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_TZ_6393-periastron.dvi MATH 3221 ADVANCED LINEAR ALGEBRA DAILY ASSIGNMENT MARCH 29, 2023 41. Let T : C2 → C2 be the operators whose matrix with respect to the standard basis is [ 1 i i...

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_TZ_6393-periastron.dvi
MATH 3221 ADVANCED LINEAR ALGEBRA
DAILY ASSIGNMENT MARCH 29, 2023
41. Let T : C2 → C2 be the operators whose matrix with respect to the standard basis is
[
1 i
i 1
]
.
Show that T is normal (though it is neither self-adjoint, skew-adoint, or unitary), and find
an orthonormal basis of C2 of eigenvectors of T .
42. Here is one characterisation of normal operators: T is normal iff T = T1 + iT2, where T1
and T2 are self-adjoint operators that commute. Show this as follows.
Let V be a complex inner-product space, and T a linear operator on V . Define
T1 =
1
2
(
T + T

)
and T1 =
1
2i
(
T − T

)
.
(a) Show that T1 and T2 are self-adjoint and that T = T1 + iT2.
(Note the analogy to the real and imaginary parts of a complex number, if we think
of matrix adjoint as analogous to complex conjugation.)
(b) Show that this decomposition is unique — that is, show that if T = U1 + iU2 with
oth U1, U2 self-adjoint, then U1 = T1 and U2 = T2.
(c) Prove that T is normal iff T1 and T2 commute.
Thus, normal operators are ones whose “real part” and “imaginary part” commute —
ut only when we interpret “real part” and “imaginary part” properly. (In particular,
this does not mean that taking the real and imaginary parts of the individual entries of
a matrix will give matrices that commute!) I don’t know if that helps with the intuition
much, but it’s interesting!
Answered Same Day Mar 31, 2023

Solution

Aditi answered on Mar 31 2023
27 Votes
Solution
1. To show that T is normal, we need to show that TT = TT, where T* is the conjugate transpose of T.
T* = 1 -i -i 1
T*T = 1 i -i 1 × 1 -i -i 1 = 2 0 0 2
TT* = 1 i i 1 × 1 -i -i 1 = 2 0 0 2
Since TT = TT, T is normal.
To find the eigenvectors of T, we need to solve the equation Tx = λx, where λ is the eigenvalue and x is the eigenvector.
(1 i) (x1) = λ(x1) (i 1) (x2) = λ(x2)
Expanding these equations gives:
x1 + ix2 = λx1 ix1 + x2 = λx2
Rea
anging these equations gives:
(1-λ)x1 + ix2 = 0 ix1 + (1-λ)x2 = 0
These equations can be written in matrix form as:
(1-λ) i x1 0 i (1-λ) 0 x2
To find the eigenvalues, we need to solve for when the determinant of this matrix is 0:
det( (1-λ) i x1 0 i (1-λ) 0 x2 ) = 0
Expanding this determinant gives:
(1-λ) (1-λ) - ix1ix2 = 0
Simplifying this equation gives:
λ2 - 2λ +...
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