Question. Let $A,B$ be real symmetric matrices of the same size, and suppose that $Q_1=\mathbf{x}^TA\mathbf{x}$ and $Q_2=\mathbf{x}^TB\mathbf{x}$ are the corresponding quadratic forms. In addition assume that $Q_1$ is positive definite.
Show that there is an invertible change of coordinates, $\mathbf{x}=P\mathbf{y}$ with $P$ not necessarily orthogonal, such that both $Q_1$ and $Q_2$ are simultaneously reduced to diagonal form.
My attempt. So I'm not really sure about what the theorem says. I know that $Q_1$ being positive definite means that $A$ is positive definite (since $A$ is also self-adjoint), i.e. all its eigenvalues are strictly positive. Now I'm really stuck and don't know where to start the proof. Would appreciate any help.
Hint: You're trying to show that there exists a $P$ such that both $P^TAP$ and $P^TBP$ are diagonal. Because $Q_1$ is a positive definite quadratic form, we know that the matrix $A$ is a positive definite (symmetric) matrix; that's all we're supposed to take from the question.
Begin by finding a $P_1$ such that $P_1^TAP_1 = I$. Then, find an orthogonal $P_2$ for which $P_2^T[P_1^TBP_1]P_2$ is diagonal. Verify that $P = P_1P_2$ is indeed a matrix for which both $P^TAP$ and $P^TBP$ are diagonal.