# Show that a set of linear maps that are not surjective, is not a subspace of all linear maps from $V$ to $W$

"Suppose $V$ and $W$ are finite-dimensional with $\dim V \ge \dim W \ge 2$. Show that $U=\{T \in \mathcal{L}(V, W): T$ is not surjective $\}$ is not a subspace of $\mathcal{L}(V, W)$. "

( $\mathcal{L}(V, W)$ is the set of all linear maps from $V$ to $W$ )

My attempt: If I assume that $U$ is a subspace of $\mathcal{L}(V, W)$ then

Let $T,S \in U$, $v\in V$ and let $T(v) = w_1, S(v) = w_2$

$$(T+S)(v)=T(v)+S(v) = w_1 + w_2 \longrightarrow T + S \in U$$ But this cannot be true because $w_1 \in \operatorname{range}(T)$ and $w_2 \in \operatorname{range}(S)$: $$\operatorname{range}(T) \neq \operatorname{range}(S)$$

$T+S\notin U$, $U$ is not a subspace.

Is this proof correct?

My second attempt:If I assume that $U$ is a subspace of $\mathcal{L}(V, W)$ then

Let $T,S \in U$, $v\in V$ and let $T(v) = w_1, S(v) = w_2$

$$(S+T)(v)=S(v)+T(v)=w_1+w_2$$ $w_1 \in \operatorname{range}(T)$ and $w_2 \in \operatorname{range}(S)$

We know that the range of any linear map $T \in \mathcal{L}(V, W)$ is a subspace of $W$. So in this case, $\operatorname{range}(T)$ and $\operatorname{range}(S)$ are subspaces of W. This means that $$\operatorname{span}(\operatorname{range}(T))=\operatorname{span}(u_1,u_2,\dots,u_m)$$ $$\operatorname{span}(\operatorname{range}(S))=\operatorname{span}(u_{m+1},u_{m+2},\dots,u_n)$$ $u_j \in W, j=1,2,\dots,n$

$w_1$ and $w_2$ can be expressed as linear combinations $$w_1 = k_1u_1 + k_2u_2 + \dots + k_mu_m$$ $$w_2 = k_{m+1}u_{m+1} + k_{m+2}u_{m+2} + \dots + k_nu_n$$

Because $u_1,u_2,\dots, u_m$ is a basis of $\operatorname{range}(T)$, there exists a unique linear map such that $$T(v_j) = u_j$$ $v_j \in V, u_j \in W, j=1,2\dots,m$

Likewise, because $u_{m+1},u_{m+2},\dots,u_{n}$ is a basis of $\operatorname{range}(S)$, there exists a unique linear map such that $$S(v_j) = u_j$$ $v_j \in V, u_j \in W, j=m+1,m+2,\dots, n$

This means that there are no $v_j \in V$ such that $$T(v_j) = S(v_j)$$ If there would exist $T,S \in U$ where $T(v_j) = S(v_j)$ then $$k_1u_1 + k_2u_2 + \dots + k_mu_m=w_2 = k_{m+1}u_{m+1} + k_{m+2}u_{m+2} + \dots + k_nu_n$$

$$\operatorname{range}(T) = \operatorname{range}(S)$$ But this is not the case. Therefore $$\operatorname{range}(T) \ne \operatorname{range}(S)$$

Because $\operatorname{range}(T) \ne \operatorname{range}(S)$, there exists $w_j \in \operatorname{range}(T), w_j \notin \operatorname{range}(S)$. Let $j=1$. Now we have

$$w_1 \in \operatorname{range}(T), w_1 \notin \operatorname{range}(S)$$ $w_2$ can either exist both in $\operatorname{range}(S)$ and $\operatorname{range}(T)$ or only in $\operatorname{range}(S)$. We test both alternatives and see the result.

Let $w_2 \in \operatorname{range}(S), w_2 \in \operatorname{range}(T)$

$$w_1+w_2 \in \operatorname{range}(T)$$ $$w_1+w_2 \notin \operatorname{range}(S)$$ This means that the range of T is closed under addition but the range of S is not closed under addition. $\operatorname{range}(S)$ can therefore not be a subspace of W. That means that $S \notin \mathcal{L}(V, W)$.

If $U=\{T \in \mathcal{L}(V, W): T$ is surjective $\}$, then $\operatorname{range}(T) = \operatorname{range}(S) = W$. But in this case $\operatorname{range}(T) \ne \operatorname{range}(S) \ne W$

Attempt 3: We know that $$\operatorname{dim}(V) \ge \operatorname{dim}(W) \ge 2$$ We know that $U=\{T \in \mathcal{L}(V, W): T$ is not surjective $\}$.

