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

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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 $$ A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{-1&0\\0&-1}. $$

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    $\begingroup$ Dear Ben: Thanks for your great help in completing my proof! The inequality you mentioned was the key to complete it successfully. I acknowledged your help in my edited posting. Thanks a lot !! 🙏🙏 $\endgroup$
    – Dr. Sundar
    Mar 31, 2022 at 14:46

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