# Reducing quadratic forms to diagonal form

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.
• this is done in Horn and Johnson, Matrix Analysis, especially Theorem 4.5.15 and accompanying table on pages 228, 229. Detail for this case (II) of the theorem are on pages 231 to 232. Short version: recipe how to do it when $A$ is nonsingular (both real symmetric) and $C = A^{-1}B$ diagonalizable over $\mathbb C,$ that is $R^{-1}A^{-1}B R = \Lambda$ is diagonal. – Will Jagy Jul 28 '19 at 19:00
• Hi @Omnomnomnom, thanks for your answer! So I have found that $P_1=PD^{-\frac{1}{2}}$ where $P$ diagonalises $A$ to a diagonal matrix $D$. Then I'm struggling a bit proving there is a $P_2$ that does the job. I'm thinking of $P_2=D^{\frac{1}{2}}$ then I got $P_2^T[P_1^TBP_1]P_2$=$P^TBP$ but I don't know how to show this is diagonal. – M. W Jul 29 '19 at 0:59
• Never mind I figured it out. Just to confirm, we can take $P_2=D^{\frac{1}{2}}P^TQ$ where $Q$ is an orthogonal diagonalising matrix for $B$. Is this correct? – M. W Jul 29 '19 at 2:39
• Sorry for sending so many questions. I don't think my $P_2$ is orthogonal though. Then I couldn't get $(P_1P_2)^TAP_1P_2$ is diagonal. So I guess my $P_2$ is still incorrect... Could you hint a bit on what $P_2$ should be? – M. W Jul 29 '19 at 2:47
• Diagonalize $P_1^TBP_1$ using the spectral theorem – Ben Grossmann Jul 29 '19 at 14:01