Linear Algebra,Conjugate Transpose

Question

Let $ M_n(\mathbb C) $ be the space of all $ n\times n $ matrices with complex entries. Prove that function $ \langle, \rangle : M_n(\mathbb C) \times M_n(\mathbb C) \to \mathbb C $ defined by $ \langle A, B\rangle = \mathrm{Tr}(A^∗B) $, is an inner product, where $ A^* $ is conjugate transpose of $ A $.

Answer 1

You need to prove that

1. $ \mathrm{Tr}(A^*A) \ge 0; $
2. $\mathrm{Tr}(A^*A) = 0 \iff A = 0; $
3. $ \mathrm{Tr}(A^*(B + C)) = \mathrm{Tr}(A^*B) + \mathrm{Tr}(A^*C); $
4. $ \mathrm{Tr}(A^*(zB)) = z\,\mathrm{Tr}(A^*B) $
5. $ \mathrm{Tr}(A^*B) = \left(\mathrm{Tr}(B^*A)\right)^* $

For (1) just notice that $ A^*A $ is positive semidefinite, thus has non negative eigenvalues and non negative trace; also, if the trace is zero all the eigenvalues need to be zero, and thus $ A^*A = 0 $, which is only possible if $ A = 0 $.

Items (3) and (4) are trivially verified because the trace is linear.

For (5), just write out the trace explicitly, apply the linearity of complex conjugation and you get the thesis.