# The Product of two Unitary Matrices is Unitary Proof

1. Sep 20, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Show that the product of two nxn unitary matrices is unitary. Is the same true of the sum of two nxn unitary matrices?

2. Relevant equations
Unitary if A†A=I
Where † = hermitian conjugate
I = identity matrix.

3. The attempt at a solution

We have the condition: (AB)†(AB)=I
I can then apply summation notation for the elements of the matrices

$(AB)^†(AB) = \Big( \sum_{k=1}^j(AB)^†_{ij}(AB)_{ij} \Big)$

Now, the idea, I suppose, is to manipulate the sum so that we see A^† *A and B^†*B and we can conclude that since A and B are unitary, then A*B is unitary.

This seems coherent, and beautiful.

I assume summation notation is needed to make this distinction. Correct?

And if this is so, I will undo the hermitian conjugate first and then manuever the pieces of the elements.
Correct?

2. Sep 20, 2015

### RJLiberator

Let me write out my proof for checking:

1. $(AB)^†(AB)_{ij} = \Big( \sum_{k=1}^j(AB)^†_{ij}(AB)_{ij} \Big)$
by element notation
2. $(AB)^†(AB)_{ij} = \Big( \sum_{k=1}^j(AB)^*_{ji}(AB)_{ij} \Big)$
by hermitian conjugate definition
3. $(AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jA^*_{ji}B^*_{ji}A_{ij}B_{ij} \Big)$
By proven earlier property that (AB)*=A*B* and earlier proposition that ABij =AijBij
4.$(AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jA^*_{ji}A_{ij}B^*_{ji}B_{ij} \Big)$
By operations, since we are dealing with components
5.$(AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jI_{ij}I_{ij} \Big)$
By earlier conditions stating these are two unitary matrices
6. Therefore I_ij = I_ij

And walouh! First part, done.

Any errors in my thinking?