Page 307 - 35Linear Algebra

17.1 Projection Matrices 307

MX = V r . We learned to find solutions to this in the previous subsection of
this book.
But here is another question, how can we determine what V r is given M
and V ? The answer is simple; suppose X is a solution to MX = V r . Then

T
T
T
T
MX = V r =⇒ M Mx = M V r =⇒ M Mx = M (V r + 0)
T
T
T
T
T
−1
T
=⇒ M Mx = M (V r +V k ) =⇒ M Mx = M V =⇒ X = (M M) M V
T
if indeed M M is invertible. Since, by assumption, X is a solution
T
−1
T
M(M M) M V = V r .
−1
T
T
That is, the matrix which projects V onto its ran M part is M(M M) M .
       
1  1 1  1 1
,
1
1
Example 154 To project   onto span    −1  = ran  1 −1  multi-
1  0 0  0 0
ply by the matrix
     −1
1 1 1 1
1 1 0 1 1 0
1 −1 1 −1
    
1 −1 0 1 −1 0
0 0 0 0
 
1 1 −1
2 0 1 1 0
=  1 −1 
0 2 1 −1 0
0 0
   
1 1 2 0 0
1 1 1 0 1
=  1 −1  =  0 2 0  .
2 1 −1 0 2
0 0 0 0 0
This gives
     
2 0 0 1 1
1
1
1
 0 2 0    =   .
2
0 0 0 1 0
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