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For an arbitrary complex matrix A show that $$A*A^\dagger$$ is Hermitian.

Where the dagger "$\dagger$" stands for the "complex conjugate and transpose" operators.

From what I understand this must mean that $$A*A^\dagger = [A*A^\dagger]^\dagger$$ But I am stuck. I don't really understand the properties of the complex conjugate function with Matrices.

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The key properties you need are the fact that hermitian-conjugating twice returns you to where you started, $$(A^\ast)^\ast=A$$ (which holds because it holds for both transposing and complex-conjugating), and the formula for the conjugate of a product, $$(AB)^\ast=B^\ast A^\ast,$$ which is inherited from the identical transpose formula and is unchanged by complex conjugation.

Of course, you can also prove both formulas from the (real) definition of the conjugate of $A$, namely that for all $u$ and $v$ you get $\langle A u,v\rangle=\langle u, A^\ast v\rangle$, as Nick points out.

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This is perhaps different than the way it's presented in physics class, but it's better and generalizes easily to infinite dimensional spaces and non-Hilbert spaces. Use the mathematicians definition of adjoint: $A^t$ is adjoint to $A$ iff $(A u,v)=(u,A^t v)$ for all possible $u,v$.

Then bringing the operators over one at a time and using associativity of matrix multiplication yields:

$$(AA^t u,v)=(A^t u,A^t v)=(u,AA^t v)$$

So by the above definition $AA^t$ is adjoint to itself.

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  • $\begingroup$ To call this different is an abuse of language--- it's the same thing in different notation. $\endgroup$ – Ron Maimon Sep 16 '12 at 8:12
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    $\begingroup$ Well I guess it depends on what the original definition is. When I took my first quantum mechanics class I was taught the definition $a^t_{ij}:=\bar{a_{ji}}$. This is fine for finite dimensions but doesn't apply to operators between infinite dimensional spaces. One can find an analog for Hilbert spaces with a countable unconditional basis, but the situation becomes untenable for spaces that are not isomorphic to their dual. The $ij'th$ entry of an operator between dual spaces - what does that even mean? $\endgroup$ – Nick Alger Sep 16 '12 at 8:23

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