# Prove that vector space and dual space have same dimension

As an exercise in my textbook, I need to prove that if $V$ is a finite dimensional vector space with dual space $V^*$ over $\mathbb{R}$, then dim$(V)$=dim$(V^*)$.

Let $\omega\in V^*$ and let $\{e_1,...,e_n\}$ be a basis for $V$. Define $e^i\in V^*$ by $e^i(e_j)=\delta_{ij}$. We show that $\{e^1,...,e^n\}$ spans $V^*$. $\omega(v)=\omega(v_1e_1+...+v_ne_n)=v_1\omega(e_1)+...+v_n\omega(e_n).$ If $\omega(e_1)=\lambda_1,...,\omega(e_n)=\lambda_n$, then $\omega(v)=v_1\lambda_1e^1(e_1)+...+v_n\lambda_ne^n(e_n)$=$\lambda_1e^1(v)+...+\lambda_ne^n(v)$.

To show $\{e^1,...,e^n\}$ is linearly independent, suppose that $0=c_1e^1+...+c_ne^n$ is the zero mapping to $\mathbb{R}$. 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 $e_j$, $2\leq j\leq n$, we see that $c_1=c_2=...=c_n=0$.

Does this proof look correct?

If the proof is correct, I have an additional small question. It seems like this proof is dependent on the fact that $V$ is a real (or complex) vector space, since we define the covectors to have the image set $\{0,1\}$. Is there a proof that works for vectors spaces over general fields?

• Every field has an additive identity (0) and a multiplicative identity (1). – Matt Samuel Feb 1 '15 at 1:52
• @MattSamuel Ahh, right. Thank you :) – user124910 Feb 1 '15 at 1:55

## 2 Answers

Looks good to me. I believe in the case of a general field you just consider {0,1} to be the corresponding identities in the general field, which exist uniquely and are distinct among other properties as seen in the very first theorems defining fields.

If you're interested in abstract algebra and linear algebra over general fields I'd check out Artin's Algebra, which covers both very nicely and thoroughly.

Your proof is correct.

By definition, a field has $0$ and $1$, and the same goes through without any difficulty (either for division rings, i.e. noncommutative fields).