Orthogonal Diagonalization

Eigenvectors orthogonality

If $A$ is a symmetric matrix, with eigenvectors $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ corresponding to two distinct eigenvalues, then $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are orthogonal.

Proof

Example $PD{P}^T$

We need to find the Eigenvectors of $A$. Skipping the computation, we get,

$$\begin{array}{rl}\overrightarrow{v_1}& =\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right],\lambda =-1\\ \overrightarrow{v_2}& =\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\lambda =1\\ \overrightarrow{v_3}& =\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right],\lambda =1\end{array}$$

Now we must find $PD{P}^T$, The columns of $P$ are the orthonormalized Eigenvectors, and $D$ is the eigenvalue.

$$\begin{array}{rl}P& =\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{-1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]\\ D& =\left[\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$$

Finding $P^T$ is trivial and left to the reader.

Properties

• If $A=PD{P}^T$, if $A$ is a symmetric matrix and it is diagonalizable.
• And the converse, $A$ is a symmetric matrix if $A=PD{P}^T$ is also true.