From rank 4 to rank 2 : a linear combination of three $4 \times 4$ symmetric matrices Consider three $4 \times 4$ symmetric matrices $A, B, C$, each of rank $4$.  Matrix $Q$ is defined as
$$ Q = x_1 A + x_2 B + x_3 C $$
where $x_1, x_2, x_3$ are scalars.
Is it always possible to find $x_1, x_2, x_3$ such that $Q$ has rank $2$ ?
What I know:  It is possible to make $Q$ of rank $3$, but it seems impossible to make it of rank $2$ for arbitrary $A, B, C$.
A detailed answer on this is highly appreciated.
 A: There is no way to ensure this, nor the rank $3$ case, so you must have made an error.
Indeed, here are four linearly independent matrices: $$A=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}, \\
B=\begin{pmatrix}1&0&0&1\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},\\
C=\begin{pmatrix}1&0&0&0\\0&1&0&1\\0&0&1&0\\0&0&0&1\end{pmatrix},\\
D= \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&1\end{pmatrix}
$$ Any non-zero linear combination of these matrices is either rank $4$ or rank $1.$
This is  because a linear combination has a constant diagonal. If the constant is non-zero, then the matrix has rank $4.$ If the constant is zero, the range of the matrix is vectors of the form $(0,0,0,z),$ which is of dimension $1,$ hence rank $1.$
Indeed, we can find six matrices where no linear combination has rank $3.$ Just add to these four the similar upper triangular examples with a single value on the third column.

If $M$ is a $4\times 4$ matrix with characteristic polynomial having no real roots, then any $aI+bM+cM^2$ will have even rank.
This is because, the eigenvalues of $M$ come in complex conjugate pairs, and the eigenvalues of $aI+bM+cM^2$ are $a+b\lambda+c\lambda^2,$ where $\lambda$ is an eigenvalue of $M.$
But if $a+b\lambda+c\lambda^2=0,$ then $a+b\overline\lambda+c\overline\lambda^2=0,$ too, so the zero eigenvalues come in pairs.
You can always get rank $2$ in this case.
Indeed, if the characteristic polynomial has no repeated roots, there will be exactly two pairs $(a,b,c)$ with $c=1$ of rank $2.$
A: The answer is negative. Consider $A=I,\,B=\operatorname{diag}(1,2,3,4)$,
\begin{aligned}
C&=\pmatrix{2&1\\ 1&2&1\\ &1&2&1\\ &&1&2},\\
Q&=xA+yB+zC=\pmatrix{x+y+2z&z\\ z&x+2y+2z&z\\ &z&x+3y+2z&z\\ &&z&x+4y+2z}.
\end{aligned}
$C$ is nonsingular because it is irreducibly diagonally dominant. Now, if $z\ne0$, then $\operatorname{rank}(Q)\ge3$ because the $3\times3$ submatrix $Q(1:3,\,2:4)$ is nonsingular. If $z=0$, since $B$ has four distinct eigenvalues, $Q=xI+yB$ has either four distinct eigenvalues (when $y\ne0$) or four repeated eigenvalues (when $y=0$); hence $\operatorname{rank}(Q)\ne2$.
