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.
linear algebramatricestrace
To prove,
$x'Ax=\mathrm{tr}(xAx')=\mathrm{tr}(Axx')=\mathrm{tr}(xx'A)$
where
Best Answer
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 $x'Ax = \mathrm{tr}(x'Ax)$.