Lecture 15 - Finishing 7C, More Applications of Linear!

Recall how we defined a positive operator:

positive operator

An operator $T \in L (V)$ is called positive if $T$ is self-adjoint and:
$⟨ T v, v ⟩ \geq 0$
for all $v \in V$ .

Recall we needed $⟨ T v, v ⟩ \in R$ and thus required that $T$ to be self-adjoint.

Like with complex numbers, we want to show how $\sqrt{}$ 's of operators stay within this positive realm.

Properties of positive operators

$T \in L (V)$ is self-adjoint, the following are equivalent:

$T$ is positive
$T$ is self-adjoint and all eigenvalues of $T$ are non-negative.
$T$ has a positive square root (we will show that this is unique later).
$T$ has a self-adjoint square root.
There exists an operator $R$ such that $T = R^{*} R$ .

Proof
We prove (a) $\to$ (b) $\to$ ... $\to$ (e) $\to$ (a). Notice that:

(d) implies (e) is pretty obvious.
(c) implies (d) comes from the fact that $T$ 's square root $R$ must be self-adjoint as $T$ is already self-adjoint.

(a) $\to$ (b): Suppose $T$ is positive. Let $T v = λ v$ for some non-zero vector $v$ . Then:

⟨ T v, v ⟩ = ⟨ λ v, v ⟩ = λ ⟨ v, v ⟩ \geq 0

via the definition. Since $⟨ v, v ⟩ = ‖ v ‖^{2} \geq 0$ so then $λ \geq 0$ and we are done.

(b) $\to$ (c): There is an orthonormal eigenbasis $β = {e_{1}, . . ., e_{n}}$ for $V$ (via the Spectral Theorem) w/ e-vals $λ_{1}, . . ., λ_{n}$ . Then:

M (T, β) = [\begin{matrix} λ_{1} & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & λ_{n} \end{matrix}] = {[\begin{matrix} \sqrt{λ_{1}} & 0 & \dots & 0 \\ 0 & \sqrt{λ_{2}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & \sqrt{λ_{n}} \end{matrix}]}^{2} = M (R, β)^{2}

where $R$ is given by $M (R, β)$ . This is allowed since each $λ_{i} \geq 0$ so taking the square root is fine here. $R$ is $T$ 's positive square root, where the positivity comes from $v = \sum_{i = 1}^{n} α_{i} e_{i}$ so then:

⟨ R v, v ⟩ = ⟨ \sum_{i = 1}^{n} α_{i} \sqrt{λ_{i}} e_{i}, \sum_{i = 1}^{n} α_{i} e_{i} ⟩ = \sum_{i = 1}^{n} | α_{i} |^{2} \sqrt{λ_{i}} \geq 0

(e) $\to$ (a): Suppose there is $R \in L (V)$ such that $T = R^{*} R$ :

⟨ T v, v ⟩ = ⟨ R^{*} R v, v ⟩ = ⟨ R v, R v ⟩ = ‖ R v ‖^{2} \geq 0

so $T$ is positive.
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Uniqueness of a positive square root of an operator

The positive square root of a positive operator is unique.

Proof
Use $R$ from the previous proof. We'll show that $R$ is unique. Suppose $R^{2} = T$ for some positive operators $R, T$ . Start with $T$ (a positive operator, thus self-adjoint and ...), given an orthonormal eigenbasis for $T$ , say $e_{1}, . . ., e_{n}$ .

We'll show that any $R e_{i} = \sqrt{λ_{i}} e_{i}$ (this is the operator defined in (b) $\to$ (c) in the previous proof). Since $R$ is positive, then it has its own orthonormal eigenbasis $f_{1}, . . ., f_{n}$ with eigenvalues $\sqrt{β_{1}}, . . ., \sqrt{β_{n}}$ (notice that these are still arbitrary and positive, per our requirement). Write:

\begin{aligned} e_{i} = \sum_{j = 1}^{n} α_{j} f_{j} \\ \Rightarrow R^{2} (e_{i}) = R^{2} (\sum_{j = 1}^{n} α_{j} f_{j}) = \sum_{j = 1}^{n} α_{j} R^{2} (f_{j}) = \underset{subtract}{\underset{⏟}{\sum_{j = 1}^{n} α_{j} β_{j} f_{j}}} = T (e_{i}) = λ_{i} e_{i} = \underset{subtract}{\underset{⏟}{\sum_{j = 1}^{n} α_{j} λ_{i} f_{j}}} \\ \Rightarrow \sum_{j = 1}^{n} (α_{j} β_{j} - α_{j} λ_{i}) f_{j} = \vec{0} \\ \Rightarrow α_{j} (β_{j} - λ_{i}) = 0 \end{aligned}

Thus $α_{j} = 0$ or $β_{j} = λ_{i}$ for each $j$ . Note that $e_{i} = \sum_{j = 1}^{n} α_{j} f_{j}$ , so throw out the terms without the $α_{j} = 0$ :

e_{i} = \underset{where β_{j} = λ_{i} as a condition}{\underset{⏟}{\sum_{j = 1}^{n} α_{j} f_{j}}}

R (e_{i}) = \sum_{j = 1}^{n} α_{j} R (f_{j}) = \underset{β_{j} = λ_{i}}{\underset{⏟}{\sum_{j = 1}^{n} α_{j} β_{j} f_{j}}} = \sum_{i = 1}^{n} α_{j} λ_{i} f_{j} = λ_{i} \sum_{i = 1}^{n} α_{j} f_{j} = λ_{i} e_{i}

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