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Given $T:V\to V$ linear and $V$ being an inner product space, we define $T^*$ by a linear operator on $V$ such that $\langle Tx,y\rangle=\langle x,T^*y\rangle$ for each $x,y\in V$.

We later see that, for finite-dimensional inner product spaces, adjoint exists and, in fact, whenever an adjoint exists is unique.

For finite-dimensional space we can draw a correspondence between conjugate transpose of a matrix and adjoint of a linear operator, taking into consideration an orthonormal basis of $V$ and representing $T$ with respect to that basis.

Now I think the motivation behind such a definition is finding a linear transformation version of the conjugate transpose. We know that if $T$ is a linear operator on inner product space $V$, and $\beta$ be an orthonormal basis of $V$, then the corresponding matrix $A_{ij}=\langle T\alpha_j,\alpha_i\rangle$.

So, naturally if $B$ is the conjugate transpose of $A$, then $B_{ij}=\overline{ A}_{ji}=\overline{\langle T\alpha_i,\alpha_j\rangle}$. So, naturally, a question arises if there exists $U$ linear on $V$ such that $B$ is the matrix of $U$ with respect to the given orthonormal basis. Then $B_{ij}=\langle T^*\alpha_j,\alpha_i\rangle=\overline{\langle T\alpha_i,\alpha_j\rangle}=\overline {A}_{ji}$.

This, I think, motivated the definition of adjoint.

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    $\begingroup$ What is the question? $\endgroup$
    – lcv
    Mar 21, 2020 at 4:56
  • $\begingroup$ What if $V$ is Banach space that is not an inner product space? $\endgroup$ Mar 21, 2020 at 4:59

2 Answers 2

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A slightly different, but essentially equivalent, way of thinking of it is that the adjoint is the linear operator on the dual space $V^*$ induced by $T$.

Consider the dual space $V^*$, consisting of linear maps $V \to \Bbb{R}$. Then there is a map $T^*$ on $V^*$ defined as follows: for $\lambda \in V^*$, $T^*(\lambda)$ is the linear functional given by: $$ \Big( T^*(\lambda) \Big)(v) = \lambda \Big( T(v) \Big) $$ If you choose a basis for $V$, for which $A$ is the matrix of $T$ in that basis, then the adjoint matrix (the transpose) is the matrix of $T^*$ in the dual basis.

The connection with your explanation is this: if you have a nondegenerate inner product on $V$ and $V$ is finite-dimensional, then there is an isomorphism between $V$ and $V^*$ using the inner product. For a vector $v$, define a linear functional $\lambda_v$ by the formula $\lambda_v(w) = \left<v,w\right>$. If you use this to identify $V$ and $V^*$, (and if the basis is orthonormal), then the basis and the dual basis are the same, and you can think of $T^*$ as an operator on $V$.

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I think it's also good to look at a proof which uses an adjoint operator. Roughly speaking, you have a linear operator $T$ bounded for $L^p \rightarrow L^p$ with $p \in (1,2]$, and the goal is to (continuously) extend this operator to $L^p \rightarrow L^p$ with $p \in (2, \infty)$. Here's a more precise formulation and a proof:

Let's say you have a linear operator $T$ which maps $L^p$ to $L^p$ for all $p \in (1, 2]$, and bounded in the sense that $\|Tf\|_p \leq C_p \|f\|_p$ for all $f \in L^p$ and $p \in (1, 2]$, where $C_p > 0$ may depend on $p$, but independent of $f$.
Let $\mathcal D \subseteq L^p$ be dense for all $p \in [1, \infty)$ (e.g. Schwartz space, $C_0^\infty$), and suppose the adjoint $T^* : L^p \rightarrow L^p$ is also bounded for $p \in (1, 2]$.

To extend $T$, it is enough to show $\|Tf\|_p \leq C_p \|f\|_p$ for all $f \in \mathcal D$ and $p \in [2, \infty)$. Let $p, q$ be Holder conjugates, where $q \in (1, 2]$. \begin{equation} \begin{split} \|Tf\|_p^p &= \int \mathrm{sgn}(Tf) |Tf|^{p-1} \overline{Tf}dx \\ &= \langle Tf, \mathrm{sgn}(Tf) |Tf|^{p-1} \rangle \\ &= \langle f, T^*(\mathrm{sgn}(Tf) |Tf|^{p-1}) \rangle \\ &\leq \|f\|_p \|T^*(\mathrm{sgn}(Tf) |Tf|^{p-1})\|_{q} \\ &\leq \|f\|_p \|T^*\|_{L^q \rightarrow L^q} \|\mathrm{sgn}(Tf) |Tf|^{p-1})\|_{q} \\ &= \|f\|_p \|T^*\|_{L^q \rightarrow L^q} \|Tf\|_p^{p/q} \end{split} \end{equation} and so $\|Tf\|_p \leq \|T^*\|_{L^q \rightarrow L^q}\|f\|_p$ as desired.

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