Answer to Question #219848 in Linear Algebra for Anuj

Question #219848

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.



1
Expert's answer
2021-07-26T09:49:21-0400

Let "\\dim K=m" and let "\\{\\beta_1,\\beta_2,\u2026,\\beta_m\\}" be a basis of "K" . Then "\\dim R=n-m" .

Since "K\\subseteq V" , it follows that vectors "\\{\\beta_1,\\beta_2,\u2026,\\beta_m\\}" can be complemented to a basis of "V" .

So, let "\\beta= \\{\\beta_1,\\beta_2,\u2026,\\beta_n\\}" be a basis of "V" .

Let "\\gamma = \\{\\gamma_1,\\gamma_2,\u2026,\\gamma_{n-m}\\}" be a basis of "R" .

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

"L\\beta_1=\u2026=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+\u2026+a_n\\beta_n\\in V" we have "Lv=a_{m+1}\\gamma_1+\u2026+a_{n}\\gamma _{n-m}"

"\\text{Im} \\ L=\\{Lv\\ |\\ v\\in V\\}=\\{a_{m+1}\\gamma_1+\u2026+a_{n}\\gamma _{n-m}|a_i\\in F\\}=R"


"Lv=a_{m+1}\\gamma_1+\u2026+a_{n}\\gamma _{n-m}=0" if and only if "a_{m+1}=\u2026=a_n=0" .

Therefore, "Lv=0" only for "v=a_1\\beta_1+\u2026+a_m\\beta_m" , where "a_i\\in F" . It means that "\\text{Ker}\\ L=K" .



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