Prove that there exists a linear transformation $L\colon V\to W$ such that $\ker L=K$ and $\mathrm{im}\, L=R$. Let $V$ and $W$ be vector spaces over $F$, where $V$ is $n$-dimensional. Let $K\subseteq V$ and $R\subseteq W$ be finite-dimensional subspaces such that $\dim K+\dim R=n$. Prove that there exists a linear transformation $L : V \to W$ such that $\ker L=K$ and $\text{Im}\,L=R$.
My work;
Let $\dim(V)=m$ and the basis of $V=\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$
Let $\dim(W)=n$ and the basis of $W=B_{1}, B_{2}, \cdots B_{n}$
Let $\dim(K)=d$ and the basis of $K=a_{1}, a_{2}, \ldots a_{d}$
Let $\therefore \operatorname{dim}(R)=m-d$ and we can take its basis as
$$
\beta_{1}, \beta_{2} \cdots \cdot \beta_{m-d} .
$$
 A: You have a bunch of typos in your work (you're given $\dim(V)=n$ in the problem, yet you start with $\dim(V)=m$; did you mean to switch $m$ and $n$?, etc.), but you have the right idea.
Let $\dim(K)=d$ and $\dim(W)=m$ and note that $\dim(R) = n-d$.
By starting with a basis for $K$ and extending to a basis for $V$, one can show there exists a basis $a_1, \ldots, a_n$ for $V$ such that $a_{n-d+1}, \ldots, a_{n-1}, a_n$ is a basis for $K$.
Similarly, there exists a basis $b_1, \ldots, b_m$ for $W$ such that $b_1, \ldots, b_{n-d}$ is a basis for $R$.
To define a linear map $L$ from $V$ to $W$, it suffices to specify $L(a_i)$ for each $i$, since by linearity $L\left(\sum_{i=1}^n c_i a_i\right) = \sum_{i=1}^n c_i L(a_i)$.
The map $L$ that satisfies $L(a_1)=b_1, L(a_2)=b_2, \ldots, L(a_{n-d})=b_{n-d}$, and $L(a_{n-d+1})=L(a_{n-d+2}) = \cdots = L(a_n)=0$ has image $R$ and kernel $K$.
A: The fact: Any basis for $K$ can be extended to a basis for $V$.
Let $\dim(K)=d$ and the basis of $K$ be $\{\alpha_1,\cdots,\alpha_d\}$, then $\{\alpha_1,\cdots ,\alpha_d,\alpha_{d+1},\cdots,\alpha_n\}$ is a basis of $V$.
Let $\dim(R)=n-d$ and the basis of $R$ be $\{\beta_{d+1},\cdots,\beta_{n}\}$, then $\{\beta_1,\cdots ,\beta_d,\beta_{d+1},\cdots,\beta_n\}$ is a basis of $W$.
Define the map $L$：
$$L(\alpha_i)=0,\text{for}\ \ \ 1\le i\le d,$$
$$L(\alpha_i)=\beta_i,\text{for}\ \ \ d+1\le i\le n.$$
Let $L$ be linear, then we prove it.
