# Adjoint of a matrix

Just a little help please...

if c.conj(AB) = conj(B)*conj(A) and if Transpose(AB) = Transpose(B)*Transpose(A) then why is Adjoint(AB) = Adjoint(B)*Adjoint(A)

It seems the two operations combined would cancel their effects in reversing order. Thanks

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c.cong=cong=complex conjugate. Adjoint is the adjoint. I didnt use "symbols" because physicist and mathematicians disagree on these set of symbols. – Squirtle Oct 5 '12 at 17:47
Are you sure that complex conjugation switches order of matrix multiplication? (And by adjoint I see you mean the conjugate transpose, the composition of the transpose with complex conjugation of entries, two operations on matrices that commute with each other.) – anon Oct 5 '12 at 17:48
yeah.... i feel like the complex conjugate should NOT switch the order of multiplication because if it did then the order would be switched twice (ie not switched at all) for adjoints. – Squirtle Oct 5 '12 at 17:53
Order may be reversed and even back again, but the operations are still being applied, so only the ordering effect is "cancelled", not the entire operation (they are not inverse operations) – adam W Oct 5 '12 at 17:53
One way to see that complex conjugation does not switch the order is to try it when $A$ is a $2\times 2$ matrix and $B$ is $2\times 1$. Then you can't switch the order and multiply! – Jason DeVito Oct 5 '12 at 17:54

## 3 Answers

As I said in the comments, one way to see that $\overline{AB} =\overline{B}\overline{A}$ can't possibly work comes from taking $A$ to be $2\times 2$ and $B$ to be $2\times 1$. Then you can't swap the order and still multiply.

I suspect the reason wikipedia got it wrong is because of the following. For complex (and real) numbers, we have $\overline{ab} = \overline{a}\overline{b} = \overline{b}\overline{a}$ since the order doesn't matter.

If one tries to generalize this to the quaternions (where order of multiplication does matter), then the correct rule is $\overline{ab} = \overline{b}\overline{a}$. From here, it's easy to guess that non commutative things swap when taking complex conjugates and then apply this reasoning to matrices (where it is wrong).

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Yes the order changes, the elements of course do need to be transposed for the multiplication to make sense, and that is why it is often called the conjugate-transpose. Hence for matrices, it does make sense, it all happens - element by element conjugation, matrix transpose, and order change. – adam W Oct 5 '12 at 19:47
@adam: I think you misread. The bars on top indicated complex conjugation alone, not the adjoint, and for just that, no order change occurs. – Jason DeVito Oct 5 '12 at 20:06
Yes, in that case I agree. – adam W Oct 5 '12 at 20:27
Why is this answer down voted? Its perfectly valid, whose the ignoramus? – Squirtle Oct 6 '12 at 3:04

The complex conjugate does not switch the order of matrix multiplication the way you've suggested above. The rules are:

$$\overline{AB} = \overline{A} \overline{B}, \quad (AB)^T = B^T A^T$$

So since the adjoint for a matrix is complex conjugation + transpose, then we indeed get

$$(AB)^{\ast} = B^{\ast} A^{\ast}$$

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I want to give the answer to Jason DeVito for his clear example of why I was wrong. But would someone also please edit the BAD wikipedia article? – Squirtle Oct 5 '12 at 18:01
Conjugate-transpose and adjoint are two different things. Both separately DO switch order of multiplication. – adam W Oct 5 '12 at 18:15
If your choice of inner product on $\mathbb{R}^n$ or $\mathbb{C}^n$ is the standard dot product, then conjugate-transpose and adjoint are the same for square matrices . In the context of matrices, I believe this is always taken to be the meaning. – Christopher A. Wong Oct 5 '12 at 18:54
Sorry, I didn't actually need for the matrix to be square. – Christopher A. Wong Oct 5 '12 at 19:02

You have c.cong() and conj()

But I believe that it is just a typo, conj(AB) = conj(B) conj(A)

The conjugate with matrices can mean element by element, but I imagine that is not the case here since your formula specifies the order of multiplication.

It is all to do with whether the elements of interest are scalars (reals or complex) or matrices (commuting or non-commuting elements). The field of reals and complex commute, so no difference need be specified, e.g. ab = ba and $\overline{ab} = \overline{b}\overline{a} = \overline{a}\overline{b}$ and there is no confusion. If the elements however are matrices, then they do not generally commute and the order does matter very much.

I think it is all a matter of what you consider to be the complex conjugate. If one starts with scalars and extends the system (most commonly the reals) to include $\sqrt {-1}$ for example, then the complex and the conjugate do not need to worry about order of multiplication. If then you do that with matrices (with complex values as elements) , or whatever other field, and want a norm that gives a scalar value within the original system (without the extension), and positive only to boot (otherwise how to have the triangle inequality or magnitude comparisons of any sort), then you end up with order and transpose. What then is the complex conjugate? The value (as a function of the input) that when multiplied by the element of interest (input) gives that norm. For matrices that means transpose is necessary. If doing only conjugate of the individual elements, then it is a different field. Is the conjugate on the scalar field or the matrix/vector field? That is the question that needs to be decided beforehand.

Keep It Simple Stupid

If the definition/formula being used requires transpose, do it. If somewhere a contradiction can be proven/found, change definitions/formulas to keep things consistent.

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yes the conjugate is a concept that may be applied to non-commutative fields. That is why the original Wikipedia article highlighted the order of elements. – adam W Oct 5 '12 at 21:06