We simply repeat the process we did in (b); for any vector u, we have that u − P 1 ( u ) {\displaystyle u-P_{1}(u)} is in the plane. Thus, the projection onto the plane is given by the linear transformation P 2 = I − P 1 {\displaystyle P_{2}=I-P_{1}} , where I is the identity matrix.
So, we compute it as follows:
P 2 = [ 1 0 0 0 1 0 0 0 1 ] − 1 3 [ 1 − 1 1 − 1 1 − 1 1 − 1 1 ] = 1 3 [ 2 1 − 1 1 2 1 − 1 1 2 ] {\displaystyle P_{2}=\left[{\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\\\end{array}}\right]-{\frac {1}{3}}\left[{\begin{array}{cccc}1&-1&1\\-1&1&-1\\1&-1&1\\\end{array}}\right]={\frac {1}{3}}\left[{\begin{array}{cccc}2&1&-1\\1&2&1\\-1&1&2\\\end{array}}\right]} .
is in the plane. Thus, the projection onto the plane is given by the linear transformation 
, where I is the identity matrix.
![{\displaystyle P_{2}=\left[{\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\\\end{array}}\right]-{\frac {1}{3}}\left[{\begin{array}{cccc}1&-1&1\\-1&1&-1\\1&-1&1\\\end{array}}\right]={\frac {1}{3}}\left[{\begin{array}{cccc}2&1&-1\\1&2&1\\-1&1&2\\\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/318b1c07cb1e708eed9efaed9e1651a76fd3c0de)
.
