Linear Map Lemma

The existence part of this result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of this result means that a linear map is completely determined by its values on a basis.

Linear Map Lemma

Suppose $v_1,\dots ,v_m$ is a basis of $V$ and $w_1,\dots ,w_n\in W$. Then there exists a unique linear map $T:V\to W$ such that

$$T{v}_k=w_k$$

for each $k=1,\dots ,n$.

Proof.

The idea of this proof is that if we know what a linear map does the basis vectors of $V$, then that linear map is completely determined for all of $V$.

First, we show the existence of the desired linear map $T$. We can explicitly define $T:V\to W$ by

$$T(c_1{v}_1+\cdots +c_n{v}_n)=c_1{w}_1+\cdots +c_n{w}_n$$

where $c_1,\dots ,c_n$ are arbitrary elements of $\mathbb{F}$. The list $v_1,\dots ,v_n$ is a basis of $V$. Thus, the equation above does indeed define a function $T$ from $V$ to $W$ (because each element of $V$ can be uniquely written in the form $c_1{v}_1+\cdots +c_n{v}_n$).

Every vector $v\in V$ can be written uniquely as a linear combination of basis vectors, so $T(v)$ is well-defined because the coefficients $c_1,\dots ,c_m$ in $v$ are uniquely determined. For one of the basis vectors $v_k$, taking $c_k=1$ and the other $c$‘s equal to $0$ in the equation above shows that $T{v}_k=c_k{w}_k=w_k$.

Now, we seek to show additivity and homogeneity (properties of a linear map) If $u,v\in V$ with $u=a_1{v}_1+\cdots +a_n{v}_n$ and $v=c_1{v}_1+\cdots +c_n{v}_n$, then

$$\begin{array}{rl}T(u+v)& =T((a_1+c_1)v_1+\cdots +(a_n+c_n)v_n)\\ & =(a_1+c_1)w_1+\dots (a_n+c_n)w_n\\ & =(a_1{w}_1+\cdots +a_n{w}_n)+(c_1{w}_1+\cdots +c_n{w}_n)\\ & =Tu+Tv\end{array}$$

Similarly, if $\lambda \in \mathbb{F}$ and $v=c_1{v}_1+\cdots +c_n{v}_n$, then

$$\begin{array}{rl}T(\lambda v)& =T(\lambda c_1{v}_1+\cdots +\lambda c_n{v}_n)\\ & =\lambda c_1{w}_1+\cdots +\lambda c_n{w}_n\\ & =\lambda (c_1{w}_1+\cdots +c_n{w}_n)\\ & =\lambda Tv\end{array}$$

Thus, $T$ is a linear map from $V$ to $W$.

To prove uniqueness, now suppose that $T\in \mathcal{L}(V,W)$ and that $T{v}_k=w_k$ for each $k=1,\dots ,n$. Let $c_1,\dots ,c_n\in \mathbb{F}$. Then, the homogeneity of $T$ implies that $T(c_k{v}_k)=c_k{w}_k$ for each $k=1,\dots ,n$. The additivity of $T$ now implies that

$$T(c_1{v}_1+\cdots +c_n{v}_n)=c_1{w}_1+\cdots +c_n{w}_n$$

Thus, $T$ is uniquely determined on $\text{span}(v_1,\dots ,v_n)$ by the equation above. Because $v_1,\dots ,v_n$ is a basis of $V$, this implies that $T$ is uniquely determined on $V$, as desired.