HW 5 - Positive Operators and Isometries

7.C: Positive Operators and Isometries

1

Question

Prove or give a counterexample: If $T \in L (V)$ is self-adjoint and there exists an orthonormal basis $e_{1}, . . ., e_{n}$ of $V$ such that $⟨ T e_{j}, e_{j} ⟩ \geq 0$ for each $j$ , then $T$ is a positive operator.

Proof
This is false via the following explanation. The proof prior shows where the proof would fail in the process.

Notice that since $e_{1}, . . ., e_{n}$ is an orthonormal basis of $V$ then any $v \in V$ has:

v = \sum_{i = 1}^{n} α_{i} e_{i}

Thus, for the condition of $T$ being positive, we already have $T$ is self-adjoint, and for positivity:

\begin{aligned} ⟨ T v, v ⟩ & = ⟨ \sum_{i = 1}^{n} α_{i} T (e_{i}), \sum_{i = 1}^{n} α_{i} e_{i} ⟩ \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} α_{i} \overset{―}{α_{j}} ⟨ T e_{i}, e_{j} ⟩ \end{aligned}

which is where the proof would have to stop. Thus we want a $⟨ T e_{i}, e_{j} ⟩ < 0$ for all $i, j$ .

Define $V = R^{2}$ and $T$ is the matrix over the standard orthonormal basis $β = {e_{1}, e_{2}}$ given by:

M (T, β) = [\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix}]

Notice $M (T^{*}, β) = M (T, β)^{*} = M (T, β)$ so $T = T^{*}$ and thus is self-adjoint. However:

⟨ T e_{1}, e_{1} ⟩ = 0 = ⟨ T e_{2}, e_{2} ⟩ \geq 0

per our requirement while:

⟨ T e_{1}, e_{2} ⟩ = - 1 = ⟨ T e_{2}, e_{1} ⟩

Thus:

⟨ T v, v ⟩ = \sum_{i = 1}^{2} \sum_{j = 1}^{2} α_{i} α_{j} ⟨ T e_{i}, e_{j} ⟩ < 0

since each $⟨ T e_{i}, e_{j} ⟩ \leq 0$ while we have at least one $⟨ T e_{1}, e_{2} ⟩ < 0$ .
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2

Question

Suppose $T$ is a positive operator on $V$ . Suppose $v, w \in V$ are such that:

T v = w

and:

T w = v

Then $v = w$ .

Proof
Since $T$ is positive, then since each positive operator has only one positive square root, then it's unique. Notice that in this specific case for $v$ :

T (T v) = T^{2} v = T w = v \Leftrightarrow (T^{2} - I) v = 0 = (T + I) (T - I) v

thus $v$ is an eigenvector of $T^{2}$ . Notice that if $v$ or $w$ are $\vec{0}$ then the other vector must be the zero vector, proving $v = w = \vec{0}$ . Now if $v, w \neq \vec{0}$ then they are valid eigenvectors of $T^{2}$ with $λ = 1$ .

Notice further that $v$ is an eigenvector of $T$ with $λ = \pm 1$ , but it can't be the negative because if it was then:

⟨ T v, v ⟩ = ⟨ - v, v ⟩ = - ⟨ v, v ⟩ \leq 0

which would be a contradiction to $T$ being positive. Thus $v$ is an eigenvector with $λ = 1$ and thus $T v = v = w$ so $v = w$ .
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4

Question

Suppose $T \in L (V, W)$ . Prove that $T^{*} T$ is a positive operator on $V$ and $T T^{*}$ is a positive operator on $W$ .

Proof
To show that we get positivity, first:

\begin{aligned} ⟨ T^{*} T v, v ⟩ & = ⟨ T v, T v ⟩ \\ \geq 0 \end{aligned}

\begin{aligned} ⟨ T T^{*} w, w ⟩ & = ⟨ T^{*} w, T^{*} w ⟩ \\ \geq 0 \end{aligned}

And then for the self-adjointness notice that:

(T^{*} T)^{*} = T^{*} (T^{*})^{*} = T^{*} T

(T T^{*})^{*} = (T^{*})^{*} T^{*} = T T^{*}

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5

Question

Prove that the sum of two positive operators on $V$ is positive.

