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Apology if this is silly or too simple. But I just realized I can't hammer out a confusion over what "adjoint" or "transpose" is supposed to mean. Suppose $X$ is an $n$-dim $k$-vector-space, and $A: X\to X$ a linear transformation. The adjoint of $A$ is defined as $A^*: V^* \to V^*$ satisfying $f \circ A=A^*(f)$ for every $f\in V^*$. If we use the so-called "natural pairing" $\langle x, f\rangle:=f(x)$ then it could be stated as $\langle A(x), f\rangle=\langle x, A^*(f)\rangle$. If equating $X$ with the $n$-dim column space and $V^*$ the $n$-dim row space, which is canonical, and if $A_n$ is the matrix for transformation $A$, then we could rewrite left-hand-side as $\langle A_nx, f\rangle=\langle x, A^*(f)\rangle$.

Now here is the confusion. Since $f\in V^*$ is a row vector, we should have $A^*(f)=fA_n$. That is, right multiplication of matrix $A_n$, without transposing it. Only if we use the isomorphism $T:V\to V^*$ sending standard unit column vectors to standard row vectors, $T(e_i):=e_i^T$, for $1\leq i\leq n$, which is matrix transposition, would we have $T^{-1}(fA_n)=T(fA_n)=A_n^T f^T$. Only now can we have the matrix transpose creeping in. But, now $f^T$ is not in $V^*$, it is in $V$!!! So we are actually left multiplying $A_n^T$ to a column vector in $V$!!! So why do we take writing $\langle A_n x, f\rangle=\langle x, A_n^T f\rangle$ for granted everywhere, when it should've been $\langle A_n x, f\rangle=\langle x, fA_n\rangle$?!

Hope it makes sense. Thanks in advance.

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There are (at least) two different ways we can express the elements of $V^*$ as arrays of numbers.

One of those ways is the one that you seem to be comfortable with: each $f\in V^*$ corresponds to a row vector, such that for all $v\in V$ we have $f(v)=fv$. (The right-hand side is a matrix multiplication of the row vector $f$ and the column vector $v$.) This representation is generally used when we don't have an inner product on $V$, and want to distinguish vectors and dual vectors.

However, we could have just as easily decided to represent $f\in V^*$ by a column vector, such that $f(v)=f\cdot v$. (This time the right-hand side is the Euclidean dot product.) This representation is usually better when we do have an inner product on $V$, or otherwise want to emphasize the choice of isomorphism between $V$ and $V^*$ by putting them on the same footing.

When you see someone write $\langle A_n x, f\rangle=\langle x, A_n^T f\rangle$, they are implicitly using the latter choice where both $V$ and $V^*$ are both represented using column vectors. Ultimately neither choice is right or wrong, and hopefully context will make it clear which representation is being used.

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  • $\begingroup$ I understand that in an inner-product space. But the context here is "natural pairing" in Banach spaces. I didn't explicitly say that because this pairing still makes sense more generally without a norm. So you are saying, $V^*$ is usually treated still as a column space, which is the same as $V$, but should be understood as another copy of $V$? That seems the same as using what I called "T", a basis-dependent isomorphism $T: V\to V^*$, which is exactly matrix transpose. $\endgroup$ Commented Aug 24, 2024 at 3:51
  • $\begingroup$ So we interpret both $v\in V$ and $f\in V^*$ as column vectors, and when we write $\langle x,f\rangle$, it's corresponding matrix form is $f^T\cdot x$, where $\cdot$ is matrix multiplication. So that all matrix multiplications happen on the left. Is that the right understanding? $\endgroup$ Commented Aug 24, 2024 at 4:01

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