Here's another old Linear Algebra comprehensive exam question:
Let A and B be two real $nxn$ matrices.
Part 1)
Show that if A and B are symmetric, then for any $\vec x \in R^n$, $$((A^2 + B^2)\vec x, \vec x) \ge((AB+BA)\vec x, \vec x)$$
Hint: look at $((A−B)^2 \vec x,\vec x)$
Part 2)
Find a counterexample to Part (1) if the assumption of symmetry is dropped.
Part 3)
If $\hat B$ is another real $nxn$ matrix, what can we say about $B-\hat B$, if for all $\vec x \in R^n$, $$(B\vec x,\vec x) = (\hat B\vec x,\vec x)$$
(do not assume that $B$ is symmetric.)
Part 4) Show that for any n×n matrix with real matrices A and B, we can find $\hat B$ so that $(B\vec x,\vec x) = (\hat B\vec x,\vec x)$ for all $\vec x \in R^n$ and $((A^2 + \hat B^2)\vec x, \vec x) \ge((A\hat B+ \hat BA)\vec x, \vec x)$ for any $x∈R^n$.
My work:
I have solved Parts 1 and 2 but am stuck on Parts 3 and 4.
(although, for Part 1, I have something slightly different: $((A^2 + B^2)\vec x, \vec x) \ge((AB+(BA)^t)\vec x, \vec x)$...I am almost certain it is $(BA)^t$...I tried Part (1) probably 10 times, just to be sure, and I still ended up with a $(BA)^t$ instead of the $(BA)$.
For Part 3, I had thought that the equation implies that $B\vec x = \hat Bx, $ so that $B\vec x -\hat Bx =0 $ and$(B-\hat B)\vec x =0 $, which implies that $(B-\hat B)$ is the zero matrix. Apparently not. The final answer gives that $(B-\hat B)$ is actually skew-symmetric. How can I derive this?
Any help with Parts 3 and 4 are welcome.
Thanks,
Part 3: Define $\tilde{B}:=B-\hat{B}$. Then $$0=x^T\tilde{B}x=\frac{1}{2}x^T(\tilde{B}+\tilde{B}^T)x\qquad\forall x\in\mathbb{R}^n$$ Since the matrix $ \tilde{B}+\tilde{B}^T$ is symmetric all eigenvalues are real and considering as $x=v_i$ the corresponding eigenvectors we obtain $\lambda_i(\tilde{B}+\tilde{B}^T)\|v_i\|^2=0$ that yields $\tilde{B}+\tilde{B}^T=0$ (all eigenvalues zero and every real symmetric matrix is diagonalizable) i.e. $\tilde{B}^T=-\tilde{B}$ (skew-symmetry property).