Proof
Consider $T_{1}, T_{2} \in L (V)$ both positive. As such $T_{1}, T_{2}$ are self-adjoint. Notice:

⟨ (T_{1} + T_{2}) v, v ⟩ = ⟨ T_{1} v + T_{2} v, v ⟩ = \underset{\geq 0}{\underset{⏟}{⟨ T_{1} v, v ⟩}} + \underset{\geq 0}{\underset{⏟}{⟨ T_{2} v, v ⟩}} \geq 0

Thus $T_{1} + T_{2}$ has the positive-definiteness. We then can show $T_{1} + T_{2}$ is self adjoint because:

\begin{aligned} ⟨ (T_{1} + T_{2}) v, v ⟩ & = ⟨ T_{1} v, v ⟩ + ⟨ T_{2} v, v ⟩ \\ = ⟨ v, T_{1} v ⟩ + ⟨ v, T_{2} v ⟩ \\ = ⟨ v, (T_{1} v + T_{2} v ⟩ \\ = ⟨ v, (T_{1} + T_{2}) v ⟩ \end{aligned}

Thus $T_{1} + T_{2}$ is self-adjoint.
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6

Question

Suppose $T \in L (V)$ is positive. Then $T^{k}$ is positive for all $k \in Z^{+}$ .

Proof
Induction over $k$ . For $k = 1$ the base case is trivial. For $k = 2$ case (first positivity then self-adjointness):

⟨ T^{2} v, v ⟩ = ⟨ T v, T v ⟩ \geq 0

⟨ T^{2} v, v ⟩ = ⟨ T v, T v ⟩ = ⟨ v, T^{2} v ⟩

Both using the fact that $T$ is positive and moving it to the other side.

Now suppose that the theorem holds for all values of $k - 1$ or less. We'll show the $k$ case. Notice first $T^{n}$ is self-adjoint because using the fact that $T$ is self-adjoint:

\begin{aligned} ⟨ T^{n} v, v ⟩ & = ⟨ T^{n - 1} T v, v ⟩ \\ = ⟨ T v, T^{n - 1} v ⟩ & Inductive Hypothesis (k - 1) \\ = ⟨ v, T T^{n - 1} v ⟩ & T is positive \\ = ⟨ v, T^{n} v ⟩ \end{aligned}

Thus $T^{n}$ is self-adjoint.

For the positivity:

\begin{aligned} ⟨ T^{k} v, v ⟩ & = ⟨ T^{k - 1} v, T v ⟩ \\ = ⟨ T^{k - 2} w, w ⟩ & T v = w \\ \geq 0 & Inductive Hypothesis k - 2 \end{aligned}

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7

Question

Suppose $T$ is a positive operator on $V$ . Then $T$ is invertible if and only if:

⟨ T v, v ⟩ > 0

for every $v \in V$ with $v \neq 0$ .

Proof
We just have to show either injective or surjective to have it be equivalent for invertibility, since $T \in L (V)$ ( $T$ is specifically an operator on one vector space). We'll do injectivity.

( $\to$ ): $T$ is invertible. Let $v \in V$ where $v \neq 0$ . Since $T$ is positive then:

⟨ T w, w ⟩ \geq 0

for every $w \in V$ . Assume for contradiction $⟨ T v, v ⟩ = 0$ . As a property, there is a (unique) operator $R \in L (V)$ where $T = R^{*} R$ where $R$ is positive (and self-adjoint). Thus:

\begin{aligned} ⟨ T v, v ⟩ & = ⟨ R^{*} R v, v ⟩ \\ = ⟨ R v, R v ⟩ \\ = ‖ R v ‖^{2} \\ = 0 \end{aligned}

Thus $R v = 0$ but then $R^{2} v = 0$ but then that means $T v = 0$ and since $T$ is invertible then that implies $v = 0$ which is a contradiction.

