Finding Diagonal and Orthogonal Diagonals

Question

Given matrix $\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}$, find a matrix $P$ such that $P^{-1}AP$ is diagonal and an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.

For the first part, I found the eigenvalues to be $\lambda = -2, 4$ and the associated eigenvectors to be $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$, so these are my basis for $\beta$ and $P = \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}$. For $Q$, I was told that I need to use the Gram-Schmidt process on the basis vectors, but I'm unsure if that's correct and if that's the right approach. Can someone clarify?

Answer 1

The problem with using $P$ as $Q$ as well is that $P$ is not orthogonal: $$ P^T P = \pmatrix{2&0\\0&2} \neq \mathbf{I} $$ But if you were to normalize each of the column vectors (so that there magnitue were $1$ instead of $2$, you would get a good $Q$. You accomplish this by dividing each by $\sqrt{2}$.

$$Q = \pmatrix{\frac1{\sqrt{2}}&0\\0&\frac1{\sqrt{2}}} $$

That normalization is the first part of Gram-Schmidt. The other part, orthogonalization, is necessary in principle, but if all the eigenvalues were distinct (as is the case here) then the eigenvectors start out orthogonal so the orthogonalization is trivial. That is why you did not recognize that you are doing Gram-Schmidt.