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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https://mathoverflow.net/q/6573 I haven't yet learned about Hilbert spaces. Also, I don't know what $\langle \: |$ and $| \: \rangle$ are.
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https://mathoverflow.net/q/6567 biadjacency matrix?
Best Answer
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:
$$\langle T(x),y \rangle=\langle x,T^*(y) \rangle \ \text{for all } x,y \iff $$
$$\iff \omega_y(T(x))=\omega_{T^*(y)}(x) \ \text{for all } x,y \iff $$
$$\iff \omega_y\circ T=\omega_{T^*(y)} \ \text{for all } y \iff $$
$$\iff T^t(\omega_y)=\omega_{T^*(y)} \ \text{for all } y \iff $$
$$\iff (T^t\circ \omega)(y)=(\omega \circ T^*)(y) \ \text{for all } y \iff $$
$$\iff T^t\circ \omega=\omega \circ T^* \iff $$
$$\iff T^*=\omega^{-1}\circ T^t \circ \omega $$