Solution to Exercise 6.4.17 from „Linear Algebra“

By Theorem 6.16 and Theorem 6.17 we get an orthonormal basis

α = {v_{1}, v_{2}, \dots, v_{n}},

where $v_{i}$ is the eigenvector with respect to the eigenvalue $λ_{i}$ , since $T$ is self-adjoint. For each vector $x$ , we may write it as

x = \sum_{i = 1}^{n} a_{i} v_{i} .

Compute

⟨ T (x), x ⟩ = ⟨ \sum_{i = 1}^{n} a_{i} λ_{i} v_{i}, \sum_{i = 1}^{n} a_{i} v_{i} ⟩

= \sum_{i = 1}^{n} | a_{i} |^{2} λ_{i} .

The value is greater than [no less than] zero for arbitrary set of $a_{i}$ ’s if and only if $λ_{i}$ is greater than [no less than] zero for all $i$ .

Denote

β

to be

{e_{1}, e_{2}, \dots, e_{n}} .

For each $x \in V$ , we may write it as

x = \sum_{i = 1}^{n} a_{i} e_{i} .

Also compute

⟨ T (x), x ⟩ = ⟨ \sum_{i = 1}^{n} (\sum_{j = 1}^{n} A_{ij} a_{j}) e_{i}, \sum_{i = 1}^{n} a_{i} e_{i} ⟩

= \sum_{i = 1}^{n} (\sum_{j = 1}^{n} A_{ij} a_{j}) \bar{a_{i}} = \sum_{i, j} A_{ij} a_{j} \bar{a_{i}} .

Since

T

is self-adjoint, by Theorem 6.16 and 6.17 we have

A = P^{∗} DP

for some matrix

P

and some diagonal matrix

D

. Now if

T

is positive semidefinite, we have all eigenvalue of

T

are nonnegative. So the

ii

-entry of

D

is nonnegative by the previous argument. We may define a new diagonal matrix

E

whose

ii

-entry is

\sqrt{D_{ii}}

. Thus we have

E^{2} = D

and

A = (P^{∗} E) (EP)

. Pick

B

to be

EP

and get the partial result.

Conversely, we may use the result of the previous exercise. If $y = (a_{1}, a_{2}, \dots, a_{n})$ is a vector in $𝔽^{n}$ , then we have

y^{∗} Ay = \sum_{i, j} A_{ij} a_{j} \bar{a_{i}}

and

y^{∗} Ay = y^{∗} B^{∗} By = {(By)}^{∗} By = ∥ By ∥^{2} \geq 0 .

Since

T

is self-adjoint, there’s a basis

β

consisting of eigenvectors of

T

. For all

x \in β

, we have

U^{2} (x) = T^{2} (x) = λ^{2} x .

If $λ = 0$ , then we have $U^{2} (x) = 0$ and so $U (x) = 0 = T (x)$ since

⟨ U (x), U (x) ⟩ = ⟨ U^{∗} U (x), x ⟩ = ⟨ U^{2} (x), x ⟩ = 0 .

By the previous arguments we may assume that $λ > 0$ . And this means that

0 = (U^{2} - λ^{2} I) (x) = (U + λI) (U - λI) (x) .

But $\det (U + λI)$ cannot be zero otherwise the negative value $- λ$ is an eigenvalue of $U$ . So we have $U + λI$ is invertible and $(U - λI) (x) = 0$ . Hence we get $U (x) = λx = T (x)$ . Finally since $U$ and $T$ meet on the basis $β$ , we have $U = T$ .

We have

T

and

U

are diagonalizable since they are self-adjoint. Also, by the fact

TU = UT

and Exercise 5.4.25, we may find a basis

β

consisting of eigenvectors of

U

and

T

. Say

x \in β

is an eigenvector of

T

and

U

with respect to

λ

and

μ

, who are nonnegative since

T

and

U

are positive definite. Finally, we get that all eigenvalue of

TU

is nonnegative since

TU (x) = λμx

. So

TU = UT

is also positive definite since they are self-adjoint by Exercise 6.4.4.

Follow the notation of Exercise 6.4.17(b) and denote

y = (a_{1}, a_{2}, \dots, a_{n})

. We have

⟨ T (\sum_{i = 1}^{n} a_{i} e_{i}), \sum_{i = 1}^{n} a_{i} e_{i} ⟩ = ⟨ \sum_{i = 1}^{n} (\sum_{j = 1}^{n} A_{ij} a_{j}) e_{i}, \sum_{i = 1}^{n} a_{i} e_{i} ⟩

\sum_{i, j} A_{ij} a_{j} \bar{a_{i}} = y^{∗} Ay = ⟨ L_{A} (y), y ⟩ .

So the statement is true.

Exercise 6.4.17

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Exercise 6.4.17

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