What is the intuition behind the kernel of $A$ being orthogonal to the range of $A^*$?

Question

We know that, if $A$ is a linear bounded operator, then $\operatorname{Ker}(A) \perp \operatorname{Range}(A^*)$, where $A^*$ is the adjoint of $A$.

I have no troubles understanding the proof of this (which can be found for example on this answer), however I cannot really understand the intuition behind it.

Is there some way to intuitively understand this result, maybe through geometrical reasoning?

Answer 1

Thinking of $T: V \to W$ as the linear operator represented by $A$, we have the adjoint map $T^{*}: W^{*} \to V^{*}$ which is represented by $A^{*}$. The adjoint map takes a covector and precomposes it with $T$, and hence every covector so obtained vanishes on the kernel of $T$, giving precisely the vector-covector notion of orthogonality.

Answer 2

That's a very interesting question.

From my point of view, the key point lies in the fact that $rank(A)=rank(A^*)$ (i.e. n° of of indipendent columns = n° of indipendent rows) .

As we know, $Ker(A)\subset \mathbb{R^n}$ acts on columns vectors $\in\mathbb{R^m}$ in such way that their combination sum up to the zero vector, thus, as $rank(A)=rank(A^*)$, its orthogonal space must necessarly coincides with ${Range}(A^*)$, then:

$$\operatorname{Ker}(A) \perp \operatorname{Range}(A^*)$$

Here is a famous picture after Prof. G. Strang book on Linear Algebra and the link to his video on the "Big Pictures":

https://www.youtube.com/watch?v=ggWYkes-n6E

I hope it can help your insight on this fact as it is for me.

Answer 3

A very simple way to understand this in finite-dimensional cases might be:

1. The kernel of $A$ is the space of vectors orthogonal to all its rows.

2. The range of $A$ is the span of its columns. The range of $A^*$ is the span of the rows of $A$, when $A$ is real. More generally, it's the span of the complex conjugates of the rows of $A$.

3. Tautologically, the vectors orthogonal to all the rows are orthogonal to the linear combinations of the rows. And this is not affected by the complex conjugate in the point above.