# Let $T,U:V\to W$ be linear transformations. Prove that if $W$ is finite-dimensional, then $\text{rank}(T+U)\leq\text{rank}(T) + \text{rank}(U)$.

Let $$T,U:V\to W$$ be linear transformations.

(a) Prove that $$R(T+U)\subseteq R(T) + R(U)$$.

(b) Prove that if $$W$$ is finite-dimensional, then $$\text{rank}(T+U)\leq\text{rank}(T) + \text{rank}(U)$$.

MY ATTEMPT

(a) If $$w\in R(T+U)$$, there exists $$v\in V$$ such that $$w = (T+U)v = T(v) + U(v)$$.

But $$T(v)\in R(T)$$ and $$U(v)\in R(U)$$.

Thus $$w\in R(T) + R(U)$$, from whence we conclude that $$R(T+U)\subseteq R(T)+R(U)$$.

(b) Since $$R(T + U)$$ is a linear subspace of $$R(T)+R(U)$$, we have that \begin{align*} \dim R(T+U) & \leq \dim(R(T) + R(U)) = \dim R(T) + \dim R(U) - \dim(R(T)\cap R(U))\\\\ & \leq \dim R(T) + \dim R(U) \end{align*}

Could someone verify if I am not doing any conceptual mistake?

• That's all correct. – Berci Apr 17 '20 at 23:20
• The equality you use in b) is worth proving. Though you don’t really need it, you can skip to the inequality. The proof of that id easier – Jonathan Hole Apr 18 '20 at 1:53

## 2 Answers

$$a)$$ looks good.

$$b)$$ follows from $$a)$$, since $$\operatorname{rank}T=\operatorname{dim}R(T)$$

(You correctly used the formula for the dimension of the sum of two subspaces.)

Concepts:

a)This shows$$R(T+U)$$ is a subspace of $$R(T)+R(U).$$

b) Note $$T+U: V \rightarrow W$$ is a linear transformation, $$rank(T)$$ and $$rank(U) ≤ dim W$$(finite). Hence,$$rank(T) + rank(U)$$ is finite.

Let's do it,

a) $$x',y'$$ are in $$R(T+U)$$ $$\rightarrow$$ there exist $$x,y$$ in$$V$$ such that $$(T+U)(x) = x'$$, $$(T+U)(y) = y'$$ $$\rightarrow x'+y'=(T+U)(x)+(T+U)(y) =(T+U)(x+y)= T(x)+U(x) + T(y)+ U(y) = T(x+y) + U(x+y)$$ $$x'+y'$$ is in $$R(T)+R(U)$$ $$x' \in R(T+U)$$ and $$c$$ is a scalar, then, there exist $$x,y \in V$$ such that $$(T+U)(x) = x'$$ $$\rightarrow (T+U)(cx)$$ $$= c(T+U)(x) = cx' = T(cx)+ U(cx)$$ for some $$x \in V$$ $$\rightarrow cx' \in R(T) + R(U)$$ This shows $$R(T+U)$$ is a subspace of $$R(T)+R(U).$$ It looks a little bit more complicated but it's the same.

b) it looks good.