Find a unitary matrix U so that for
A = [ 2, 3-3i] [3+3i, 5],A = UDU^H, where D is diagonal
Explain why the determinant of A has to have real determinant
note that A is hermetian, so A = A*.
note as well that A* = (UDU*)* = U**D*U* = UD*U*.
this means that D* = U*A*U = U*AU = D.
now det(A) = det(UDU*) = det(D). since D = diag(d_{11},d_{22}),
D* = diag(d_{11}*,d_{22}*), and det(D) = det(D*) thus means:
d_{11}d_{22} = (d_{11})*(d_{22})* = (d_{11}d_{22})*
hence det(A) = det(D) is real.
in this particular case, of course, one can just compute det(A):
det(A) = (2)(5) - (3+3i)(3-3i) = 10 + 9 - 9i^{2} = 10 + 9 - (-9) = 10 + 18 = 28
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the above proof can be adapted easily to any nxn Hermitian matrix.
so how do we find such a matrix U?
first, lets find the eigenvalues for A:
det(xI - A) =
so the eigenvalues of A are -1 and 8. solving (A + I)u = 0 we get:
u = (-1+i,1) as an eigenvector. normalizing u we get the unit vector ((-1+i)/√3,1/√3).
solving (A - 8I)v = 0, we get:
v = (1-i,2), which normalizes to the unit vector ((1-i)/√6,2/√6). note that <u,v> = 0, so we may take:
i leave it in your capable hands to show that UAU* =
which is, of course, diagonal.