# Dual Space Isomorphism

If $V$ is a finite dimensional real vector space. Let $$V^* = \{f: V \to \mathbb{R} : f ~\text{is linear}\}$$ (Note $V^*$ is called the dual space of $V$.)

Prove the vector spaces $V$ and $V^*$ are isomorphic.

I have attempted this question by trying to prove the dimensions of both vector spaces are equal. $$V = Span(v_1,\ldots,v_n)$$ $f_i$ is an element of $V^*$ and $f_i(v_j) = 1$ when $f_i = v_i$ and $0$ otherwise and now I want to show $V^*$ is spanned by $\{f_1,\ldots,f_n\}$.

• V* is a vector space with addition defined by (f + g)(v) := f(v) + g(v) and scalar multiplication (rf)(v) := r(f(v))
– Sam
Aug 16, 2015 at 2:34
• This is the standard proof. $\{f_i\}$ is called the dual basis to $\{v_i\}$. You just need to show that it is linearly independent and spans all of $V^*$. For independence, consider $f_i(v_k)$. For the span, recall that every element of $V$ has to be a linear combination of $\{v_i\}$, and use this to figure out how to express an arbitrary linear functional in terms of $\{f_i\}$ Aug 16, 2015 at 2:43
• @Henry Welcome to MSE! You'll notice that I've improved the formatting of your question by using TeX. If you intend to be a continued user of MSE, it may be worth learning the basics of TeX. You can see how I made the changes by attempting to edit your question. Aug 16, 2015 at 2:47

You're on the right track. Let's say $\beta = \{ v_1, \dots , v_n \}$ is a basis for $V$ then we define $f_j: V \rightarrow \mathbb{F}$ by $f_j(v_i) = \delta_{ij}$ extended linearly. Now, show that $\beta^*$ forms a basis for $V^*$ and you're done since that shows the dimensions of $V$ and $V^*$ are the same (which shows $V$ and $V^*$ are isomorphic). So, what do you need to show $\beta^*$ is a basis ?

1. linear independence (LI)
2. spanning

I can show you both, but I stop here for now. Ok, so for LI consider scalars $c_1, \dots , c_n \in \mathbb{F}$ for which: $$c_1f_1+ \cdots + c_nf_n = 0$$ this is a sum of $\mathbb{F}$-valued functions on $V$. In particular, we can evaluate at $v_i$ to find: $$c_1f_1(v_i)+ \cdots +c_jf_j(v_i)+ \cdots + c_nf_n(v_i) = 0(v_i) = 0$$ which yields $c_i=0$ for arbitrary $i$ since only the $i=j$ term survives. But, this shows that $c_1 = 0, \dots , c_n=0$ hence $\beta^*$ is LI. To prove spanning consider $\alpha: V \rightarrow \mathbb{F}$ and $v \in V$ where $v = x_1v_1+ \cdots + x_nv_n$, $$\alpha(v) = \alpha(x_1v_1+ \cdots + x_nv_n) = x_1 \alpha(v_1) + \cdots + x_n \alpha(v_n).$$ Now, here's the neat thing: $x_j = f_j(v)$ (prove it for yourself if you doubt me) thus, $$\alpha(v) = \alpha(x_1v_1+ \cdots + x_nv_n) = \alpha(v_1)f_1(v) + \cdots + \alpha(v_n)f_n(v) = \left( \alpha(v_1)f_1 + \cdots + \alpha(v_n)f_n \right)(v).$$ Therefore, $\alpha \in \text{span}(\beta^*)$ and this completes the proof that $\beta^*$ is a basis for $V^*$.

Let me just add a bit more on this construction. Notice, our construction depends on the choice of $\beta$. Fortunately, we can show the number of vectors in a basis is independent of the choice (that is, dimension is well-defined). Hence, this choice is harmless. That said, if we could construct an isomorphism which did not rely on such a choice then that isomorphism would be a canonical isomorphism. In finite dimensions, $V$ and $V^{**}$ are canonically isomorphic by the rule $\alpha(v) = \tilde{v}(\alpha)$ where $\tilde{v} \in V^{**}$ and $\alpha \in V^*$ and the isomorphism is $v \mapsto \tilde{v}$. In contrast, for $V$ and $V^*$ we have $G(v_i)=f_i$ extended linearly gives the isomorphism, but, this construction depends on our choice of $\beta$. However, if we have an inner-product $g: V \times V \rightarrow \mathbb{F}$ then $g$ allows the construction of an isomorphism of $V$ and $V^*$ independent of basis choice; $\Psi(v)(w) = g(v,w)$ for all $v,w \in V$ defined $\Psi(v) \in V^*$ and you can check $\Psi: V \rightarrow V^*$ is an isomorphism.

• oops, I used $v^j$ for your $f_j$. I'll change to your notation when I edit the rest of the answer. Aug 16, 2015 at 2:46
• To show $\{e^1,...,e^n\}$ is linearly independent, can we suppose that $0=c_1e^1+...+c_ne^n$ is the zero mapping to $\mathbb{R}$ and consider the image of $e_1:$ $0(e_1)=c_1*1+...+c_n*0=c_1$ Hence, $c_1=0$. Repeating the procedure for $e_j$, $2\leq j\leq n$, and see that $c_1=c_2=...=c_n=0$?
– Sam
Aug 16, 2015 at 2:58
• @Henry exactly. Of course, I'm lazy so I just do the argument for $e_i$ well $v_i$ since I thought that was the basis we're working with... sometimes $e_i$ is reserved for the basis of $\mathbb{F}^n$ (but this is by no means universal) Aug 16, 2015 at 3:06
• Thankyou!! All of this has helped so much!
– Sam
Aug 16, 2015 at 3:08