Solution 3b 7

Suppose $V$ and $W$ are finite-dimensional with $2 \le \dim V \le \dim W$ .
Show that $\{T \in \mathcal L(V,W) : T \text{ is not injective}\}$ is not a subspace of $\mathcal L(V,W)$ .

This is the same as 3b/4, non-injectivity isn't closed under addition.

Let $v_1,\dots,v_m$ be a basis for $V$ and $w_1,\dots,w_n$ be a basis for $W$ . Define (note $a_2$ exists as $\dim V \ge 2$ )

\begin{aligned} T(a_1v_1 + \dots + a_mv_m) &= a_2w_2 + \dots + a_mw_m \\ S(a_1v_1 + \dots + a_mv_m) &= a_1w_1 \end{aligned}

Both $T$ and $S$ are not injective since $Tv_1 = 0$ and $Sv_2 = 0$ . But

(S+T)(a_1v_1+\dots+a_mv_m) = a_1w_1 + \dots + a_mw_m

is injective, thus the given set is not a subspace as it isn't closed under addition.