A question about a sequence of matrices with infinite operator norm limit Suppose $A_n$ are symmetric non-negative definite $d\times d$ matrices with real entries such that $||A_n|| \rightarrow \infty$ as $n \rightarrow \infty$. (The norm is the operator norm, but all norms are equivalent so it doesn't really matter). Does there exist a single $z \in \mathbb{R}^d$ such that 
$$ \langle z, A_n z \rangle \rightarrow \infty \quad \textrm{as} \quad n \rightarrow \infty?$$
Many thanks in advance.
Does this hold more generally? (I.e. in infinite dimensions, or when the matrices aren't non-negative definite).
 A: Since $A_n$ is symmetric and non-negative definite, there is a (unique) symmetric non-negative definite matrix $\sqrt{A_n}$ with $A_n = \sqrt{A_n} \sqrt{A_n}$ (use that $A_n$ can be diagonalized using an orthonormal basis).
If there does not exist a $z$ with the desired property, then
$$
\langle z, A_n z\rangle = \langle \sqrt{A_n} z, \sqrt{A_n} z \rangle = \Vert \sqrt{A_n} z \Vert^2
$$
would be a bounded sequence for each $z \in \Bbb{R}^d$. But (since we are in a finite dimensional setting), this easily implies that $(\sqrt{A_n})_n$ is a bounded sequence, whence also $(A_n)_n$ is a bounded sequence, contradiction.
We can also do the same in infinite dimensions (if the space under consideration is complete, i.e. a Hilbert space): By the argument above, if there is no such $z$, then $(\sqrt{A_n} z)_n$ is a bounded sequence for every $z \in H$, where $H$ is the Hilbert space under consideration. Hence, by the uniform boundedness principle, $(\Vert\sqrt{A_n}\Vert)_n$ is a bounded sequence, whence also $(A_n)_n$ is a bounded sequence, contradiction.
If you do not assume the matrices under consideration to be symmetric (and you allow real Hilbert spaces instead of complex ones, which I think you do, since you consider $\Bbb{R}^d$), then consider
$$
A_{n}=\left(\begin{matrix}0 & n\\
-n & 0
\end{matrix}\right)
$$
Here, we have $\langle z, A_n z\rangle = 0$ for all $z \in \Bbb{R}^d$, but $\Vert A_n \Vert \to \infty$.
Now, if you assume the $A_n$ to be symmetric, but not necessarily non-negative definite, I am not sure at the moment.
