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?



Existence of double eigenvalue
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.


