To Prove $x'Ax=\mathrm{tr}(xAx')=\mathrm{tr}(Axx')=\mathrm{tr}(xx'A)$ To prove,


  $x'Ax=\mathrm{tr}(xAx')=\mathrm{tr}(Axx')=\mathrm{tr}(xx'A)$


where 


$A$ is a square matrix.
$x'$ is the transpose of $x$.
For each $x,x'$ are column vector, row vector.
Hint: note that so long as $A$ and $B$ are compatible (that is, $A$ is $m \times n$ and $B$ is $n \times m$), we have trace$(AB) = $ trace$(BA)$, so that
$$
\mathrm{tr}(x'(Ax)) = \mathrm{tr}((Ax)x') \\
\mathrm{tr}((xx')A) = \mathrm{tr}(A(xx'))
$$
All you have to show then is that $x'Ax = \mathrm{tr}(x'Ax)$.
$$xAx'=\sum_{i=1}^{n}\sum_{j=1}^{n}x_{i}x_{j}a_{ij}=(x_{i}x_{j}) \cdot (a_{ij})=tr(xx'A)$$