Clever way to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$?

Question

How to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$? (inner product of matrices)

1. $xx^T$ is rank one.

The following is one way to prove it:

$$\langle A,X\rangle=\text {tr}(AX)$$

Therefore, listing the trace of $AX$ and compare it to $x^TAx$ is one way.

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Is there any other clever way to prove it?

The following is what I am thinking:
$\langle A,X\rangle = \langle A,xx^T \rangle$
and
$\langle x,Ax\rangle =x^TA^Tx=x^TAx$

Therefore, can I say that
$\langle A,xx^T \rangle=\langle Ax,x^T \rangle $ ? and why?
(It is something related to adjoint?)

Answer 1

In general, $\text{tr}(AB) = \text{tr}(BA)$, so $\text{tr}(Axx^T) = \text{tr}((Ax)x^T) = \text{tr}(x^TAx) = x^TAx$.