Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 6.6.5

Exercise 6.6.5

Answers

1.

Since

T

is an orthogonal projection, we may write

V = R (T) \oplus R {(T)}^{⊥}

. So for each

x \in V

we could write

x = u + v

such that

u \in R (T)

and

v \in R {(T)}^{⊥}

. So we have

∥ T (u + v) ∥ = ∥ u ∥ \leq ∥ u + v ∥ = ∥ x ∥ .

The example for which the in which the inequality does not hold is $T (a, b) = (a + b, 0)$ , since we have

∥ T (1, 1) ∥ = ∥ (2, 0) ∥ = 2 > ∥ (1, 1) ∥ = \sqrt{2} .

Finally, if the equality holds for all $x \in V$ , then we have $∥ u ∥ = ∥ u + v ∥$ . Since $u$ and $v$ are orthogonal, we have

∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2} .

So the equality holds only when $v = 0$ . This means that $x$ is always an element in $R (T)$ and so $R (T) = V$ . More precisely, $T$ is the identity mapping on $V$ .

2.

T

is a projection on

W

along

W^{'}

, we have

V = W \oplus W^{'}

. So every vector

x \in V

could be written as

x = u + v

such that

u \in W

and

v \in W^{'}

. If

W^{'} \neq W^{⊥}

, we may find some

u \in W

and

v \in W^{'}

such that they are not orthogonal. So

⟨ u, v ⟩

is not zero. We may pick

t = \frac{2 ∥ v ∥^{2}}{2 ℜ ⟨ u, v ⟩}

and calculate that

∥ T (tu + v) ∥^{2} = ∥ tu ∥^{2} .

But now we have

∥ tu + v ∥^{2} = ∥ tu ∥^{2} + 2 ℜ ⟨ tu, v ⟩ + ∥ v ∥^{2}

= ∥ tu ∥^{2} - ∥ v ∥^{2} < ∥ T (tu + v) ∥^{2} .

So $T$ must be an orthogonal projection.

jephianlin

2011-06-27 00:00

Exercise 6.6.5

Answers

Comments

Add answer