Let \(W\subseteq \R^3\) be the plane passing through the origin with normal vector \(\boldn=(1,1,1)\) (with respect to the dot product): i.e.,
\begin{equation*}
W=\{(x,y,z)\in \R^3\colon x+y+z\}\text{.}
\end{equation*}
Consider the orthogonal projection transformation
\(\operatorname{proj}_W\colon \R^3\rightarrow \R^3\text{.}\) With respect to the standard basis
\(B=(\bolde_1, \bolde_2, \bolde_3)\) we know from
Theorem 5.2.3 that
\([\operatorname{proj}_W]_B=[\operatorname{proj}_W]_B^B\) is just the matrix
\(A\) such that
\(\operatorname{proj}_W=T_A\text{.}\) Using the general formaul for projection onto a plane derived in
Example 4.3.17, we conclude:
\begin{equation*}
[T]_B=A=\frac{1}{3}\begin{amatrix}[rrr]
2\amp -1\amp -1\\
-1\amp 2\amp -1\\
-1\amp -1\amp 2
\end{amatrix}\text{.}
\end{equation*}
Now consider a nonstandard basis \(B'\) of \(\R^3\) constructed with an eye toward the geometry involved in the definition of this projection operator. Namely, we begin with an orthogonal basis of the plane \(W\) and extend to an orthogonal basis of \(\R^3\text{.}\) The vectors \((1,-1,0), (1,1,-22)\) clearly form an orthogonal basis of \(W\text{.}\) To complete the basis we simply add the normal vector \(\boldn\text{:}\)
\begin{equation*}
B'=\left( \underset{\boldv_1}{(1,-1,0)}, \underset{\boldv_2}{(1,1,-2)}, \underset{\boldv_3}{(1,1,1)} \right)\text{.}
\end{equation*}
Let
\(A'=[T]_B'\text{,}\) and for all
\(1\leq i\leq 3\) let
\(\boldc_i\) be the
\(i\)-th column of
\(A\text{.}\) Using
(5.2.1) we compute
\begin{align*}
\boldc_1 = [\proj{\boldv_1}{W}]_{B'}\amp =[\boldv_1]_{B'} \amp (\boldv_1\in W)\\
\amp =(1,0,0) \amp (\boldv_1=1\boldv_1+0\boldv_2+0\boldv_3)\\
\boldc_2 = [\proj{\boldv_2}{W}]_{B'}\amp =[\boldv_2]_{B'} \amp (\boldv_2\in W)\\
\amp =(0,1,0) \amp (\boldv_2=0\boldv_1+1\boldv_2+0\boldv_3)\\
\boldc_3 = [\proj{\boldv_3}{W}]_{B'}\amp =[\boldzero]_{B'} \amp (\boldv_3\in W^\perp)\\
\amp =(0,0,0) \amp (\boldzero=0\boldv_1+0\boldv_2+0\boldv_3)\text{.}
\end{align*}
Thus
\begin{equation*}
[T]_{B'}=A'=\begin{amatrix}[rrr]1\amp 0\amp 0\\ 0\amp 1\amp 0\\ 0\amp 0\amp 0 \end{amatrix}\text{.}
\end{equation*}
Wow,
\(A'\) is way simpler! How can the two very different matrices
\(A\) and
\(A'\) represent the same linear transformation? A useful way of thinking about this is to consider
\(A\) and
\(A'\) as two matrix formulas for
\(\operatorname{proj}_W\) with respect to two different
coordinate systems. This can be made precise by using the defining properties (
(5.2.2)) of
\(A=[T]_B\) and
\(A'=[T]_{B'}\text{.}\) For
\(A=[T]_B\) we have
\begin{equation}
A[\boldx]_B=[\proj{\boldx}{W}]_B \implies A\boldx=\proj{\boldx}{W}\text{,}\tag{5.2.4}
\end{equation}
where we have used the fact that for the
standard basis \(B\) we have
\([\boldy]_B=\boldy\) for any
\(\boldy\in \R^3\text{.}\) Thus we can compute
\(\proj{\boldx}{W}\) directly with
\(A\) as
\(A\boldx\text{.}\) For
\(A'=[T]_{B'}\text{,}\) on the other hand, property
(5.2.2) reads as
\begin{equation}
A'[\boldx]_{B'}=[\proj{\boldx}{W}]_{B'}\text{.}\tag{5.2.5}
\end{equation}
Formula
(5.2.5) indicates that we cannot use
\(A'\) directly to compute
\(\proj{\boldx}{W}\text{.}\) Rather, we must first compute the
\(B'\)-coordinates of
\(\boldx\) and then multiply by
\(A'\text{,}\) at which point the
\(B'\)-coordinates of
\(\proj{\boldx}{W}\) are returned. In other words,
\(A'\) describes the operation of
\(\operatorname{proj}_W\) with respect
\(B'\)-coordinates of vectors in
\(\R^3\text{.}\) As such,
\(A\) may be more useful to us in terms of computing
\(\operatorname{proj}_W\) directly. However,
\(A'\) has the advantage of giving us a clear
conceptual picture of projection. For example, we see directly that
\(A'\) has nullity one and rank 2, and thus the same is true of
\(\operatorname{proj}_W\text{.}\) Furthermore, understanding that the columns of
\(A'\) describe how
\(\operatorname{proj}_W\) acts on the basis
\(B'\text{,}\) we see easily that
\(\operatorname{proj}_W\) acts as the identity on the 2-dimensional space spanned by the first two basis elements, and sends the subspace spanned by the third basis element to
\(\boldzero\text{.}\) In other words,
\(A'\) neatly encodes the conceptual picture of
\(\operatorname{proj}_W\) as an operator that fixes the plane
\(W=\Span\{(1,-1,0),(1,1,-2)\}\) and sends everything in
\(W^\perp=\Span\{(1,1,1)\}\) to
\(\boldzero\text{.}\)
