How to interpret $Rank(T+U)\leq Rank(T)+Rank(U).$

Question

In linear algebra,I encountered a theorem which states that for two matrices $m\times n$ namely $A,B$ the following holds, $Rank(A+B) \leq Rank(A)+Rank(B)$.Similarly if we say in terms of linear transformations $T,U:\mathbb F^n \to \mathbb F^m$ then $Rank(T+U)\leq Rank(T)+Rank(U)$.How to understand it geometrically,I have proved the result in terms of matrices formally,but I want to understand geometrically in terms of linear transforamtions why this should hold i.e. I want to get myself convinced geometrically.

Answer 1

Hint: Everything in the image of $T+U$ is a linear combination of elements from the image of $T$ and the image of $U$.

Answer 2

The rank of a transformation T is the dimension of the image.

We can think of 3 spaces. That subspace of $\mathbb R^m$ which is in the image of $U$ but not in the image of $V.$ The subspace in the image of $V$ but not $U.$ And the subspace in the image of both. $U+V$ will map onto all 3 of these subspaces.

$Rank (U) + Rank (V)$ would count the overlapping space twice.

But if there is no overlap then $Rank (U) + Rank (V) = Rank (U+V)$