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_TZ_6393-periastron.dvi MATH 3221 ADVANCED LINEAR ALGEBRA DAILY ASSIGNMENT MARCH 15, 2023 My apologies for posting this so late! I was trying to make the numbers work out on the “find the adjoint...

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_TZ_6393-periastron.dvi
MATH 3221 ADVANCED LINEAR ALGEBRA
DAILY ASSIGNMENT MARCH 15, 2023
My apologies for posting this so late! I was trying to make the numbers work out on the
“find the adjoint of the derivative operator D : P3 → P3” problem, and I just couldn’t get it to
work, so I’m not going to assign it. My apologies, if you were looking forward to seeing how
it worked out, but it seemed to be a lot uglier than the dual space ev2 one. If I can make it
tractable we might revisit it on a future problem set, but for now I’ve just given you a (easy)
theoretical problem instead.
35. Fix an invertible n×n matrix P , and define TP : Mn(C) → Mn(C) by TP (A) = P
−1AP . Find
the adjoint of TP . (Recall that the inner product on Mn(C) is given by 〈A,B〉 = tr(A
∗B).)
Under what conditions will TP be self-adjoint?
36. Let V be a complex inner-product space, not necessarily finite-dimensional.
(a) If T is a self-adjoint operator on V , prove that 〈Tv, v〉 is real for every v ∈ V .
Use this to show that all eigenvalues of T are real. (Recall that λ is an eigenvalue of T
if there exists a nonzero vector v with Tv = λv.)
(b) If T is a unitary operator on V , prove that all eigenvalues of T have absolute value 1.
Note that the possibility of V being infinite-dimensional means you have to work with
the operator directly, not with a matrix.
(c) If V is finite-dimensional, prove that the converse to (a) holds: If T : V → V satisfies
〈Tv, v〉 ∈ R for all v ∈ V , then T is self-adjoint.
Answered Same Day Mar 16, 2023

Solution

Dr. Saloni answered on Mar 16 2023
33 Votes
1
Math 3221
35.
To find the adjoint of TP , we need to find the operator T∗P such that hTP (A), Bi = hA, T∗P (B)
for all A, B ∈ Mn(C). Using the definition of TP , we have:
hTP (A), Bi = hP−1AP, Bi = tr((P−1AP)∗B) = tr(P−1A∗PB)
= tr(A∗PBP−1) = hA, PBP−1i.
Thus, we need T∗P = PBP−1, where B is the adjoint of P−1, i.e., P−1B = BP−1 = adj(P−1).
If TP is self-adjoint, then we have T∗P = TP , which implies that PBP−1 = P−1AP for all A ∈ Mn(C).
Multiplying both sides by P on the left and P−1 on the right, we get BP−1 = P−1A(adj(P))P−1.
Taking adjoints on both sides, we get B = P−1(adj(P))P−1A∗. Thus, B is uniquely determined by
P and is...
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