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 $\text{rank}(T+U)\leq\text{rank}(T) + \text{rank}(U)$.


(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?

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    $\begingroup$ That's all correct. $\endgroup$ – Berci Apr 17 '20 at 23:20
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    $\begingroup$ 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 $\endgroup$ – Jonathan Hole Apr 18 '20 at 1:53

$a)$ 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.)



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.


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