Let be $A$ and $B$ two real matrices of $n \times n$. And $\left \langle , \right \rangle$ denotes the usual inner product in $\mathbb{R}^{n}.$

Prove that if $A$ and $B$ are symmetric then $\forall x \in \mathbb{R}^{n}$ it satisfies:

\begin{align*} \left \langle (A^{2}+B^{2})x,x \right \rangle\geq \left \langle (AB+BA)x,x \right \rangle \end{align*} Hint: Consider $\left \langle (A-B)^{2}x,x \right \rangle$

What I think I can do is to note that:

\begin{align*} \left \langle A^{2}+B^{2})x,x \right \rangle&=\left \langle A^2x,x \right \rangle + \left \langle B^2x,x \right \rangle\\\left \langle AB+BA)x,x \right \rangle&=\left \langle AB,x \right \rangle + \left \langle BA,x \right \rangle \end{align*}

And then try to prove in general that:

\begin{align*} \left \langle A^2x,x \right \rangle&\geq\left \langle ABx,x \right \rangle\\ \end{align*}

Neverthless, I don't know how to use the hint and the fact that the matrices are symmetric. Can you help me please? I would really appreciate it.

  • $(A-B)^2 = (A^2 + B^2) - (AB + BA)$
  • The eigenvalues of $(A-B)^2$ are nonnegative so $\langle (A-B)^2 x, x \rangle \ge 0$ for any $x$.
  • $\begingroup$ Thank you! But, where do I have to use the fact that the matrices are symmetric? @angryavian $\endgroup$
    – luisegf
    Oct 19 '20 at 2:09
  • 1
    $\begingroup$ $A$ and $B$ being symmetric implies that $A-B$ is symmetric, so $(A-B)^T(A-B) = (A-B)^2$, which implies the second point in the answer. $\endgroup$
    – PkT
    Oct 19 '20 at 2:18

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