Suppose we have $V$ and $W$ be $2$ $K$-vectorspaces and $f:V\rightarrow W$ is a linear map then:

  1. For all $\varphi\in W^\ast$ one has $\varphi\circ f\in V^*$
  2. The map $f^\ast:W^\ast\rightarrow V^\ast:\varphi\mapsto\varphi\circ f\;$ is a linear map, which is also called a dual map.

Where $V^\ast$, the dual space, is defined as: \begin{equation} V^\ast:= Hom_K(V,K) = \{\;f:V\rightarrow K\;|\;f\;\text{is a linear map}\;\} \end{equation}

Where $K$ is a $1$-dimensional $K$-vector space and $V$ a $K$-vector space

But I don't really get what this means, could anyone give a clear explanation because I have a hard time understanding this. Especially with that extra $\varphi$ function


The set $V^*$ of all linear maps from a vector space $V$ to its base field $K$ is, itself, a vector space over $K$: this is easy to show, with the operations being $(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda(f(x))$, for all $f, g\in V^*$ and all $\lambda \in K$.

Now, given any $\varphi \in W^*$, and any $f: V \to W$, $\varphi\circ f$ is a linear map from $V$ to $K$ (since it's a composition of linear maps), so is an element of $V^*$ (you have a typo in your question: it is not an element of $V$).

Thus, we define a new function, $f^*$, that takes as its input an element $\varphi$ of $W^*$, and spits out an element of $V^*$, by composing $\varphi$ with $f$. This, it turns out, is linear, and this, too, is not hard to check: if $\varphi,\psi\in W^*$, and $v \in V$, then

\begin{align*}f^*(\varphi+\psi)(v) &= (\varphi + \psi)\circ f(v) \\&= (\varphi+\psi)(f(v)) \\&= \varphi(f(v)) + \psi(f(v)) \\&= \varphi\circ f(v) + \psi \circ f(v) \\&= (f^*\varphi + f^*\psi)(v),\end{align*}

hence $f^*(\varphi + \psi) = f^*\varphi + f^*\psi$. Similarly, for any $\lambda \in K$, we have

\begin{align*}f^*(\lambda\varphi)(v) &= (\lambda\varphi)\circ f(v) \\&= (\lambda\varphi)(f(v))\\&=\lambda(\varphi(f(v))) \\&= \lambda(\varphi \circ f)(v)) = \lambda f^*(\varphi)(v),\end{align*}

hence $f^*(\lambda\varphi) = \lambda f^*(\varphi)$, and $f^*$ is, indeed, linear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.