Solution to Exercise 6.5.20 from „Linear Algebra“

If it’s a complex inner product space, we have

⟨ U (x), U (y) ⟩ = \sum_{k = 1}^{4} i^{k} ∥ U (x) + iU (y) ∥^{2}

\sum_{k = 1}^{4} i^{k} ∥ U (x + iy) ∥^{2} = \sum_{k = 1}^{4} i^{k} ∥ x + iy ∥^{2} = ⟨ x, y ⟩ .

If it’s a real inner product space, we may also use the above equality but take the summation only over $k = 2, 4$ .

Since

U (x + y) = U (x)

if

x \in W

an

y \in W^{⊥}

, we know that

R (U) = U (W)

. We get the desired result by applying the previous argument.

Just extend the set

{v_{1}, v_{2}, \dots, v_{k}}

to be an orthonormal basis

γ = {v_{1}, v_{2}, \dots, v_{n}}

for $V$ , where $n = \dim (V)$ . First we know that $U (v_{j}) = 0$ if $j > k$ . So the $j$ -th column of ${[U]}_{γ}$ is zero. On the other hand, if we write $A = {[U]}_{γ}$ we have

U (v_{j}) = \sum_{i = 1}^{n} U_{ij} v_{i} .

So the first $k$ columns is orthonormal since

0 = ⟨ U (v_{s}), U (v_{t}) ⟩ = ⟨ \sum_{i = 1}^{n} U_{is} v_{i}, \sum_{i = 1}^{n} U_{it} v_{i} ⟩

= \sum_{i = 1}^{n} U_{is} \bar{U_{it}} = {(A e_{t})}^{∗} A e_{s}

and

1 = ⟨ U (v_{s}), U (v_{s}) ⟩ = ⟨ \sum_{i = 1}^{n} U_{is} v_{i}, \sum_{i = 1}^{n} U_{is} v_{i} ⟩

= \sum_{i = 1}^{n} U_{is} \bar{U_{is}} = {(A e_{s})}^{∗} A e_{s} .

Since

V

is finite-dimensional inner product space, we have

R {(U)}^{⊥} \oplus R (U) = V

. And so by Exercise 6.5.20(b) we get the desired result.

First,

T

is well-defined since the set

β

defined in the previous question is a basis. To show that

T = U^{∗}

, it’s sufficient to check that

⟨ U (x), y ⟩ = ⟨ x, T (y) ⟩

for all $x$ and $y$ in $β$ by Exercise 6.1.9. We partition $β$ into two parts $X$ and $Y$ , who consist of all $U (v_{i})$ ’s and $w_{i}$ ’s.

If $x = U (v_{i}), y = U (v_{j}) \in X$ , we have
$⟨ U (v_{i}), T (U (v_{j})) ⟩ = ⟨ U (v_{i}), v_{j} ⟩ = ⟨ U^{2} (v_{i}), U (v_{j}) ⟩$

by Exercise 6.5.20(a).

If $x = U (v_{i}) \in X$ and $y = w_{j} \in Y$ , we have

⟨ U (v_{i}), T (w_{j}) ⟩ = ⟨ U (v_{i}), 0 ⟩

= 0 = ⟨ U^{2} (v_{i}), w_{j} ⟩ .

If $x = w_{i} \in X$ and $y = U (v_{j}) \in Y$ , we have
$⟨ w_{i}, T (U (v_{j})) ⟩ = ⟨ w_{i}, v_{j} ⟩ = ⟨ U (w_{i}), U (v_{j}) ⟩ .$
If $x = w_{i}, y = w_{j} \in Y$ , we have
$⟨ w_{i}, T (w_{j}) ⟩ = ⟨ w_{i}, 0 ⟩ = 0 = ⟨ U (w_{i}), w_{j} ⟩ .$

Take the subspace

W^{'}

to be

R (U)

. Thus we have

T ({(W^{'})}^{⊥}) = {0}

by the definition of

T

. Also, we may write an element

x

in

R (U)

to be

x = \sum_{i = 1}^{k} a_{i} U (v_{i}) .

Since the set of $U (v_{i})$ ’s is orthonormal, we have

∥ x ∥^{2} = \sum_{i = 1}^{k} | a_{i} |^{2} = ∥ \sum_{i = 1}^{k} a_{i} v_{i} ∥^{2} = ∥ T (x) ∥^{2} .

Exercise 6.5.20

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Exercise 6.5.20

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