Let $T \in U$. $T$ is not surjective, which implies that $$\operatorname{range}(T) \ne W$$ But because $T \in \mathcal{L}(V, W)$, $\operatorname{range}(T)$ is a subspace of $W$. Let $w_1, w_2, w_3, ... , w_n$ span $\operatorname{range}(T)$. $$\operatorname{span}(\operatorname{range}(T)) = \operatorname{span}(w_1, w_2, \dots, w_n)$$ Now because $\operatorname{range}(T)$ is a subspace of $W$. $w_1, w_2,\dots, w_n$ can be expanded to span $W$. $$\operatorname{span}(W) = \operatorname{span}(w_1, w_2, \dots , w_n, w_{n+1}, w_{n+2}, \dots , w_{m})$$ This implies that $\operatorname{dim}(\operatorname{range}(T)) < \operatorname{dim}(W)$, which means that $$\operatorname{dim}(V) > \operatorname{dim}(\operatorname{range}(T)) \ge 2$$ We can check if $U$ is injective by checking if $\operatorname{null}(T) = \{ 0 \}$. In order for that to be true then $T(u) = T(v), u = v, u,v \in V$.

Let $$\operatorname{span}(w_1, w_2, \dots , w_m) = \operatorname{span}(Tv_1, Tv_2, \dots , Tv_m), Tv_j = w_j, j=1,2\dots,m, v_j \in V, w_j \in W$$ This means that $$T(v) = k_1w_1 + k_2w_2 + \dots + k_nw_n + k_{n+1}w_{n+1} + \dots + k_mw_m = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n + k_{n+1}Tv_{n+1} + \dots + k_mTv_m$$

But we know that $T(v) \in \operatorname{range}(T)$ and because of that $$T(v) = k_1w_1 + k_2w_2 + \dots + k_nw_n = k_1 = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n$$

This means that $$k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n + k_{n+1}Tv_{n+1} + \dots + k_mTv_m$$ $$k_{n+1}Tv_{n+1} + \dots + k_mTv_m = 0$$ $$T(k_{n+1}v_{n+1}) + \dots + T(k_mv_m) = 0$$ $$T(k_{n+1}v_{n+1} + \dots + k_mv_m ) = 0$$ Which implies that $$k_{n+1}v_{n+1} + \dots + k_mv_m \in \operatorname{null}(T)$$ This means that $\operatorname{null}(T) \ne \{0\}$ which means that T is not injective. So for all $T \in \mathcal{L}(V, W)$, $T$ is not injective.

Now, let $T,S \in \mathcal{L}(V, W)$ and $u,v \in V$ such that $T(v) = T(u), u \ne v$ $$(T+S)(v) = T(v) + S(v) = T(u) + S(v) \ne (T+S)(v)$$

• Why does it matter that the ranges are different? Why can you conclude the sum isn't in $U$? – Michael Burr Jul 10 '18 at 11:28
• What makes you think that the ranges are different? – José Carlos Santos Jul 10 '18 at 11:28
• General hint. When a problem begins "Show that ... is not true" you should probably start by looking for a relatively explicit single counterexample rather than a more abstract argument. You can see that strategy in the two answers. – Ethan Bolker Jul 10 '18 at 11:55
• @E.Bob You're welcome. Life is a little trickier when the problem says "Prove or disprove ...". Then perhaps you start looking for a counterexample. If you can't find one you look for a proof. If you can't find one you go back to the search for a counterexample ... until you finally understand it all. The best kind of question. – Ethan Bolker Jul 10 '18 at 12:08
• The updated answer is still not correct. Some issues: The ranges could overlap (even if they’re not equal), so you can’t get distinct basis elements. Also, these maps are not injective, so you can’t guarantee the uniqueness of the $v_j$’s. Moreover, the range argument doesn’t work because closure doesn’t tell you anything when you’re adding something not in the range. The condition on $U$ is non-surjectivity and you’ve never really used surjectivity in the entire answer (this hints at the fact that there’s something missing). – Michael Burr Jul 10 '18 at 21:09

## 2 Answers

Pick a basis $v_1,...,v_m,...,v_n$ of $V$ and a basis $w_1,...,w_m$ of $W$. Consider the linear maps $Tv_i=w_i$, for $i=1,...,m-1$ and $Tv_i=0$, for $i\geq m$, while $Sv_m=w_m$ and $Sv_i=0$ for $i\neq m$. When you add them you get a surjective linear map.

You want to find two linear transformations from $V \to W$ that are not surjective but their linear combination is surjective.

Suppose that $\dim V = \dim W = 2$ for now to see the main idea more clearly.

Consider $T$ and $S$ as maps from $\mathbb{R}^2 \to \mathbb{R}^2$ sending $(x,y) \to (x,0)$ and $(x,y) \to (0,y)$ respectively. None of these transformations are surjective on their own, yet $T+S$ is surjective.

Now, similarly generalize this argument to other cases where $\dim V \geq \dim W \geq 2$.