_TZ_6393-periastron.dviMATH 3221 ADVANCED LINEAR ALGEBRADAILY ASSIGNMENT MARCH 29, 202341. Let T : C2 → C2 be the operators whose matrix with respect to the standard basis is[1 ii...

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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 both 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 — but 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 DayMar 31, 2023

Answer To: _TZ_6393-periastron.dviMATH 3221 ADVANCED LINEAR ALGEBRADAILY ASSIGNMENT MARCH 29, 202341. Let...

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 trans
pose 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
Rearranging 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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