# Why is the canonical formulation of a dual basis actually a basis of the dual space?

Let $B=\{v_{i}\}$ be the basis for $V$, and $B^{*}=\{f_{i}\}$ be our basis for $V^{*}$. The typical formulation of $B^{*}$ is:

$$f_{i}(v_{j}) = \delta_{j}^{i} = 1 \textrm{ (if i = j), otherwise } = 0$$

However, I'm having a hard time understand why exactly this is a basis of $V^{*}$. That is, suppose I have a new linear functional $f'$ that I want to represent as a linear combination of dual basis vectors. How would I do so? How do I show they are linearly independent & span $V^{*}$? I have a feeling this is very simple, but I'm having a tough time wrapping my head around it.

• The picture to have in mind is that if $V$ is a vector space of $n \times 1$ matrices, then $V^*$ is a vector space of $1 \times n$ matrices. The standard bases on both are the vectors that have a $1$ in one position and $0$ in the others. (and these bases are dual) – user14972 Dec 10 '16 at 14:05
• Thanks, this ^ is helpful. – sir_thursday Dec 10 '16 at 14:40

• Suppose you want $$f'=\sum b_jf_j$$ Evaluating at $v_i$ on both sides, you have a linear equation on the coefficients $b_i$.
• To show the linear independence of $\{f_j\}$, write $$\sum b_jf_j=0.$$ You want to show $b_j=0$. But the right hand side is the zero function on $V$. What do you get by taking the action on ${v_i}$ on both sides?
• Hmmm, I see, so each $f'(v_{j})$ just ends up equally $b_{i}$. For the second bullet, you get that each $b_{j} =$. Damn, I get it, still a little abstract but I'll keep mulling over it. Thanks. – sir_thursday Dec 10 '16 at 14:37