Linear Algebra-Linear Operator

Question

How do i construct a linear operator of rank=2 from R^3 to itself. Is it sufficient if i find a linear operator and deduce that it is of rank 2 and it fulfills the conditions of being a linear operator?

Answer 1

Define it on the basis vectors of $R^3$. Let $v_1,v_2, v_3$ be a basis of $R^3$. Define $T : R^3 \to R^3$ by $$Tv_1 = v_1\\ Tv_2 = v_2\\ Tv_3 = v_2$$ Then range $T$ is a two dimensional subspace of $R^3$ as $Tv_1,Tv_2,Tv_3$ spans $\text{range } T$ and reducing this spanning list to a basis gives us $Tv_1, Tv_2$.

Here I am using the easy to prove theorem that if $v_1,\dots,v_n$ is a basis of $V$, then $Tv_1,\dots,Tv_n$ spans $\text{range }T$ for a linear map $T : V\to W$.