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I'm having a bit of trouble understanding this fact from Linear Algebra Done Right

Let T $\in \mathcal{L}(V, W)$ Suppose $e_{1}, \dots, e_{n}$ is an orthonormal basis of $V$ and $f_{1}, \ldots, f_{m}$ is an orthonormal basis of $W$. Then then adjoint $T^{*}$ is the conjugate transpose of $T$ in $\langle T v, w\rangle=\left\langle v, T^{*} w\right\rangle$

I went through the proof and did an example but I can't get a clearer picture as to why this must be true intuitively or geometrically. What is so special about a transpose that makes the transformation act in this bridge-like fashion?

Also, I am also not able to appreciate the fact that we have an orthonormal basis, while it is important in the proof, what would go wrong if we didnt?

Here's the entire proof for reference:enter image description here

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  • $\begingroup$ In short, it's $$\langle u, A^*v\rangle =u^*A^*v=(Au)^*v=\langle Au, v\rangle$$ $\endgroup$ – Berci Jul 21 at 15:59
  • $\begingroup$ How come you made no reference to the need of orthonormal basis like the theorem? $\endgroup$ – Rahul Deora Jul 21 at 16:05
  • $\begingroup$ What is $\mathcal{M}$? In any case this book is rather dubious, since the fact you're asking about is only true for the Euclidean inner product, and not in general. $\endgroup$ – user7530 Jul 21 at 16:07
  • $\begingroup$ M is the matrix representation of the linear map in the first argument. OK so then answer for this special case will ya? $\endgroup$ – Rahul Deora Jul 21 at 16:12
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    $\begingroup$ If you are looking for intuition, I would suggest reading Weyl, specifically chapter 1 of The Theory of Groups and Quantum Mechanics (freely available here link). He writes very differently from modern authors, but if you can follow the first 17 pages (where he introduces the conjugate transpose), you'll have a very geometric understanding of the conjugate transpose and it's relation to the dual space. $\endgroup$ – ChainedSymmetry Jul 22 at 1:00
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Any inner product $\langle v, w\rangle$ on a real vector space has a representation as a symmetric positive-definite matrix $M$, with $$\langle v, w\rangle = v^T M w.$$

Now if $T$ and $T^*$ are adjoint you must have $$\langle Tv, w\rangle = v^T T^TMw = v^T MT^* w = \langle v, T^*w\rangle.$$ Since $v$ and $w$ are arbitrary, it follows that $$T^TM = MT^*.$$

In the special case of the Euclidean inner product, $M=I$ and the above reduces to $T^T=T^*$. Notice that this relationship is not true for general inner products $M$.

Now if you have two different inner product spaces $V$, $W$ with inner products $$\langle v_1, v_2\rangle_{M_v}, \langle w_1, w_2\rangle_{M_w}$$ and $T: V\to W$, an identical calculation shows that $T$ and $T^*$ satisfy $$T^TM_{w} = M_v T^*$$ with $T^T = T^*$ in the special case that both inner products are Euclidean.

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  • $\begingroup$ The explanation above is not correct. First, there is the confusion that $T$ is used to denote the linear map and is also used to denote what I assume is supposed to be the transpose, leading to the expression $T^T$ in the explanation. Second, if $v$ is an element of the inner product space (which is not necessarily $\mathbf{R}^n$), then what does $v^T$ mean? $\endgroup$ – Sheldon Axler Jul 21 at 17:31
  • $\begingroup$ @SheldonAxler 1) Sure, it's not ideal, but the OP names the map $T$. I trust it will not overly strain the reader to infer from context that one is the map, and one is the transpose operator. 2) I don't understand your objection. $v^T$ is the transpose of $v$. Take the entries of $v$ in whatever basis you have chosen for $V$, and rotate the numbers by 90 degrees... $\endgroup$ – user7530 Jul 21 at 18:58

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