Answer to Question #219805 in Linear Algebra for wisdom

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

Answer:-

Let $\dim K=m$ and let $\{\beta_1,\beta_2,…,\beta_m\}$ be a basis of $K$ . Then $\dim R=n-m$ .

Since $K\subseteq V$ , it follows that vectors $\{\beta_1,\beta_2,…,\beta_m\}$ can be complemented to a basis of $V$ .

So, let $\beta= \{\beta_1,\beta_2,…,\beta_n\}$ be a basis of $V$ .

Let $\gamma = \{\gamma_1,\gamma_2,…,\gamma_{n-m}\}$ be a basis of $R$ .

Now we define a linear transformation $L:V\rightarrow W$ for basis vectors:

$L\beta_1=…=L\beta_m=0$ and $L\beta_{m+1}=\gamma_1$ , … , $L\beta_n=\gamma_{n-m}$ .

For $v =a_1\beta_1+a_2\beta_2+…+a_n\beta_n\in V$ we have $Lv=a_{m+1}\gamma_1+…+a_{n}\gamma _{n-m}$

$\text{Im} \ L=\{Lv\ |\ v\in V\}=\{a_{m+1}\gamma_1+…+a_{n}\gamma _{n-m}|a_i\in F\}=R$

$Lv=a_{m+1}\gamma_1+…+a_{n}\gamma _{n-m}=0$ if and only if $a_{m+1}=…=a_n=0$ .

Therefore, $Lv=0$ only for $v=a_1\beta_1+…+a_m\beta_m$ , where $a_i\in F$ . It means that $\text{Ker}\ L=K$ .