Have I proven that every matrix is symmetric? I've been given this as an assignment:
"From definitions of different classes of matrices, prove the following claims:
A) Positive => Symmetric…"
There is also a hint: If $<Au,u> = <Bu,u>$ for all $u$, then $A=B$.
The definitions have been given: A matrix is symmetric if $A^*=A$ and positive if $<Au,u>$ is positive for all $u$. The adjugate $A^*$ is defined as the matrix which satisfies $<Ax,y>=<x,A^*y>$.
I tried this:
$$<Au,u>=<u,A^*u>$$
The inner product is commutative, so:
$$<Au,u>=<A^*u,u>$$
So using the hint:
$$A^*=A$$
But for this proof, I didn't need the matrix $A$ to be positive. I seem to have proven that every matrix is symmetric. This probably isn't true, so what was my mistake?
 A: There's a straightforward way to figure out what went wrong in a situation like this, when you prove a statement you know is false. You find a counterexample to the statement you know is false, and substitute the counterexample into your proof to figure out when you first wrote down a false statement about your counterexample. 
So: it's clearly false that every matrix is symmetric. For example, the matrix $A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$ isn't symmetric. So what happens when you put it into your proof? It's true that
$$\langle Au, u \rangle = A^{\ast} u, u \rangle$$
for all $u \in \mathbb{R}^2$; explicitly, if $u = (x, y)$, then both sides are equal to $xy$. However, it's not true that $A = A^{\ast}$, since $A^{\ast} = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$. So the hint is false. That's all. 

That was straightforward. Here are some facts which are also true but which are less straightforward. If $\langle Au, u \rangle = \langle Bu, u \rangle$ all you can conclude is that $A - B$ is skew-symmetric, which is the case above. It's also false that positive matrices are symmetric, with the given definitions. (The correct definition, to my mind, is that positive should only be applied to symmetric matrices.) 
