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 4.2.3 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 5.3.17, 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 a basis of the plane \(W\) and extend to a full basis of \(\R^3\) by adding the normal vector \((1,1,1)\) defining \(W\text{.}\) Note that since \(W\) is a 2-dimensional subspace, a basis by formed by any two linearly independent members. As such a basis of \(W\) is easily produced by inspection. We choose \(\boldv_1=(1,-1,0)\) and \(\boldv_2=(1,0,-1)\) as our basis for \(W\text{.}\) Adding the normal vector \(\boldv_3=(1,1,1)\) then yields the ordered basis
\begin{equation*}
B'=\left( \underset{\boldv_1}{(1,-1,0)}, \underset{\boldv_2}{(1,0,-1)}, \underset{\boldv_3}{(1,1,1)} \right)
\end{equation*}
of \(\R^3\text{.}\)
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
(4.4) 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 (
(4.5)) 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{4.7}
\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
(4.5) reads as
\begin{equation}
A'[\boldx]_{B'}=[\proj{\boldx}{W}]_{B'}\text{.}\tag{4.8}
\end{equation}
Formula
(4.8) 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 to
\(B'\)-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
\(W=\Span\{(1,-1,0),(1,0,-1)\}\) and sends everything in
\(W^\perp=\Span\{(1,1,1)\}\) to
\(\boldzero\text{.}\)
