# Surjectivity of $T$ is not a necessary condition for $\operatorname{range}ST = \operatorname{range}S$

I was asked to demonstrate that given the finite dimensional vector spaces $V,W$ and $U$ and the linear maps $T:V\to W$, $S:W\to U$ such that $\operatorname{range}ST = \operatorname{range}S$ it may be possible that $T$ is not surjective.

Here is my example is it correct?

Let $V,W$ and $U$ be vector spaces such that $\dim V = \dim W = \dim U = 3$. so that $v_1,v_2,v_3$ and $w_1,w_2,w_3$ and $u_1,u_2,u_3$ are bases for $U$, $V$ and $W$ respectively.

Now let $T$ and $S$ be defined as follows $$T(c_1v_1+c_2v_2+c_3v_3) = (c_1+c_2)w_2+c_3w_3$$ $$S(d_1w_1+d_2w_2+d_3w_3) = (d_1+d_2)u_2+u_3u_3$$

evidently $T$ is not surjective since $\forall \alpha\in V(T\alpha\neq w_1)$. Just to be sure i also verified that $\operatorname{range}S \subseteq \operatorname{range}ST$ since $\operatorname{range} ST\subseteq \operatorname{range}S$ is obvious.

Let $u = (x_1+x_2)u_2+x_3u_3$ be an arbitrary vector in $\operatorname{range}S$ and let $v = (x_1+x_2)v_2+x_3v_3$ we show that $STv =u$. $$ST(v) = ST((x_1+x_2)v_2+x_3v_3) = S((x_1+x_2)w_2+x_3w_3) = (x_1+x_2)u_2+x_3u_3 = u$$

A simpler example would consist in taking $V=W=\mathbb{R}^2$, $U=\mathbb R$, $T(x,y)=(x,x)$ and $S(x,y)=x$.
• Simpler still, $S=T=0$. Commented Feb 16, 2018 at 21:45