[proofplan] If $V=\{0\}$, the estimate is immediate. Otherwise, choose a finite basis of $V$ and compare the given norm on $V$ with the coordinate maximum norm determined by that basis. Linearity and the triangle inequality bound $\|T v\|_W$ by a fixed finite constant times the coordinate norm of $v$, and finite-dimensional norm equivalence converts this into a bound by $\|v\|_V$. [/proofplan] [step:Handle the zero-dimensional case] Assume first that $\dim_{\mathbb{F}} V=0$. Then $V=\{0\}$, and for the unique vector $0\in V$ one has $T0=0$ by linearity. Hence \begin{align*} \|T0\|_W=0\leq 0\cdot \|0\|_V. \end{align*} Thus $T$ is bounded with bounding constant $C=0$. [/step] [step:Choose coordinates and define the coordinate norm] Assume now that $\dim_{\mathbb{F}} V=n\geq 1$. Choose an ordered basis $(e_1,\ldots,e_n)$ of $V$. For each $v\in V$, let $a_1(v),\ldots,a_n(v)\in\mathbb{F}$ denote the unique scalars such that \begin{align*} v=\sum_{i=1}^{n} a_i(v)e_i. \end{align*} Define the coordinate maximum norm $\|\cdot\|_{\infty,e}:V\to [0,\infty)$ by \begin{align*} \|v\|_{\infty,e}=\max_{1\leq i\leq n}|a_i(v)|. \end{align*} This is a norm on the [finite-dimensional vector space](/page/Finite-Dimensional%20Vector%20Space) $V$. By the finite-dimensional norm-equivalence theorem (citing a result not yet in the wiki: All norms on a finite-dimensional [vector space](/page/Vector%20Space) are equivalent), applied to the two norms $\|\cdot\|_{\infty,e}$ and $\|\cdot\|_V$ on $V$, there exists a constant $K>0$ such that \begin{align*} \|v\|_{\infty,e}\leq K\|v\|_V \end{align*} for every $v\in V$. [guided] Because $V$ is finite-dimensional and nonzero, we may choose an ordered basis $(e_1,\ldots,e_n)$ of $V$, where $n=\dim_{\mathbb{F}}V\geq 1$. The purpose of choosing a basis is to reduce an arbitrary vector estimate to finitely many estimates involving the fixed vectors $e_i$. For every $v\in V$, the basis property gives unique scalars $a_1(v),\ldots,a_n(v)\in\mathbb{F}$ satisfying \begin{align*} v=\sum_{i=1}^{n} a_i(v)e_i. \end{align*} Using these coordinates, define the coordinate maximum norm $\|\cdot\|_{\infty,e}:V\to [0,\infty)$ by \begin{align*} \|v\|_{\infty,e}=\max_{1\leq i\leq n}|a_i(v)|. \end{align*} This is a genuine norm: positivity and homogeneity follow from the corresponding properties of the absolute value on $\mathbb{F}$, and the triangle inequality follows from the coordinate identity $a_i(u+v)=a_i(u)+a_i(v)$ for each $i\in\{1,\ldots,n\}$. The proof must ultimately bound $\|T v\|_W$ in terms of the given norm $\|v\|_V$, not merely in terms of coordinates. This is exactly where finite-dimensionality is used. By the finite-dimensional norm-equivalence theorem (citing a result not yet in the wiki: All norms on a finite-dimensional vector space are equivalent), applied to the norms $\|\cdot\|_{\infty,e}$ and $\|\cdot\|_V$ on the same finite-dimensional vector space $V$, there exists a constant $K>0$ such that \begin{align*} \|v\|_{\infty,e}\leq K\|v\|_V \end{align*} for every $v\in V$. [/guided] [/step] [step:Estimate the image of a vector from the images of the basis vectors] Define the finite constant $M\geq 0$ by \begin{align*} M=\sum_{i=1}^{n}\|T e_i\|_W. \end{align*} Let $v\in V$. Using the coordinate expansion of $v$ and the linearity of $T$, \begin{align*}T v=T\left(\sum_{i=1}^{n} a_i(v)e_i\right)=\sum_{i=1}^{n} a_i(v)T e_i.\end{align*} Applying the triangle inequality in the [normed space](/page/Normed%20Space) $W$ and then homogeneity of $\|\cdot\|_W$, we obtain \begin{align*}\|T v\|_W\leq \sum_{i=1}^{n}\|a_i(v)T e_i\|_W.\end{align*} Hence \begin{align*}\|T v\|_W\leq \sum_{i=1}^{n}|a_i(v)|\,\|T e_i\|_W.\end{align*} Since $|a_i(v)|\leq \|v\|_{\infty,e}$ for every $i\in\{1,\ldots,n\}$, it follows that \begin{align*}\|T v\|_W\leq M\|v\|_{\infty,e}.\end{align*} [/step] [step:Convert the coordinate estimate into a boundedness estimate] Combining the estimate from the previous step with the norm-comparison constant $K>0$, we get, for every $v\in V$, \begin{align*}\|T v\|_W\leq M\|v\|_{\infty,e}.\end{align*} Therefore \begin{align*}\|T v\|_W\leq MK\|v\|_V.\end{align*} Define $C=MK\geq 0$. Then \begin{align*}\|T v\|_W\leq C\|v\|_V\end{align*} for every $v\in V$. This is precisely the boundedness of the [linear map](/page/Linear%20Map) $T:V\to W$ with respect to the given norms on $V$ and $W$. [/step]