Exercise 1906Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

The dual basis

#dual space#dual basis#linear forms

Statement

Let $\mathcal{B}=(e_1,e_2,e_3)$ be a basis of $\mathbb{R}^3$ . Define the dual basis $(\varphi_1,\varphi_2,\varphi_3)$ by $\varphi_i(e_j)=\delta_{ij}$ . Show that $(\varphi_1,\varphi_2,\varphi_3)$ is a basis of the dual space $E^*$ .

Hint

Show independence: if $\sum\lambda_i\varphi_i=0$ , evaluate at each $e_j$ . Show spanning: any $\varphi$ equals $\sum\varphi(e_i)\varphi_i$ .

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Solution

Independence. Suppose $\sum_{i=1}^3\lambda_i\varphi_i=0$ (the zero linear form). Evaluating at $e_j$ :
$0=\left(\sum_i\lambda_i\varphi_i\right)(e_j)=\sum_i\lambda_i\varphi_i(e_j)=\lambda_j.$
So all $\lambda_j=0$ , and the family is independent.

Spanning. Let $\varphi\in E^*$ . Define $\psi=\sum_{i=1}^3\varphi(e_i)\varphi_i$ . For each $j$ ,
$\psi(e_j)=\sum_i\varphi(e_i)\varphi_i(e_j)=\varphi(e_j).$
So $\psi$ and $\varphi$ agree on the basis, hence $\psi=\varphi$ .

Conclusion. $(\varphi_1,\varphi_2,\varphi_3)$ is a basis of $E^*$ , called the dual basis of $\mathcal{B}$ .

The dual basis

Statement

Solution

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