Proving $\mathbf x$ is a $n\times 1$ vector and $\mathbf A$ an $n\times n$ matrix, then $\mathbf x'\mathbf A \mathbf x = \text{tr} (\mathbf {Axx}')$

Question

If $\mathbf x$ is a $n\times 1$ vector and $\mathbf A$ an $n\times n$ matrix, then $\mathbf x'\mathbf A \mathbf x = \text{tr} (\mathbf {Axx}') (\mathbf A'=transpose A) $

Answer 1

Let $x=(x_1 x_2...x_n)^t$ and $A=(a_{ij})_{n\times n}$

Then $x^tAx=\displaystyle \sum_{i=1}^n \sum_{k=1}^n (x_ka_{ik})x_i$

And $ij$th element of $Axx^t$ is $ \displaystyle \sum_{k=1}^n (a_{ik}x_k)x_j=\displaystyle \sum_{k=1}^n (x_ka_{ik})x_j$

So Trace($Axx^t$)= $\displaystyle \sum_{i=1}^n \sum_{k=1}^n (x_ka_{ik})x_i=x^tAx$

Answer 2

Trace is invariant under cyclic permutations, $\DeclareMathOperator{\tr}{tr} x^tAx=\tr(x^tAx)=\tr(Axx^t).$