How to prove that $A(v^TAv) =Avv^TA$?

Question

How to prove that

$$A(v^TAv) =Avv^TA$$

where $v$ is an $n \times 1$ column vector and $A$ is an $n \times n$ matrix?

Answer 1

It's not true. Consider $A = \begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}$, and $v = \begin{pmatrix} 1\\ 0 \end{pmatrix}$

Then $$\text{LHS: } A(v^TAv)=\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix} \cdot \left(\begin{pmatrix} 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}\right)=\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix} \cdot 1 = \begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}$$

$$\text{RHS: } Avv^TA=\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}\begin{pmatrix} 1\\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2\\ 2 & 1 \end{pmatrix}^2=\begin{pmatrix} 5 & 4\\ 4 & 5 \end{pmatrix}$$

Answer 2

You can't, LHS is a scalar and RHS is an $n \times n$ matrix.

Assuming real matrices, You might want to prove $v^TAv = v^TA^Tv$, of which case, the hint is they are scalar.

Edit:

Well, it is still not true, Let $A=I$, then the left hand side is a diagonal matrix with diagonal entries $\|v\|^2$ while the right hand side is $vv^T$ which need not be diagonal, for example, let $v$ be the all one vector.