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Exercise 6.2.16

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1.

Let

W = span (S)

. If

u

is an element in

W

, who is finite-dimensional, then by Exercise 6.2.15(a) we know that

∥ u ∥^{2} = \sum_{i = 1}^{n} | ⟨ u, v_{i} ⟩ |^{2} .

Now for a fixed $x$ , we know that $W^{'} = span (W \cup {x})$ is finite-dimensional. Applying Exercise 6.2.10, we have $T (x) \in W$ and $∥ T (x) ∥ \leq ∥ x ∥$ . This means

∥ x ∥^{2} \geq ∥ T (x) ∥^{2} = \sum_{i = 1}^{n} | ⟨ T (x), v_{i} ⟩ |^{2}

by our discussion above. Ultimately, by the definition of $T$ , we have $x = T (x) + y$ for some $y$ who is orthogonal to all the elements in $W$ . Thus we have

⟨ x, v_{i} ⟩ = ⟨ T (x), v_{i} ⟩ + ⟨ y, v_{i} ⟩ = ⟨ T (x), v_{i} ⟩ .

So the inequality holds.

2.

We’ve explained it.

jephianlin

2011-06-27 00:00

Exercise 6.2.16

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