Question

TRUE OR FALSE

If a $\mathbf{3}\mathbf{\times }\mathbf{3}$matrix A represents the orthogonal projection onto a plane V in ${\mathit{R}}^{\mathbf{3}}$, then there must exists an orthogonal $\mathbf{3}\mathbf{\times }\mathbf{3}$matrix S such that ${\mathit{S}}^{\mathbf{T}}\mathit{A}\mathit{S}$is diagonal.

Short answer

The given statement is true.

Step-by-step solution

Explanation of the solution

Consider an orthogonal basis for the subspace V, say $\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3\right\}$and ${\overrightarrow{v}}_3$is a vector perpendicular to the plane V.

Since A represents the orthogonal projection onto the plane V as follows:

role="math" localid="1659629274695" $$\begin{gathered} A\left[{\overrightarrow{v}}_1 {\overrightarrow{v}}_2 {\overrightarrow{v}}_3\right]=\left[{\overrightarrow{v}}_1 {\overrightarrow{v}}_2 0\right] \\ AS=\left[{\overrightarrow{v}}_1 {\overrightarrow{v}}_2 0\right] where S=\left[{\overrightarrow{v}}_1 {\overrightarrow{v}}_2 {\overrightarrow{v}}_3\right] \\ AS=\left[{\overrightarrow{v}}_1 {\overrightarrow{v}}_2 {\overrightarrow{v}}_3\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ AS=S\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} AS=S\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ {S}^{-1}AS=S^{-1}S\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Since S is orthogonal is $S^T=S^{-1}$.

$S^{T}AS=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

This implies that$S^{T}AS$is a diagonal matrix.

Thus, the given statement there exits an orthogonal $3\times 3$matrix such that $S^{T}AS$is diagonal” is true.