Question

Let Abe the matrix of an orthogonal projection. Find${\mathit{A}}^{\mathbf{2}}$ in two ways:

a.Geometrically. (Consider what happens when you apply an orthogonal projection twice.)

b.By computation, using the formula given in Theorem 5.3.10

Short answer

$A^2=A$

Step-by-step solution

Consider for part (a).

Observe that, in the figure below the brown vector is the orthogonal projection of the vector $\stackrel{\perp }{v}$onto the space W. so, if we project onto a subspace W of $R^n$, then for any $\stackrel{I}{x}\in R^n,A\stackrel{I}{x}\in W$. Now, again taking the orthogonal projection won’t do anything as the vector $A\stackrel{\perp }{x}$is already there on the subspace W and therefore, $A\left(A\stackrel{I}{x}\right)=A\stackrel{I}{x}=A^2\stackrel{I}{x}$

Since it is true for any $\stackrel{I}{x}\in R^n$thus,

$A^2=A$

Consider the diagram below.

Consider for part (b).

Observe that the matrix of the orthogonal projection is given by .

As the ijth entry of the matrix is given by

$\stackrel{r}{u_i}\stackrel{r}{u_j}=\left\{\begin{array}{cc}0& i\ne j\\ 1& i\ne j\end{array}\right.$

So, consider, localid="1659500792037" $A=Q{Q}^T\Rightarrow A^2$

$$\begin{gathered} A=Q{Q}^T\Rightarrow A^2 \\ =Q\left(Q^{T}Q\right)Q^T \\ =\left(Q{Q}^T\right)\left(Q{Q}^T\right) \\ =Q{Q}^T \\ =A \end{gathered}$$

Thus, it gives$A^2=A$