Lecture 14 - Starting Positive Operators and Isometries

We start with a definition. Keep in mind we keep talking about these in the finite-dimensional case.

positive operator

An operator $T \in L (V)$ is called positive if $T$ is self-adjoint and:

⟨ T v, v ⟩ \geq 0

for all $v \in V$ .

This is sometimes called positive-semidefinite. If the $\geq$ becomes a $>$ then it's just called positive-definite. Notice that we need $T$ to be self-adjoint because then we can make these comparisons as then $⟨ T v, v ⟩ \in R$ .

Note

If the field is $R$ then you can drop the part requiring $T$ being self-adjoint as then $⟨ T v, v ⟩ \in R$ is true no matter what.

Example

Given a self-adjoint operator $T$ , I claim that the operator $T^{2} + I$ is positive. We can check this for any $v \in V$ :

\begin{aligned} ⟨ (T^{2} + I) v, v ⟩ & = ⟨ T^{2} v, v ⟩ + ⟨ v, v ⟩ \\ = ⟨ T v, T v ⟩ + ⟨ v, v ⟩ & (T = T^{*}) \\ = ‖ T v ‖ + ‖ v ‖ \\ \geq 0 \end{aligned}

where the last part comes from the norm being non-negative.

We can actually show that if we have any $p (T)$ , that this process works no matter what. This is similar to the process we did for Chapter 7 - Operators on Inner Product Spaces#^94dc7b.

Example

Let $T \in L (R^{2})$ defined by:

T (x_{1}, x_{2}) = [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]

we'll show $T$ is positive:

\begin{aligned} ⟨ T (x_{1}, x_{2}), (x_{1}, x_{2}) ⟩ & = ⟨ (x_{1} - x_{2}, - x_{1} + x_{2}), (x_{1}, x_{2}) ⟩ \\ = x_{1} (x_{1} - x_{2}) + x_{2} (x_{2} - x_{1}) \\ = x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2} \\ = (x_{1} - x_{2})^{2} \\ \geq 0 \end{aligned}

since the square of any number is positive.

We'll talk a bit about the results of positive operators. But one definition:

square root of an operator

An operator $R$ is called a square root of an operator $T$ if $R^{2} = T$ .

Example

Rotation by $π / 4$ radians is the square root of the $π / 2$ radian rotation (over the same direction). Notice that it is also the square root of a rotation by $5 π / 4$ rotations.

This idea is similar to how we do $\sqrt{4} = 2$ . We normally don't write $\sqrt{4} = \pm 2$ because there's no context of $x$ that we are plugging in (the $\sqrt{}$ symbol implies non-negative outputs).