The Product of two Unitary Matrices is Unitary Proof

[RJLiberator]

Homework Statement

Show that the product of two nxn unitary matrices is unitary. Is the same true of the sum of two nxn unitary matrices?

Homework Equations

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

The Attempt at a Solution

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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?

 

[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?