Exercise 5.3.13

Question

Suppose $P$ is the orthogonal projection onto an subspace $E$, and $Q$ is the orthogonal projection onto the orthogonal complement $E^{\perp}$.
\begin{enumerate}[label = \alph*)]
    \item  What are $P + Q$ and PQ?
    \item Show that $P - Q $ is its inverse.
\end{enumerate}

Jing Huang · Accepted Answer

\begin{proof}
a) $P + Q = I$ since $(P + Q) \mathbf{x} = P\mathbf{x} + Q\mathbf{x} = P_E\mathbf{x} + Q_{E^{\perp}}\mathbf{x} = \mathbf{x}$.  
ewline
$PQ = 0_{n 	imes n}$ as $\mathbf{x^*}PQ\mathbf{x} = \mathbf{x^*}P^*Q\mathbf{x} = (Q\mathbf{x}, P\mathbf{x}) = 0, \, \forall \mathbf{x}$ (using $P$ is self-adjoint shown in Problem 3.11).  \par

b) $(P-Q)^2 = (P-Q)(P-Q) = P^2 - PQ - QP + Q^2 = P^2 + Q^2 = P^2 + Q^2 + PQ + QP = (P+Q)^2 = I^2 = I$ (using $PQ = QP = 0$). i.e., $(P-Q)^{-1} = P-Q$.
\end{proof}

Exercise 5.3.13

Answers

Comments

Exercise 5.3.13

Answers

Comments

Add answer