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For a $3\times 3$ matrix $A$ over the field of real numbers, let $r(A)$ denote its rank. Do there exist two $3\times 3$ matrices $B,C$ such that:

  • $r(B) = 2, r(C) = 1$ and
  • $r(B + C) = 3, r(B - C) = 2$?

I know intuitively that this is not possible (have tried to contruct a counterexample to no avail) but I cannot get a formal proof out of this. For example we have: $C=(c_1 \; \lambda_1 c_1 \; \lambda_2 c_1)$ where $c_1$ is the independent column vector of the $C$ matrix and that $\sum_{i=1}^3 \lambda_i(b_i+c_i)=0 \Rightarrow \lambda_i=0$ because of the fact that $r(B + C) = 3$ but I can't find anything else. Can anyone help me?

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Suppose towards a contradiction that this is possible. Let $v$ be a non-zero vector with $B(v) = C(v)$, using the rank of $B - C$. By the rank of $B + C$, we cannot have $B(v) = C(v) = 0$. But $C$ has rank 1, which means that every vector in its image is a multiple of $C(v)$. Now for any vector $w$, let $\lambda_wC(v) = C(w)$. Then we have that $$ (B+C)(w) = B(w) + \lambda_wC(v) = B(w+\lambda_wv). $$ This means that the image of $B+C$ is contained in that of $B$, contradicting that $r(B+C) > r(B)$.

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    Can you please explain a little further: $rank(B+C)=2 \Rightarrow (B+C)(r)=0,r \neq 0$. How to we get that we cannot have: $B(u)=C(u)=0$? – 2017-02-09
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    Sorry I meant: $rank(B+C)=3 \Rightarrow (B+C)(r)=0,r=0$. How to we get that we cannot have: $B(u)=C(u)=0$? – 2017-02-09
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    $B+C$ has full rank, and thus is injective. Hence, $B(v) + C(v) = (B+C)(v) = 0 \implies v = 0$. But by assumption, $v$ was non-zero. – 2017-02-09