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Suppose $B$ has a full column rank. Does it hold that $\operatorname{rank}(AB)=\operatorname{rank}(A)$?


I found a similar post. But it does not prove or disprove the above statement.

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The statement is only true for $B$ square otherwise, as counterexample, we can consider $A_{n\times n} $ full rank and $B_{n\times 1} $ such that rank$(AB)=1$

$$A=\begin{pmatrix} 1&1 \\1&-1 \end{pmatrix}, B=\begin{pmatrix} 1\\0 \end{pmatrix} \implies AB=\begin{pmatrix} 1\\1 \end{pmatrix}$$

which leads to rank$(AB)=1$ with rank$(A)=2$.

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  • $\begingroup$ why $Av$ does not equal to 0? $\endgroup$
    – mxdxzxyjzx
    Nov 16, 2019 at 9:10
  • $\begingroup$ @mxdxzxyjzx I've revised with a simpler way to see that, indeed the first one wasn't completely correct and enlightening. $\endgroup$
    – user
    Nov 16, 2019 at 9:37

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