Bound on the trace of a product of a symmetric positive definite matrix and a nil-potent matrix. Let $A$ a matrix with the following form:
$A = vu^{T}$
where $u$ and $v$ are orthogonal vectors and $B$ be a symmetric positive definite matrix. Is there a way to upper bound $Trace(AB)$? I know that the eigen-values of $A$ are all zero but I can't seem to find a way to use this information.
 A: We have
$$
\operatorname{tr}(AB) = \operatorname{tr}(uv^TB) = \operatorname{tr}(v^TBu) = v^TBu.
$$
On the other hand, the Cauchy interlacing theorem tells you that if $u,v$ are orthogonal unit vectors, $\lambda_1\leq\cdots\leq\lambda_n$ are the eigenvalues of $B$ and $\mu_1 \leq \mu_2$ are the eigenvalues of the matrix
$$
C := \pmatrix{u & v}^TB\pmatrix{u & v} = \pmatrix{u^TBu & u^TBv\\ v^TBu & v^TBv},
$$
then $\lambda_1\leq \mu_1 \leq \lambda_{n-1}$ and $\lambda_2 \leq \mu_2 \leq \lambda_n$, from which we can conclude that the condition numbers of $B$ and $C$ satisfy $\kappa(C) \leq \kappa(B) = \lambda_{\max}(B)/\lambda_{\min}(B)$. From there, this post gives us the formula
$$
\kappa(C) = \frac{1 + \frac{|u^TBv|}{\sqrt{u^TBu\cdot v^T Bv}}}{1 - \frac{|u^TBv|}{\sqrt{u^TBu\cdot v^TBv}}} \leq \kappa(B).
$$
Rearranging this inequality yields
$$
\frac{|u^TBv|}{\sqrt{u^TBu\cdot v^TBv}} \leq \frac{\kappa(B) - 1}{\kappa(B) + 1}.
$$
With that, we can state that
$$
|\operatorname{tr}(AB)| = |u^TBv| \leq \frac{\kappa(B) - 1}{\kappa(B) + 1} \cdot \sqrt{u^TBu\cdot v^TBv}.
$$
If we use the inequality $u^TBu \leq \lambda_{\max}(B) \|u\|^2$, then we get the weaker upper bound
$$
|\operatorname{tr}(AB)| \leq \frac{\kappa(B) - 1}{\kappa(B) + 1} \cdot \lambda_{\max}(B) \cdot \|u\| \cdot \|v\|,
$$
which is a strict improvement over the bound $|\operatorname{tr}(AB)| \leq \lambda_{\max}(B) \cdot \|u\| \cdot \|v\|$ that holds without the $u^Tv = 0$ condition.
We could also get a slightly better bound using the inequality
$$
\begin{align}
\sqrt{u^TBu\cdot v^TBv} &\leq \frac 12 \left(u^TBu + v^TBv\right)  
\\ & = \frac 12 \operatorname{tr}(C)  = \frac 12 (\mu_1 + \mu_2) \leq \frac 12 (\lambda_{n-1} + \lambda_n).
\end{align}
$$
