We now examine the solutions for the previous problems:
1) Show that the matrix M =
(1 + i.....2)
(2..........5i)
is not Hermitian
Solution: Recall that a matrix is called Hermitian if it has complex numbers, and if: A = A* i.e. the matrix is found equal to its conjugate transpose
The complex conjugate of matrix M is M' =
(1 - i.....2)
(2..........-5i)
The transpose is: t^M' =
(1 - i.....2)
(2..........-5i)
But on inspection we see M does not equal t^M' , hence M is not Hermitian.
2) Determine whether the matrix Y =
(1.....(1+ 1i).......5)
((1- i).......2... ...i)
(5..........-i...........7)
is Hermitian or not.
Solution: First write the conjugate matrix, or Y' =
(1..... (1- i).....5)
((1+ i).......2... ..- i)
(5.......... i...........7)
The transpose of this 3 x 3 matrix is: t^Y' =
(1..... (1+ i).....5)
((1- i).......2... ..- 2)
(5.......... i...........7)
So by inspection the matrix Y is not Hermitian.
3) Determine whether the matrix, X =
(-i...1)
(1.....i)
is unitary or not.
Solution: Recall that we say a matrix is unitary if: A^-1 = A* i.e. if the inverse of the matrix is equal to its conjugate transpose. This also implies that we have:A A* = A* A = I
In this case, the conjugate of X is X' =
(i......1)
(1.....-i)
The transpose of this is: t^X' = X* =
(i......1)
(1.....-i)
and we see t^X = X* does not equal X, hence X is not unitary. We can check this is so, since if unitary we would expect A* A = I (the identity matrix) we therefore take: X* X =
(i......1) (-i.....1)
(1 ...-i)(1.......i) =
(2......-2i)
(2i........2)
Which is not the identity matrix.
4) Let A and B be 2 x 2 Hermitian matrices. Show that (A + B) is Hermitian.
Solution: Let A =
(a......bi)
(-bi.....c) and B =
(a......-bi)
(bi.......c)
Then: A + B =
(2a........0)
(0.........2c)
And the transpose t^(A + B) =
(2a........0)
(0.........2c)
So: t^(A + B) = A + B, and the sum is Hermitian
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