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  "body": "\n## Proof\n\nWe show that $T$ is bounded and then use the equivalence of boundedness and continuity.\n\n1. If $V = \\{0\\}$, then $T = 0$, and $\\|Tx\\|_W \\le 0 \\cdot \\|x\\|_V$. Let $V \\neq \\{0\\}$.\n2. Take a [[basis|06MRNP4BNT]] $e_1, \\ldots, e_n$ of the [[finite-dimensional|018W3V3TCT]] space $V$. Write $x = \\sum_j a_j(x) e_j$, with linear coordinates $a_j$ by [[this result|059THXBRBB]].\n3. Let $\\|x\\|_\\infty = \\max_j |a_j(x)|$. It is a norm on $V$: it vanishes only at $0$, it is homogeneous, and $|a_j(x + y)| \\le |a_j(x)| + |a_j(y)|$.\n4. By the [[equivalence of norms in finite dimension|03MC70ZRNR]] and the definition of [[equivalent norms|0J3578562S]], there is $c \u003e 0$ with $c \\|x\\|_\\infty \\le \\|x\\|_V$ for all $x \\in V$.\n5. By linearity, $Tx = \\sum_j a_j(x) \\, T e_j$. By the [[triangle inequality|0FJZMBX0ZZ]], $\\|Tx\\|_W \\le \\sum_j |a_j(x)| \\, \\|T e_j\\|_W \\le K \\|x\\|_\\infty$, where $K = \\sum_j \\|T e_j\\|_W$.\n6. With step 4, $\\|Tx\\|_W \\le (K / c) \\|x\\|_V$ for all $x \\in V$. So $T$ is bounded.\n7. By [[bounded linear maps are continuous|0F3K4C7XYW]], $T$ is a [[continuous linear map|0S3SMRH0Z5]].\n\n## Idea\n\nA linear map is fixed by its values on a basis, so $\\|Tx\\|$ is at most a constant times the largest coordinate of $x$. In finite dimension, the largest coordinate is controlled by the norm of $x$, because all norms are equivalent. So $T$ is bounded, hence continuous.\n",
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