MAT247 - Sec 6.5 - Q.4 – Conversation in Mathematics

Conversation started Jan 19, 2023 at 1:27.

D.C. the III

Jan 19, 2023 01:27

For $z \in \mathbb{C}$, define $T_{z}: \mathbb{C} \to \mathbb{C}$ by $T_{z}(u) = zu$. Characterize those $z$ for which $T_z$ is normal, self adjoint, or unitary.

For $T_z$ to be unitary means $\|T_z(u)\| = \|u\|$. Therefore $\|zu\| = \|u\|$. From which I work out:

$$\|zu\| = \langle zu,zu \rangle = z\overline{z}\langle u,u \rangle$$

Which would mean that $|z|^2\langle u,u \rangle = \langle u,u \rangle$ can only occur iff $|z| = 1$.

What I'm having trouble with is showing normal and self-adjointedness. I know for a linear operator to be normal means $TT^* = T^{\star}T$ . I have linear o…

Goku カカロット

@shintuku I do it just so I can laugh

D.C. the III

as an aside, I had to write T^{\star} to get the star to appear and I couldn't just use the * symbol by itself. Is that something new?

leslie townes

certainly outside of math mode, * sometimes gets trapped for emphasis. could be that happens in math mode, too.

D.C. the III

yeah it seemed like that was happening

leslie townes

is your inner product linear in the first component or the second?

D.C. the III

Jan 19, 2023 01:29

it was fidgety. some stars appeared some didn't

leslie townes

you got the unitaries right, by the way.

D.C. the III

No specific inner produc defined

leslie townes

ok. what is the adjoint of T_z, then? you need a definition there.

you're writing <u,u> up above. what is <a,b> when a and b are not the same?

that's my question. there are two conventional answers, it doesn't really matter which one you pick, except for how you might write things out in ways that don't use < >

D.C. the III

leslie townes

presumably the inner product on C (or perhaps C^n?) would be specified earlier in that text somewhere.

D.C. the III

Jan 19, 2023 01:33

So I guess the standard inner product on C. i.e $\Sigma a_i \overline{b_i}$

leslie townes

i'm not trying to being deliberately obtuse or nitpicky here, but this is exactly the question raised by "seeing somebody somewhere do [blah]." they're implicitly using whatever the definition of the inner product actually is.

okay. so for any z, a, and b, <T_z a, b> = za conj(b) = a conj(conj(z) b) = <a, conj(z) b> = <a, T_{conj(z)} b>. using that definition of the inner product. i'm writing conj( ) because i don't want to overline or use tex or worry about * ruining my text.

so the adjoint of T_z is T_{conj(z)}.

if you were using stars, you could just pop the * from a superscript on the T and move it down to a superscript on the z.

Ted Shifrin

No, $T^*$ works fine.

D.C. the III

Ok let me write this out on paper to see it clearer

D.C. the III

Jan 19, 2023 01:52

$T_zT_{\overline{z}}^*(u) = z\overline{z}u$, likewise $T_{\overline{z}}^*T_z = \overline{z}zu$. So then this would mean that this operator is normal iff $z \in \mathbb{R}$.

because $z\overline{z} = |z|$ which is a real number

leslie townes

you've got one too many bars or stars there. T_z star, or T_{z bar}, but not both the star and the bar.

note that "is T normal?" asks only if those operators are equal, not anything about what the operator is if they happen to be equal.

so while i agree that z z-bar is real for every z, that seems somewhat unrelated to the previous line. is $z \bar{z} u = \bar{z} z u$ for all $u$?

D.C. the III

hmmmm...

I want to say yes, but you asking me this makes it feel like it is a trick question

leslie townes

Jan 19, 2023 02:13

no tricks here. just multiplication of complex numbers.

D.C. the III

Then I say yes they are equal for all $u$

Ted Shifrin

What are indices? Also, avoid $i$ ;)

Oh, it vanished.

D.C. the III

this doesn't pertain to me does it?

Ted Shifrin

You erased it.

leslie townes

dc: so T_z is normal no matter what z is.

D.C. the III

Jan 19, 2023 02:17

no it can only be normal if $z \in \mathbb{R}$

leslie townes

why? you didn't prove that up above. you remarked that whatever z is, T_z T_z* is multiplication by a real number. so what?

T_i T_i^* = T_i^* T_i is the identity. it has to be, because you already told us that T_z was unitary if |z| = 1. so there's a normal T_z where z isn't real.

D.C. the III

Jan 19, 2023 02:34

So if I'm understanding you correct, you're saying because I showed that $T_z$ was unitary then as a consequence I have already that $T_z$ is normal (which is a part of a theorem in the text showing the equivalent statements for something being unitary), now for this all to be true relies on $|z| = 1$ so $z$ can be complex or real just as long as $|z| = 1$ then the condition is satisfied.

Ted Shifrin

Every hermitian, shew-hermitian, or unitary operator is always normal.

D.C. the III

Aye. I agree with such

Ted Shifrin

Such agrees with you.

D.C. the III

I happened to show unitary first because I didn't know how to show normality first. WHat would I have done in the case where I hadn't shown it being unitary first but tried to show normality?

Ted Shifrin

Jan 19, 2023 02:51

That’s what leslie was talking about directly above.

leslie townes

there was a calculation up above showing that for any $z$, the operators $T_z T_z^*$ and $T_z^* T_z$ are equal, and in fact equal to $T_{|z|^2}$.

so if you hadn't done the unitary part first, you'd then ask, ok, when is $T_{|z|^2}$ the identity. that's when you get a unitary.

D.C. the III

got it.

Conversation ended Jan 19, 2023 at 2:54.

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