Exercise 5.3.11

Question

Let $P = P_E$ be the matrix of an orthogonal projection onto a subspace $E$. Show that 
\begin{enumerate}[label = \alph*)]
    \item The matrix $P$ is 	extit{self-adjoint}, meaning that $P^* = P$.
    \item $P^2 = P$.
\end{enumerate}

Jing Huang · Accepted Answer

\begin{remark}
The above $2$ properties completely characterize orthogonal projection.
\end{remark}

\begin{proof}
a) From the orthogonality, we have $(\mathbf{x}, \mathbf{x} - P\mathbf{x}) = (\mathbf{x} - P\mathbf{x}, \mathbf{x}) = 0, \, \forall \mathbf{x}$. $(\mathbf{x}, \mathbf{x} - P\mathbf{x}) = (\mathbf{x} - P\mathbf{x})^* \mathbf{x} = (\mathbf{x}^* - \mathbf{x}^* P^*) \mathbf{x} = \mathbf{x^*x} - \mathbf{x^*}P^*\mathbf{x} = 0$. On the other hand, $(\mathbf{x} - P\mathbf{x}, \mathbf{x}) = \mathbf{x^*} (\mathbf{x} - P\mathbf{x}) = \mathbf{x^*x} - \mathbf{x^*} P \mathbf{x} = 0$. Subtract two equalities, $\mathbf{x^*}(P - P^*) \mathbf{x} = 0, \, \forall \mathbf{x}$. Then $P - P^* = 0_{n 	imes n}, P = P^*$.  \par

b) Consider $(P\mathbf{x}, \mathbf{x}-P\mathbf{x}) = (\mathbf{x} - P\mathbf{x})^* P\mathbf{x} = (\mathbf{x^*} - \mathbf{x^*}P^*) P\mathbf{x} = \mathbf{x^*} (P - P^*P)\mathbf{x} = 0$. Thus $P = P^*P = P^2$ since $P = P^*$.
\end{proof}

Exercise 5.3.11

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Exercise 5.3.11

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