Notice that since $v \neq 0$ then

( $\leftarrow$ ): Suppose $⟨ T v, v ⟩ > 0$ for any $v \in V ∖ {\vec{0}}$ . So $T v \neq 0$ so then $v \notin null (T)$ . Thus since $v$ was arbitrary non-zero vector then only $null (T) = {\vec{0}}$ so $T$ is injective and thus invertible.
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8

Question

Suppose $T \in L (V)$ . For $u, v \in V$ define ${⟨ u, v ⟩}_{T}$ by:

{⟨ u, v ⟩}_{T} = ⟨ T u, v ⟩

Prove that ${⟨ \cdot, \cdot ⟩}_{T}$ is an inner product on $V$ iff $T$ is an invertible positive operator (with respect to the original inner product $⟨ \cdot, \cdot ⟩$ )

Proof
( $\to$ ): First we know ${⟨ \cdot, \cdot ⟩}_{T}$ is an inner product. Notice that we only need to show injectivity for invertibility since $T$ is an operator. Let $v \in V$ be arbitrary vector. Let $T v = 0$ thus $v \in null (T)$ . We'll show $v = 0$ because:

{⟨ v, v ⟩}_{T} = ⟨ T v, v ⟩ = 0 \Rightarrow {⟨ v, v ⟩}_{T} = 0

so since ${⟨ \cdot, \cdot ⟩}_{T}$ is an inner product we must have $v = 0$ . Thus $null (T)$ is trivial and thus $T$ is injective and thus invertible.

For positivity, showing that the inner product is always positive-definite is easy because:

{⟨ v, v ⟩}_{T} = ⟨ T v, v ⟩ \geq 0

because ${⟨ \cdot, \cdot ⟩}_{T}$ has the property of positive-definiteness. For $T$ being self-adjoint:

\begin{aligned} {⟨ u, v ⟩}_{T} & = \overset{―}{{⟨ v, u ⟩}_{T}} \\ ⟨ T u, v ⟩ & = \overset{―}{⟨ T v, u ⟩} \\ = ⟨ u, T v ⟩ \end{aligned}

Thus $T$ is self-adjoint.

( $\leftarrow$ ): Suppose $T$ is invertible and positive. We show ${⟨ \cdot, \cdot ⟩}_{T}$ is an inner product:

Positive-Definiteness: If $v \neq 0$ then:

{⟨ v, v ⟩}_{T} = ⟨ T v, v ⟩ > 0

via HW 5 - Positive Operators and Isometries#7. Clearly if $v = 0$ then $⟨ T v, v ⟩ = 0$ .

Linearity:

\begin{aligned} {⟨ α v + β u, w ⟩}_{T} & = ⟨ T (α v + β u), w ⟩ \\ = ⟨ α T v + β T u, w ⟩ \\ = α ⟨ T v, w ⟩ + β ⟨ T u, w ⟩ \\ = α {⟨ v, w ⟩}_{T} + β {⟨ u, w ⟩}_{T} \end{aligned}

Conjugate Symmetry:

\begin{aligned} {⟨ u, v ⟩}_{T} & = ⟨ T u, v ⟩ \\ = \overset{―}{⟨ v, T u ⟩} \\ = \overset{―}{⟨ T v, u ⟩} & T is self-adjoint \\ = \overset{―}{{⟨ v, u ⟩}_{T}} \end{aligned}

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10

Question

Suppose $S \in L (V)$ . Then prove that the following are equivalent:

$S$ is an isometry
$⟨ S^{*} u, S^{*} v ⟩ = ⟨ u, v ⟩$ for all $u, v \in V$
$S^{*} e_{1}, . . ., S^{*} e_{m}$ is an orthonormal list for every orthonormal list of vectors $e_{1}, . . ., e_{m}$ in $V$
$S^{*} e_{1}, . . ., S^{*} e_{n}$ is an orthonormal basis for some orthonormal basis $e_{1}, . . ., e_{n}$ of $V$ .

Proof
Suppose (a), so $S$ is an isometry. Then by the characterization of isometries, then $S^{*}$ must be an isometry. As such, the following are equivalent:

By (b) of the lemma, then $⟨ S^{*} v, S^{*} v ⟩ = ⟨ u, v ⟩$ for all $u, v \in V$
By (c) of the lemma, then $S^{*} e_{1}, . . ., S^{*} e_{m}$ is northonormal for every orthonormal list of vectors $e_{1}, . . ., e_{m}$ in $V$
By (d) of the lemma, then $S^{*} e_{1}, . . ., S^{*} e_{m}$ is an orthonormal basis for some orthonormal basis $e_{1}, . . ., e_{m}$ of $V$ .

