For $A,B,C,D$ $3\times 3$ symmetric complex matrices there exists a non trivial linear combination of $A,B,C,D$ whose rank is $\leq1$. 
Let $A,B,C$ and $D$ be $3\times 3$ symmetric matrices over $\mathbb C$. Then have to show that there exist complex numbers $\alpha,\beta,\gamma$ and $\delta$, not all zero, such that the matrix $M=\alpha A+\beta B+\gamma C+\delta D$ has rank less than or equal to $1$.

If the matrices are linearly dependent or one of the matrices has rank less than equal to $1$ then we are done. I need some hint to proceed in the remaining cases. Thanks.
 A: (As this question comes from a qualifying exam, there should be some simpler solutions, but the following is what immediately comes to mind.)
Consider the linear subspace $V$ spanned by the leftmost columns of $A,B,C$ and $D$.
If $\dim V\le2$, we may assume that the leftmost columns of $A$ and $B$ are zero. As $A$ and $B$ are symmetric, we must have $A=0\oplus A_1$ and $B=0\oplus B_1$ for some symmetric $2\times2$ matrices $A_1$ and $B_1$. Hence some non-trivial linear combination of $A_1$ and $B_1$ is singular. In turn, the linear combination of $A$ and $B$ with the same coefficients has rank $\le1$.
If $\dim V=3$, we may assume that the first columns of $A,B,C$ are linearly independent and the first column of $D$ is zero. Hence $D=0\oplus D_1$ for some symmetric $2\times2$ matrix $D_1$. If $D_1$ is singular, $D$ already has rank $\le1$ and we are done.
If $D_1$ is nonsingular, by a congruence transform (and by Takagi factorisation) we may assume that $D_1=I_2$. Since the leftmost columns of $A,B,C$ are linearly independent, we may assume that they are the standard vectors. By adding some suitable multiples of $D$ to the other three matrices, we may also assume that $a_{22}=b_{22}=c_{22}=0$. Hence
\begin{aligned}
S&:=\alpha A+\beta B+\gamma C+\delta D\\
&=
\alpha\pmatrix{1&0&0\\ 0&0&a_{23}\\ 0&a_{23}&a_{33}}
+\beta\pmatrix{0&1&0\\ 1&0&b_{23}\\ 0&b_{23}&b_{33}}
+\gamma\pmatrix{0&0&1\\ 0&0&c_{23}\\ 1&c_{23}&c_{33}}
+\delta\pmatrix{0&0&0\\ 0&1&0\\ 0&0&1}.
\end{aligned}
If $S$ is a non-trivial linear combination and it has rank $\le1$, we must have $\alpha\ne0$ and $\operatorname{rank}(S)=1$. Hence we may let $\alpha=1$ and the rank-one matrix that $S$ is supposed to be equal to must be $(1,\beta,\gamma)^\top(1,\beta,\gamma)$. You may continue from here.
