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Okay, so it really isn't that strange but I haven't come into contact with anything like it before, so I want to make sure I am interpreting it correctly.

Let $V$ be a finite dimensional real inner product space. Define a function $\Phi: V\rightarrow V^*$ from the vector space $V$ to its dual space $V^*$ by: $$\Phi(v) = \{w\mapsto\langle w,v \rangle\}$$

Now I know that the dual space of $V$ is defined to be the set of linear transformations from the vector space to its field of scalars; so in this case $V^* = \{T \: \vert \: T:V\rightarrow \mathbb{R}\}$

Moreover, we just did a theorem in class that said for any linear transformation $T:V\rightarrow \mathbb{F}$ there exists a unique vector $y\in V$ such that $T(x) = \langle x,y\rangle$. Initially I read the above as $w$ is representing the unique vector $y$ for the generic linear transforms. That is to say, if $T:V\rightarrow \mathbb{R}$, then $w \equiv T(x) = \langle w,x\rangle$.

This all seems quite clumsy to me and I wanted to make sure I have definition of the function right in my head before I continue with the question. Thanks for your help!

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  • $\begingroup$ It's cases like these that make me wish lambda notation were standard mathematical notation. e.g. $\Phi := \lambda (v \in V) . [\lambda (w \in V) . \langle w, v \rangle]$. And the theorem would be: there exists a unique vector $y \in V$ such that $T = \lambda (x \in V) . \langle x, y \rangle$. $\endgroup$ – Daniel Schepler Sep 22 '17 at 20:32
  • $\begingroup$ Here, $\Phi(v)\in V^*$ is the map $\Phi(v):V \to V$ such $\Phi(v)(w) = \langle{w, v}\rangle$. $\endgroup$ – anomaly Sep 22 '17 at 23:20
  • $\begingroup$ @DanielSchepler As a logician, I ought to approve of the lambda notation, but then $\mapsto$ notation used by most mathematicians looks much more natural to me. $\endgroup$ – Andreas Blass Sep 22 '17 at 23:52
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First of all, $V^*=\{T\colon V\longrightarrow\mathbb{R}\,|\,T\text{ is linear}\}$.

Besides, it is not correct to write that $w\equiv T(x)$. What happens here is that, for every linear map $T\colon V\longrightarrow$, there is one and only one vector $w$ such that $T=\langle w,\cdot\rangle$, which means that$$(\forall v\in V):T(v)=\langle w,v\rangle.$$

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The notation $w\mapsto\langle w,v\rangle$ means that the function in question (in this case $\Phi(v)$), when applied to an arbitrary input $w$, produces the output $\langle w,v\rangle$.

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