Is $V^TAV$ diagonal in general? I want to know if the following is a true statement:

If $A$ is real a symmetric and $V$ is orthogonal then $V^TAV=D$ where $D$ is a diagonal matrix. 

To prove it I can say  
$$(V^TAV)^T=V^TA^TV = V^TAV.$$
So $V^TAV$ is itself symmetric. So if I know $V^TAV$ is triangular then I'm done. How do I know $V^TAV$ is triangular (if it is)?
 A: It is false: take $V$ to be the identity and $A$ to be symmetric but not diagonal.
Then $D=V^TAV=A$ so $D$ is not diagonal.
A: You're probably misunderstanding what your textbook or instructor is trying to tell you.

If $A$ is any real square matrix with all real eigenvalues, there exists an orthogonal matrix $V$ such that
  $$
V^TAV
$$
  is triangular.

This is done by induction on the order of $A$; choose an eigenvalue $\lambda$ of $A$, and complete a norm $1$ eigenvector to an orthonormal basis; the matrix $V_0$ having these basis vector as columns has the property that $V_0^TAV_0$ has the block form
\begin{bmatrix}
\lambda & r_1 \\
0 & A_1
\end{bmatrix}
where $r_1$ is some row. By the induction hypothesis, there is an orthogonal matrix $V_1$ such that $V_1^TA_1V_1$ is upper triangular. Note that the matrix $A_1$ has the same eigenvalues as $A$ (except for $\lambda$ having multiplicity one less), so also $A_1$ has all real eigenvalues. Now the orthogonal matrix
$$
V=V_0\begin{bmatrix} 1 & 0 \\ 0 & V_1\end{bmatrix}
$$
has the required property that $V^TAV$ is upper triangular.
If $A$ is symmetric (so its eigenvalues are real), then
$$
V^TAV=V^TA^TV=(V^TAV)^T
$$
is also lower triangular, so it is diagonal.
Hence the theorem is

If $A$ is a real symmetric matrix, there exists an orthogonal matrix $V$ such that $V^TAV$ is diagonal.

