# 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). Feb 1 '15 at 1:52
• @MattSamuel Ahh, right. Thank you :) Feb 1 '15 at 1:55

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