Basically, a square matrix (one with the same number of rows and columns) is called Hermitian if it has complex numbers, and if:

*A = A**i.e. the matrix is found equal to its conjugate transpose. We already saw how to obtain the transpose of a matrix for normal, non-complex matrices, e.g. if X=

(-1.....2)

(-2.....3)

Then: t^(X) =

(-1.....-2)

(2.......3)

Now, we need to reckon in the conjugate transpose such that: A* = t^(A') (Which we will soon see in some examples). Meanwhile, 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*where I denotes the identity matrix. For example, for a 2 x 2 matrix, I =

(1...0)

(0....1)

Example: Show that the matrix M =

(2.....i)

(-i....5)

is Hermitian

We first get the complex conjugate of the matrix or, M' =

(2.....-i)

(i......5)

Then we obtain the transpose of that, or M* = t^(M') =

(2.....i)

(-i......5)

and since M = M* = t^(M')

the matrix is Hermitian.

Example (2). Show that the matrix A =

(cos Θ ..... sin Θ)

(-sin Θ ..... cos Θ)

is a unitary matrix.

By definition, the matrix A (a real matrix, not complex) will be unitary if: A ( t^A) = 1

We note that the transpose t^A =

(cos Θ ..... -sin Θ)

(sin Θ ..... cos Θ)

and: A (t^A) =

(cos Θ ..... sin Θ) (cos Θ ..... -sin Θ)

(-sin Θ ..... cos Θ)(sin Θ ..... cos Θ)

= 1

Hence, the matrix is unitary.

Problems:

1) Show that the matrix M =

(1 + i.....2)

(2..........5i)

is not Hermitian

2) Determine whether the matrix Y =

(1.....(1+ 1i).......5)

((1- i).......2... ...i)

(5..........-i...........7)

is Hermitian or not.

3) Determine whether the matrix, X =

(-i...1)

(1.....i)

is unitary or not.

4) Let A and B be 2 x 2 Hermitian matrices. Show that (A + B) is Hermitian.

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