# Deriving $\left( A | v \rangle \right)^\dagger = \langle v | A^\dagger$ without using $A^\dagger=\left(A^* \right)^T$

From Nielsen & Chuang (10th edition), page 69:

Suppose $$A$$ is any linear operator on a Hilbert space, $$V$$. It turns out that there exists a unique linear operator $$A^\dagger$$ on $$V$$ such that for all vectors $$|v\rangle$$, $$|w\rangle \in V$$,

$$(|v, A|w\rangle)=(A^\dagger|v\rangle, |w\rangle). \tag{2.32}$$

This linear operator is known as the adjoint or Hermitian conjugate of the operator $$A$$. From the definition it is easy to see that $$(AB)^\dagger = B^\dagger A^\dagger$$. By convention, if $$|v\rangle$$ is a vector, then we define $$|v\rangle^\dagger \equiv \langle v|$$. With this definition it is not difficult to see that $$(A|v\rangle)^\dagger = \langle v|A^\dagger$$.

Well, to me it is difficult to see that $$(A|v\rangle)^\dagger = \langle v|A^\dagger \tag1\label1$$

at least without invoking $$A^\dagger=\left(A^* \right)^T \tag2\label2$$ which I don't want to do because the book haven't introduced \eqref{2} at this point!

I realise that by using the definition $$|v\rangle^\dagger \equiv \langle v|$$ and right multiplying it by $$A^\dagger$$ I get:

$$|v\rangle^\dagger A^\dagger = \langle v| A^\dagger \tag3$$

This is pretty close to \eqref{1} and I only need to show that

$$|v\rangle^\dagger A^\dagger = (A|v\rangle)^\dagger \tag4\label4$$

My first instinct was to use $$(AB)^\dagger = B^\dagger A^\dagger$$ here. However, this does not feel quite right because $$A$$ and $$B$$ are both linear operators but in \eqref{4} I'm dealing with a linear operator and a vector. I tried getting around this by going to the matrix representation of linear operators and extending the vector $$|v \rangle$$ into a matrix such as:

$$B = \begin{bmatrix} \vert & \vert & \dots & \vert \\ |v \rangle & 0 & \dots & 0 \\ \vert & \vert & \dots & \vert \\ \end{bmatrix}$$

Then I could invoke $$(AB)^\dagger = B^\dagger A^\dagger$$ but I'm not sure what to do with this because at this point in the book we don't know that $$B^\dagger = \left(B^* \right)^T$$. Therefore, we don't know that the first row of $$B^\dagger$$ will be $$\langle v|$$. Does anybody know how to proceed?

The application of $$(AB)^\dagger = B^\dagger A^\dagger$$ directly is indeed not quite right.
At first note that $$(A|v\rangle,|w\rangle) = (|w\rangle, A|v\rangle)^* = (A^\dagger|w\rangle, |v\rangle)^* = (|v\rangle , A^\dagger|w\rangle)$$ BTW, from this you can immediately deduce $$(A^\dagger)^\dagger = A$$.
Now for all $$|w\rangle$$ we have $$(A|v\rangle)^\dagger |w\rangle = (A|v\rangle,|w\rangle) = (|v\rangle,A^\dagger|w\rangle) =$$ $$= \langle v | \big(A^\dagger|w\rangle\big) = \langle v | A^\dagger|w\rangle = \big(\langle v | A^\dagger \big)|w\rangle$$ Since it's true for all $$|w\rangle$$ we can deduce the required $$(A|v\rangle)^\dagger= \langle v | A^\dagger$$.