Question: Suppose that $\Lambda$ is an $N × N$ real-valued diagonal matrix and $Q$ is real symmetric. Suppose that $tr\ \Lambda \neq 0$ and $tr\ Q \neq 0$. Prove there is a vector $v \in \mathbb{R}^N$ such that $$v^T\Lambda v = tr\ \Lambda, v^TQv = tr\ Q $$
My thought:
Decompose the $\Lambda$ as $\Lambda = U^T\Lambda U, \ \forall\ U^TU = I $ and $Q = U^T\Lambda_1 U$.
So $$v^T\Lambda v = (Uv)^T\Lambda (Uv)= tr\ \Lambda$$ and $$v^TQ v = (Uv)^T\Lambda_1 (Uv) = tr \ Q = tr\ \Lambda_1$$ Intutively,the structure is same so they could preserve the same property. But how could I prove the $v$ in two formulas is the same one?
Could anyone help me out? Thank in advance!
The assumption that $\Lambda$ is diagonal is a premature simplification. We can just assume that $\Lambda$ is real symmetric. Also, by scaling $\Lambda$ and $Q$ if necessary, we may assume that both matrices have traces $1$. Thus we want to solve $v^T\Lambda v=v^TQv=1$ when $\Lambda$ and $Q$ are real symmetric matrices of traces $1$.
Let $D=\Lambda-Q$. By a change in orthonormal basis, we may assume that $D$ is a traceless diagonal matrix. Let $S\subset\mathbb R^N$ be the set of all vectors whose entries belong to $\{-1,1\}$. Then $\frac1{2^N}\sum_{v\in S}v^TQv=\operatorname{tr}Q>0$ and hence $v^TQv>0$ for some $v\in S$. Since $D$ is a traceless diagonal matrix, we also have $v^TDv=\operatorname{tr}D=0$.
Therefore $v^T\Lambda v=v^T(Q+D)v=v^TQv>0$. Now we can scale $v$ to make $v^T\Lambda v=v^TQv=1$.