Notice if you wanted to show the other directions, that one of the other points being true implies (a), that you can just use the characterization of isometries of any of these to imply that $(S^{*})^{*} = S$ must be an isometry.
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11

Question

Suppose $T_{1}, T_{2}$ are normal operators on $L (F^{3})$ and both operators have $2, 5, 7$ as eigenvalues. Then prove there exists an isometry $S \in L (F^{3})$ such that $T_{1} = S^{*} T_{2} S$ .

Proof
Since $dim (F^{3}) = 3$ and we have 3 eigenvalues, then we have an orthonormal eigenbasis $f_{1}, f_{2}, f_{3}$ where $λ_{1} = 2$ and $λ_{2} = 5$ and $λ_{3} = 7$ for $T_{1}$ , while $T_{2}$ has the same eigenvalues with the eigenbasis $e_{1}, e_{2}, e_{3}$ (with the same eigenvalues).

Construct $S$ such that it takes in any vector from $F^{3}$ and gives a basis representation in terms of the $f$ 's from the $e$ 's. Namely:

S f_{i} = e_{i}

Notice then that this is an isometry because since $v = \sum_{i = 1}^{3} α_{i} f_{i}$ :

\begin{aligned} ‖ S v ‖^{2} & = ⟨ S v, S v ⟩ \\ = ⟨ S (\sum_{i = 1}^{3} α_{i} f_{i}), S (\sum_{i = 1}^{3} α_{i} f_{i}) ⟩ \\ = ⟨ \sum_{i = 1}^{3} α_{i} S (f_{i}), \sum_{i = 1}^{3} α_{i} S (f_{i}) ⟩ \\ = ⟨ \sum_{i = 1}^{3} α_{i} e_{i}, \sum_{i = 1}^{3} α_{i} e_{i} ⟩ \\ = \sum_{i = 1}^{3} \sum_{j = 1}^{3} α_{i} \overset{―}{α_{j}} \underset{= χ (i = i) = 1}{\underset{⏟}{⟨ e_{i}, e_{i} ⟩}} \\ = \sum_{i = 1}^{3} \sum_{j = 1}^{3} α_{i} \overset{―}{α_{j}} ⟨ f_{i}, f_{i} ⟩ \\ = ⟨ \sum_{i = 1}^{3} α_{i} f_{i}, \sum_{i = 1}^{3} α_{i} f_{i} ⟩ \\ = ⟨ v, v ⟩ \\ = ‖ v ‖^{2} \end{aligned}

Thus $S$ is an isometry. We'll show $T_{1} = S^{*} T_{2} S$ . Notice that since $S$ is an isometry then $S^{*} = S^{- 1}$ . As a result:

S^{- 1} e_{i} = f_{i} = S^{*} e_{i}

Then take any arbitrary $v = \sum_{i = 1}^{3} α_{i} f_{i}$ and then notice:

\begin{aligned} T_{1} v & = T_{1} (\sum_{i = 1}^{3} α_{i} f_{i}) \\ = \sum_{i = 1}^{3} α_{i} T_{1} (f_{i}) \\ = \sum_{i = 1}^{3} α_{i} λ_{i} f_{i} \end{aligned}

while:

\begin{aligned} S^{*} T_{2} S v & = S^{*} T_{2} (S v) \\ = S^{*} T_{2} (\sum_{i = 1}^{3} α_{i} S (f_{i})) \\ = S^{*} T_{2} (\sum_{i = 1}^{3} α_{i} e_{i}) \\ = S^{*} \sum_{i = 1}^{3} α_{i} T_{2} (e_{i}) \\ = S^{*} \sum_{i = 1}^{3} α_{i} λ_{i} e_{i} \\ = \sum_{i = 1}^{3} α_{i} λ_{i} S^{*} (e_{i}) \\ = \sum_{i = 1}^{3} α_{i} λ_{i} f_{i} \end{aligned}

Thus since $v$ was arbitrary it follows $T_{1} = S^{*} T_{2} S$ .
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