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Recently I've been reviewing linear algebra. The definition of adjoint of a linear map on an inner product space seems not really natrual. Looks like people use this to define normal and self-adjoint and prove the spectral theorem. But from the result of the spectral theorem (diagonolizability wrt an orthogonal basis) I can see nothing to do with adjoint. Why do we define adjoint and what makes it important?

jiefu hou
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  • Compare the adjoint of a linear map $\mathbb{C}^n\to\mathbb{C}^m$ to the conjugate transpose of its associated matrix in $\mathcal{M}_{m\times n}(\mathbb{C})$ with respect to the canonical basis – Lorago Dec 02 '22 at 16:49
  • Do you care about only the inner product space case or do you also want to think about the notion of dual space? – Ian Dec 02 '22 at 19:36
  • Anyway, to me the main elementary motivation (beyond "it lets you prove stuff you want to prove") is the four fundamental subspaces, which is the main topic of Strang chapter 2. – Ian Dec 02 '22 at 19:45
  • Adjoint is used for theoretical arguments in infinite dimensions, e.g. for orthogonal maps, orthogonal projections, eigen-decompositions. Also $A^TB = (\langle a_i, b_j \rangle)$, so an the adjoint of a map can encode the operation of taking inner products. – Mason Dec 02 '22 at 21:07
  • Actually the reason why i add a dual-map tag is that someone relate these two object, introducing one right after the other. So i wonder whether there is a historical sequencial order between them, so to speak. @Ian – jiefu hou Dec 03 '22 at 17:25

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Adjoints are useful in many cases. I will just give one, very important example: solving linear equations.

For instance, Consider a general bounded operator $L$ from a Hilbert-space $H_1$ to another one $H_2$ and let $\langle\cdot,\cdot\rangle_{H_1}$ and $\langle\cdot,\cdot\rangle_{H_2}$ be the associated inner-products. Then, one has

$$\langle Lu,v\rangle_{H_2}=\langle u,L^*v\rangle_{H_1}$$

for all $u\in H_1$ and $v\in H_2$, and where $L^*$ is the adjoint of $L$.

Now, assume that we would like to solve $Lx=y$. Assuming that $LL^*$ is invertible, a solution is given by $x=L^*(LL^*)^{-1}y$.

In the finite-dimensional case, that is, $H_1=\mathbb{R}^n$ and $H_2=\mathbb{R}^m$, then the operator $L$ can be represented as a matrix $M$ and we have $y=Mx$. The solution of which is given by $x=M^*(MM^*)^{-1}y$ where we have assumed that $MM^*$ is invertible (equivalently, $M$ is full row rank). In fact, the expression $M^*(MM^*)^{-1}$ is nothing else but the Moore-Penrose pseudoinverse of the matrix $M$.

KBS
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