Answer to Question #220099 in Linear Algebra for Anuj

Question

Let V and W be vector spaces over F, where V is n-dimensional. Let K ≤ V and R ≤ W

be finite-dimensional subspaces such that dim K + dim R = n. Prove that there exists a linear

transformation L : V −→ W such that ker L = K and Im L = R.

Answer 1

Let $r=\dim R$, then $\dim K=n-r$.

Let $e_1, ..., e_r$ be any basis for $R$.

This basis can be extended to a basis $e_1, \dots, e_r, e_{r+1}, \dots, e_m$ for $W$, where $m=\dim W$.

Let $f_{r+1}, ..., f_{n}$ be any basis for $K$.

This basis can be extended to a basis $f_1, \dots, f_{r}, f_{r+1}, \dots,f_n$ for $V$.

Define a linear transformation $L:V\to W$ by the rule:

$L(a_1f_1+a_2f_2+\dots+a_nf_n)=a_{1}e_1+\dots +a_{r}e_{r}$

1) Show that ${\rm Im}\,L=R$:

Since $e_1, ..., e_{r}$ is a basis for $R$,

$L(a_1 f_1+\dots+a_n f_n)=a_1e_1+\dots +a_{r}e_{r}\in R$ and ${\rm Im}\,L\subset R$

For any $y\in R$ if $y=y_1e_1+\dots + y_{r}e_{r}$ then, evidently, $y=L(y_1f_1+\dots + y_{r}f_{r})\in {\rm Im}\,L$.

Therefore, ${\rm Im}\,L=R$.

2) Show that ${\rm Ker}\ L=K$:

For any $x\in K$ if $x=x_{r+1}f_{r+1}+\dots+x_nf_n$ then $L(x)=0$, i.e. $x\in{\rm Ker}\ L$ and, hence, $K\subset {\rm Ker}\ L$.

Conversely, if $L(x)=0$ and $x=x_{1}f_{1}+\dots+x_nf_n$ then $L(x)=x_{1}e_1+\dots +x_{r}e_{r}=0$. Since Since $e_1, ..., e_{r}$ is a basis for $R$, they are linear independent, therefore $x_1=\dots=x_r=0$.

This implies that $x=x_{r+1}f_{r+1}+\dots+x_nf_n\in K$ and, hence, ${\rm Ker}\ L\subset K$.

Therefore ${\rm Ker}\ L= K$.