Prove that if $A$ is a linear transformation on a finite dimensional vector space $V$, then $V\cong \mathrm{Range}(A)\oplus \mathrm{Nullspace}(A)$.

Question

In order to prove this do I just need to show that the dimensions are the same?

Prove that if $A$ is a linear transformation on a finite dimensional vector space $V$, then $V\cong \mathrm{Range}(A)\oplus \mathrm{Nullspace}(A)$.

Answer 1

Take any basis of null space and extend to the basis of whole space V.

Let $B'=\{\alpha_1,\alpha_2,\cdot\cdot\cdot ,\alpha_k\}$ be any basis of null space then there exsist vectors $\alpha_{k+1},\cdot\cdot\cdot,\alpha_n$ such that $B''=\{\alpha_1,\alpha_2,\cdot\cdot\cdot,\alpha_k,\alpha_{k+1},\cdot,\alpha_n\}$ Now prove that $B'''=\{A(\alpha_{k+1}),\cdot\cdot\cdot,A(\alpha_n)\}$ is basis of range of $A$.

Answer 2

$\newcommand{\Range}{\mathrm{Range}} \newcommand{\Nullspace}{\mathrm{Nullspace}}$Let $\{v_{1},...,v_{n}\}$ be a base of $N(A):=\Nullspace(A)$. Complete $\{v_{1},...,v_{n}\}$ to a base of $V$. If $\{u_{1},...,u_{m},v_{1},...,v_{n}\}$ is the base, lets prove that $\{A(u_{1}),...,A(u_{m})\}$ is a base of $R(A):=\Range(A)$.

For all $A(u)\in R(A)$, write $$u=a_{1}u_{1}+...+a_{m}u_{m}+b_{1}v_{1}+...+b_{n}v_{n}.$$ Therefore $$A(u)=a_{1}A(u_{1})+...+a_{m}A(u_{m})+b_{1}A(v_{1})+...+b_{n}A(v_{n})=b_{1}A(v_{1})+...+b_{n}A(v_{n}).$$ Therefore $R(A)$ is generate by $\{A(u_{1}),...,A(u_{m})\}$. Now, if $$0=c_{1}A(v_{1})+...+c_{n}A(v_{n})=A(c_{1}v_{1}+...+c_{n}v_{n})$$ then $c_{1}v_{1}+...+c_{n}v_{n}\in N(A)$ and so $$c_{1}v_{1}+...+c_{n}v_{n}=d_{1}u_{1}+...+d_{m}u_{m}$$ i.e., $$d_{1}u_{1}+...+d_{m}u_{m}-c_{1}v_{1}-...-c_{n}v_{n}=0$$ Since $\{u_{1},...,u_{m},v_{1},...,v_{n}\}$ is LI, we have $c_{1}=...=c_{n}=d_{1}=...=d_{m}=0$. Thus, we prove that $\{A(u_{1}),...,A(u_{m})\}$ is a base of $R(A)$. Therefore $$\dim(V)=m+n=\dim(N(A))+\dim(R(A))$$ and $V\cong N(A)\oplus R(A).$