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  "body": "\n## Proof\n\nWe build a basis of $V \\times W$ from bases of $V$ and $W$.\n\n1. Let $n = \\dim V$ and $m = \\dim W$. Let $(v_1, \\dots, v_n)$ be a [[basis|0KCQ9Y3QKA]] of $V$ and $(w_1, \\dots, w_m)$ a basis of $W$.\n2. In the [[product|0EX2HPXPJ6]] $V \\times W$, consider the family $\\mathcal{F}$ of the $n + m$ vectors $(v_1, 0), \\dots, (v_n, 0), (0, w_1), \\dots, (0, w_m)$.\n3. The family $\\mathcal{F}$ [[spans|088RYKAWK9]] $V \\times W$. Let $(v, w) \\in V \\times W$. Since the bases span, $v = \\sum_{i=1}^n a_i v_i$ and $w = \\sum_{j=1}^m c_j w_j$ for some scalars $a_i, c_j \\in F$.\n4. By the componentwise operations of the product, $\\sum_i a_i (v_i, 0) + \\sum_j c_j (0, w_j) = \\left(\\sum_i a_i v_i, \\sum_j c_j w_j\\right) = (v, w)$.\n5. The family $\\mathcal{F}$ is [[linearly independent|0CRMGPAH83]]. Suppose $\\sum_i a_i (v_i, 0) + \\sum_j c_j (0, w_j) = (0, 0)$ with $a_i, c_j \\in F$.\n6. By the computation of step 4, this says $\\sum_i a_i v_i = 0$ in $V$ and $\\sum_j c_j w_j = 0$ in $W$.\n7. The family $(v_i)$ is linearly independent, so all $a_i = 0$. The family $(w_j)$ is linearly independent, so all $c_j = 0$.\n8. By steps 3 to 7, $\\mathcal{F}$ is a basis of $V \\times W$. It is finite, so $V \\times W$ is [[finite-dimensional|0ZHWKQMVX5]].\n9. A basis of $V \\times W$ has $n + m$ vectors, so $\\dim(V \\times W) = n + m = \\dim V + \\dim W$ by the definition of [[dimension|03GSZ8RTZJ]].\n\n## Idea\n\nA pair $(v, w)$ splits as $(v, 0) + (0, w)$. So a basis of $V$ placed in the first slot, together with a basis of $W$ placed in the second slot, gives a basis of the product. The two bases do not interact, so their sizes add.\n",
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