Why is ' the conjugate transpose of a matrix?

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This seems to me to be making things more complicated than necessary. As I said above, the underlying reason for conjugate-transposition stems from how you define an inner product of complex vectors.

You want an inner (“dot”) product of a vector with itself to be a norm², which should always be ≥ 0 (and = 0 only if the vector is zero). But for complex vectors this means you need a conjugation.

Consider the “column” vector z = [1, i]. Then:

z^T z = \begin{pmatrix}1 & i \end{pmatrix} \begin{pmatrix}1 \\ i \end{pmatrix} = 1^2 + i^2 = 0

does not make sense as a norm because z \ne 0. Whereas if you conjugate:

\overline{z^T} z = \begin{pmatrix}1 & -i \end{pmatrix} \begin{pmatrix}1 \\ i \end{pmatrix} = 1^2 - i^2 = 2

you get a reasonable “Euclidean” length² ≥ 0.

In general, for an n-component complex vector z \in \mathbb{C}^n, you have

\overline{z^T} z = \Vert z \Vert^2 = \sum_{k=1}^n |z_k|^2 \, ,

which is clearly ≥ 0 for all z, and = 0 only if z = 0.

Once you accept that complex conjugation is necessary for inner products of complex vectors, then it follows that you want the conjugate transpose for matrices too (to move the matrix from one side to the other of an inner product).

(It is for the same reason that you want the complex-transposition to be recursive for matrices of matrices, in order that x'y is an inner product for vectors of vectors.)

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