Solution to Linear Algebra Problems

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