# Proof of matrix conjugate (for the complex numbers)

1. Sep 6, 2009

### philnow

1. The problem statement, all variables and given/known data

Supposing that A*B is defined (where A and B are both matrices in the field of the complex numbers), show that the conjugate of matrix A * the conjugate of matrix B is equal to the conjugate of A*B.

2. Relevant equations

None.

3. The attempt at a solution

I'm stuck. I've already shown that for 2 complex numbers z1 and z2, the conjugate of z1 + the conjugate of z2 is equal to the conjugate of (z1+z2). I've also shown that the conjugate of z1 * the conjugate of z2 = the conjugate of (z1*z2). My prof says to use the above to help with the proof.

I'm quite inexperienced with proofs, so any hint or tip would be extremely appreciated. Thanks.

2. Sep 6, 2009

### lanedance

you could try writing out th sum as elements

Ie say you have

C = AB

then for and elemnt of C at row i, & column j, each cij is given by the sum
cij = (sum over k) aikbkj

This reduces the matrix multiplication to addition & multiplication of individual complex numbers

3. Sep 7, 2009

### HallsofIvy

Staff Emeritus
How, exactly, is your "conjugate" defined? The conjugate of a linear operator, A, on an innerproduct space over the complex numbers is defined as the linear operator A* such that, for all vectors u, v, <Au, v>= <u, A*v> where < , > is the inner product. It is easy to show that if A and B are linear operators, <ABu, v>= <Bu, A*v>= <u, B*A*v> so that B*A*= (AB)*.

If you have defined the conjugate of a matrix as "the matrix you get by swapping rows and columns and taking the complex cojugate of the matrix" (the complex conjugate of the transpose), then it would be useful to prove that <Au, v>= (Au)v= u(A*v)= <u, A*v> for row vector u and column vector v.
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