Page 268 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes 2 2 2
= E E

2
( E )
with strict inequality when – E . Therefore, – E is the best approximation to by vectors
in W .

3
Example 14: Given R the standard inner product. Then the orthogonal projection of
(– 10, 2, 8) on the subspace W that is spanned by (3, 12, –1) is vector
(( 10, 2, 8)|(3, 12, 1))
= (3, 12, 1)
9 144 1

14
= (3, 12, 1).
154
3
The orthogonal projection of R on W is the linear transformation E defined by
3x 12x x
,
(x x x ) 1 2 3 (3, 12, 1).
1 2 3
154
The rank of E is clearly 1; hence its nullity is 2. On the other hand,
E ( , x x ) (0, 0, 0)
x
,
1 2 3
if and only if 3x + 12x – x = 0. This is the case if and only if (x , x , x , is in W . Therefore, W .
1 2 3 1 2 3
is the null space of E, and dim W 2. Computing
3x 12x x
x , x x 1 2 3 3, 12, 1
1 2 3
154
we see that the orthogonal projection of R on W is the linear transformation I – E that maps the
3
vector (x , x , x ) onto the vector
1 2 3
1
145x 1 36x 2 3x 3 36x 1 10x 2 12x 3 , 3x 1 12x 2 153x 3 .
154
The observations made in Example 14 generalize in the following fashion.
Theorem 5: Let W be a finite-dimensional subspace of an inner product space V let E be the
orthogonal projection of V on W. Then E is an idempotent linear transformation of V onto W,
W is the null space of E, and

V = W W .
Proof: Let be an arbitrary vector in V. Then E is the best approximation to that lies in W. In
particular, E = when is in W. Therefore, E(E ) = E for every in V; that is, E is idempotent:
2
E = E. To prove, that E is a linear transformation, let and be any vectors in V and c an
arbitrary scalar. Then, by Theorem 4, – E and – E are each orthogonal to every vector in W.
Hence the vector

( c E ) ( E ) (c ) (cE E )

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