# A quick proof for $\mbox{rank}(A + B) \leq \mbox{rank}(A) + \mbox{rank}(B)$

A recent question in this forum led me to recall the linear algebra result: $$\mbox{rank}(A + B) \leq \mbox{rank}(A) + \mbox{rank}(B), \ \ \left( A, B \in \mathbf{R}^{n \times n} \right)$$

(I corrected my first posting after Ben Grossmann corrected it and cited an important inequality in linear algebra. I like to thank him first!)

Define $$U = \mbox{Range}(A)$$, $$V = \mbox{Range}(B)$$.

Then $$U$$ and $$V$$ are subspaces of $$\mathbf{R}^n$$.

Clearly, $$\mbox{rank}(A) = \mbox{dim}(U)$$, $$\mbox{rank}(B) = \mbox{dim}(V)$$.

Also, $$\mbox{rank}(A + B) \leq \mbox{dim}(U + V)$$.

(Thanks to Ben Grossmann for correcting my original statement!)

We know the theorem from linear algebra: $$\mbox{dim}(U + V) = \mbox{dim}(U) + \mbox{dim}(V) - \mbox{dim}(U \cap V)$$ which implies that $$\mbox{dim}(U + V) \leq \mbox{dim}(U) + \mbox{dim}(V).$$

Thus, $$\mbox{rank}(A + B) \leq \mbox{dim}(U + V) \leq \mbox{rank}(A) + \mbox{rank}(B)$$.

You have made one mistake: it is not necessarily the case that $$\operatorname{rank}(A + B) = \dim(U + V)$$. However, it is true that $$\operatorname{Range}(A + B) \subseteq U + V$$, which means that $$\operatorname{rank}(A + B) \leq \dim(U + V)$$.
Otherwise, your proof is correct and complete. After changing $$\operatorname{rank}(A + B) = \dim(U + V)$$ to $$\operatorname{rank}(A + B) \leq \dim(U + V)$$, the rest of your proof works as is.
A quick counterexample to $$\operatorname{rank}(A + B) = \dim(U + V)$$: consider $$A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{-1&0\\0&-1}.$$