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  "body": "\n## Proof\n\nWe show that the dual basis is a basis of $V^*$.\n\n1. Let $n = \\dim V$ and let $(b_1, \\dots, b_n)$ be a [[basis|0KCQ9Y3QKA]] of $V$. Write $\\delta_{ij} = 1$ if $i = j$ and $\\delta_{ij} = 0$ if $i \\ne j$.\n2. For each $i$, [[the extension of values on a basis|0R8X9XA83A]] gives a [[linear form|07YQ1C78G0]] $b_i^* \\colon V \\to F$ with $b_i^*(b_j) = \\delta_{ij}$ for all $j$. These form the [[dual basis|0ZD2CF6ZS0]] $(b_1^*, \\dots, b_n^*)$ in the [[dual space|07JBSZE0V4]] $V^*$.\n3. The family $(b_1^*, \\dots, b_n^*)$ is [[linearly independent|0CRMGPAH83]]. Suppose $\\sum_{i=1}^n \\lambda_i b_i^* = 0$ with $\\lambda_i \\in F$.\n4. The operations of $V^*$ are pointwise. Evaluating at $b_j$ gives $0 = \\sum_{i=1}^n \\lambda_i b_i^*(b_j) = \\sum_{i=1}^n \\lambda_i \\delta_{ij} = \\lambda_j$. This holds for each $j$.\n5. The family $(b_1^*, \\dots, b_n^*)$ [[spans|088RYKAWK9]] $V^*$. Let $g \\in V^*$, and let $h = \\sum_{i=1}^n g(b_i)\\, b_i^*$.\n6. For each $j$, $h(b_j) = \\sum_{i=1}^n g(b_i) \\delta_{ij} = g(b_j)$. So $g$ and $h$ are linear maps $V \\to F$ that agree on a basis.\n7. By the uniqueness part of [[the same result|0R8X9XA83A]], $g = h$. So $g$ lies in the span of the $b_i^*$.\n8. By steps 3 to 7, $(b_1^*, \\dots, b_n^*)$ is a basis of $V^*$ with $n$ elements.\n9. So $V^*$ is [[finite-dimensional|0ZHWKQMVX5]], and its [[dimension|03GSZ8RTZJ]] is $n = \\dim V$.\n\n## Idea\n\nThe dual basis vector $b_i^*$ reads off the $i$-th coordinate. A linear form is fixed by its values on the basis, so it is the combination of the coordinate functions with those values as coefficients. This gives exactly $n$ basis vectors for $V^*$.\n",
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