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Let $V$ be a finite-dimensional inner product space, and let $T$ be a linear operator on $V$. Then $T^*$ ($T$ adjoint) is defined as the unique function such that $\langle T(x), y \rangle = \langle x, T^*(y) \rangle$ for all $x, y \in V$. Furthermore, $T^*$ is linear.

I know how to manipulate the adjoint algebraically, but I'm not sure how to interpret it geometrically.

This has been asked before, but the previous questions did not suit my needs.

Definition of adjoint operator (asking for intuition) I'm not asking about $T^*$'s relation to $A$'s conjugate transpose.

Geometric intuition of adjoint I'm asking about intuition about $T^*$, not $\text{Ker}(T^*)=(\text{Im}(T))^\perp$.

https://mathoverflow.net/q/6552 The answers to this question feel too complicated to me.

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    $\begingroup$ In a finite-dimensional inner product space, Riesz representation theorem simply means that every linear functional $f\in V^\ast$ can be expressed as an inner product with a fixed vector. That is, there is some $v\in V$ such that $f(x)$ is identically equal to $\langle v,x\rangle$. One of the linked answers in your question uses this fact to explain why the adjoint of a rank-one linear operator exists. Since a rank-$r$ linear operator is just a sum of $r$ rank-one linear operators, it follows that the adjoint of a linear operator always exists. $\endgroup$
    – user1551
    Commented Dec 14, 2021 at 20:38

1 Answer 1

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This is not a geometric intuition. Nevertheless I find it pretty enlightning, so I hope to be helpful.

Given a linear operator $T:V\to V$, you can define the transpose operator:

$$T^t:V^*\to V^*$$ $$\ \ \ \ \ \ \ \ \ \varphi \mapsto \varphi\circ T$$

The transpose operator is essentially the simplest way you can imagine to construct an operator on $V^*$ using an operator $T$ on $V$. If you understand this then the adjoint is straight forward! In fact the idea is to use the fact that $V$ is isomorphic to its dual $V^*$. Normally there aren't natural isomorphism between a space and its dual, but in this case you are in a inner product (finite dimensional ) space! So you can construct a canonical isomorphism(actually if the field is $\mathbb{C}$, this is conjugate-linear):

$$\omega:V\to V^*$$ $$\ \ \ \ \ \ \ \ \ v \mapsto \langle \bullet,v\rangle=\omega_v$$

(Basically you associate to a vector $v$ the linear functional $w\mapsto \langle w,v \rangle$).

Now the adjoint is simply the transpose "re-interpreted" through this natural isomorphism:

$$T^*=\omega^{-1}\circ T^t \circ \omega:V\to V$$

As you can see the adjoint is simply a natural reinterpretation of the transpose, so if you find the transpose intuitive, you should also find the adjoint intuitive.

Clearly I should prove that my definition of $T^*$ coincides with yours, and this is straightforward:

$$\langle T(x),y \rangle=\langle x,T^*(y) \rangle \ \text{for all } x,y \iff $$

$$\iff \omega_y(T(x))=\omega_{T^*(y)}(x) \ \text{for all } x,y \iff $$

$$\iff \omega_y\circ T=\omega_{T^*(y)} \ \text{for all } y \iff $$

$$\iff T^t(\omega_y)=\omega_{T^*(y)} \ \text{for all } y \iff $$

$$\iff (T^t\circ \omega)(y)=(\omega \circ T^*)(y) \ \text{for all } y \iff $$

$$\iff T^t\circ \omega=\omega \circ T^* \iff $$

$$\iff T^*=\omega^{-1}\circ T^t \circ \omega $$

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    $\begingroup$ Very good answer! (Assuming the OP knows what $V^*$ and its elements are.) But I think you don't need the fact that $V$ and $V^*$ are isomorphic. The first 6 lines (up until 'straight forward!') would entail an entirely fine answer by itself in my opinion, even (or especially) in a hypothetical (or infinite dimensional) world where $V*$ is not isomorphic to $V$. $\endgroup$
    – Vincent
    Commented Dec 14, 2021 at 23:00
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    $\begingroup$ Hi @Vincent , thank you for your kindness. Truly, I studied this topic only in the realm of finite dimensional spaces, so I don't really know deeply how infinite-dimensional spaces work. For sure in this broader setting, $\omega$ may not be an isomorphism, so this construction fails. I don't think that reducing the answer to the first 6 lines is a good idea, since the OP refers to the adjoint as an operator on $V$ and not on its dual(the question isn't about the transpose). $\endgroup$
    – Kandinskij
    Commented Dec 14, 2021 at 23:15
  • $\begingroup$ @Eureka $\omega$ is an isomorphism if $V$ is a Hilbert space, though if $V$ is a complex vector space, $\omega$ will be conjugate linear rather than linear. This is the content of the Riesz representation theorem for Hilbert spaces. $\endgroup$
    – Mason
    Commented Dec 15, 2021 at 5:03
  • $\begingroup$ Right, good point. $\endgroup$
    – Vincent
    Commented Dec 15, 2021 at 8:14
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    $\begingroup$ Stupid question: why do we care at all about dual spaces? $\endgroup$ Commented Jan 10, 2022 at 2:21

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