I would like a concrete example of the determination of a dual space.

How to frame the example is up to you, but if you wish me to frame it, then consider the space spanned by the columns of

\begin{equation}V = \left[\begin{array}{rr}1 & 0 \\ 0 & 1 \\ 0 & 2\end{array}\right].\end{equation}

Working in real numbers (or in complex numbers if you prefer), what is the dual space of $V$, and why?

Am I just to take the transpose (adjoint? no, transpose, I think) of the Moore-Penrose pseudoinverse of $V$, or is there something more to it? If just the Moore-Penrose, then the space spanned by the columns of

\begin{equation}[V^+]^T=\left[(V^\dagger V)V^\dagger\right]^T=\left[\begin{array}{rr}1 & 0 \\ 0 & \frac 1 5 \\ 0 & \frac 2 5\end{array}\right],\end{equation}

where $V^+$ symbolizes the Moore-Penrose pseudoinverse and $V^\dagger$ the adjoint, would be the dual space.

Or is my question just confused?

Even if my answer were right (I suppose that it isn't), I do not get the point of the exercise. I do not understand what it is all about. Do you?

Incidentally, I know what a kernel is, if that helps.

(Reason for the question: I am an electrical engineer trying to learn differential geometry, Hamiltonian mechanics and general relativity; the dual space arises in this study; and I do not yet grasp the concept.)

See also this answer to a related question.


In light of the accepted answer, I now believe that the columns of $[V^+]^T$ are covectors rather than the basis of a dual space. Apparently, I had my definitions confused.

The belief might lead to a new question regarding the relationship, if any, between covectors and the dual space, insofar the books I am reading seem to imply that such a relationship exists; but that would be a question for another day.

Incidentally, for the benefit of future readers, I have modified the notation of the question as originally asked to match the answerer's notation.

  • $\begingroup$ What do you mean by "determination of the dual space" ? The definition of a dual space ? $\endgroup$ – NAC Jan 13 '18 at 3:37
  • $\begingroup$ @NAC: Your inquiry implies that my question is ill formed. You're probably right. I'm confused. Let me edit the question to say what I think the answer is—but even if my answer were correct, I don't get the point of it. $\endgroup$ – thb Jan 13 '18 at 4:24
  • $\begingroup$ @NAC: I have edited the question to say what I think the answer is. $\endgroup$ – thb Jan 13 '18 at 4:36
  • $\begingroup$ @QiaochuYuan: It is. That makes sense. That's quite helpful, actually. $\endgroup$ – thb Jan 13 '18 at 4:53
  • $\begingroup$ Cool, I'll write it up in an answer. $\endgroup$ – Qiaochu Yuan Jan 13 '18 at 6:17

Suppose $T : V \to W$ is a linear transformation between finite-dimensional vector spaces. Then the image $\text{im}(T)$ is a subspace of $W$, and accordingly the dual of the image is naturally a quotient space of $W^{\ast}$.

Furthermore, suppose $V$ and $W$ are inner product spaces. Then we can identify the duals of subspaces of $W$ with their orthogonal complements, hence we can think of the "dual of $\text{im}(T)$" as the orthogonal complement $\text{im}(T)^{\perp} \subseteq W$.

An alternative characterization of this orthogonal complement is that it is the kernel $\text{ker}(T^{\dagger})$ of the adjoint of $T$. Concretely: the orthogonal complement of the column space of a real matrix $M$ is the null space of its transpose $M^T$. It's a nice exercise to show this.


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