly independent.

Must there exist a nonzero real linear combination $aA + bB$ which has a repeated eigenvalue?

$\endgroup$
23
$\begingroup$

The answer is 'no'. The generic pair $A$ and $B$ of $4$-by-$4$ Hermitian symmetric matrices will not have any nonzero real linear combination that has a double eigenvalue.

For a specific example, take $$ A = \begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&2\end{pmatrix} \quad \text{and}\quad B = \begin{pmatrix}0&i&0&0\\-i&0&0&0\\0&0&0&2i\\0&0&-2i&0\end{pmatrix}. $$ Then $$ \det(aA+bB - tI_4) = (t^2-a^2-b^2)(t^2-4a^2-4b^2), $$ and the roots of this polynomial in $t$ are distinct unless $a=b=0$. (Recall that we are assuming that $a$ and $b$ are real, which, of course, implies that $t$ is real.)

Added Remark: To see the claim that this property holds for a generic linearly independent pair of Hermitian symmetric $4$-by-$4$ matrices $A$ and $B$, it is only necessary to observe the following: The question is whether, for a generic such pair $A$ and $B$ in the $16$-dimensional real vector space $\mathcal{H}_4$ consisting of $4$-by-$4$ Hermitian symmetric matrices, the (real) span of $A$, $B$, and $I_4$ contains a nonzero element of rank at most $2$. Now, it is not difficult to show that the cone $C_2$ of elements in $\mathcal{H}_4$ that have rank at most $2$ is a closed algebraic cone of dimension $12$ (one that is singular along the $7$-dimensional locus of elements of rank at most $1$). Hence the generic $3$-dimensional subspace of $\mathcal{H}_4$ will meet this cone only at the zero matrix. It is also an open (though not dense) condition on a $3$-dimensional subspace that it contain a positive definite element. Since the standard $\mathrm{GL}(4,\mathbb{C})$-action on $\mathcal{H}_4$ acts transitively on the space of positive definite elements, it follows that the generic pair $A$, $B$, together with $I_4$ will span a $3$-plane that meets $C_2$ only at the origin, which is what was to be shown.

$\endgroup$
  • 5
    $\begingroup$ Thank you. Federico's reference points out that the answer is generically yes if complex coefficients are allowed, but unfortunately I really needed real coefficients... $\endgroup$ – Nik Weaver Jan 1 '18 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.