Rank of product of matrices with full column rank

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Let $widetildeZ = ZA$ where $Z$ is a $N$ by $K$ matrix and $A$ is a $K$ by $M$ matrix with full column rank (with $K>M$). Also let $X$ be a $N$ by $M$ matrix. Can we say anything about the rank of $widetildeZ$? Is the product $widetildeZ'X$ invertible?



I know that in general, if we have matrices $A$ ($m$ by $n$), $C$ ($l$ by $m$) of rank $m$, then $rank(CA) = rank(A)$, but that doesn't seem to help here.







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  • Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
    – LinAlg
    Jul 25 at 14:33














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down vote

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Let $widetildeZ = ZA$ where $Z$ is a $N$ by $K$ matrix and $A$ is a $K$ by $M$ matrix with full column rank (with $K>M$). Also let $X$ be a $N$ by $M$ matrix. Can we say anything about the rank of $widetildeZ$? Is the product $widetildeZ'X$ invertible?



I know that in general, if we have matrices $A$ ($m$ by $n$), $C$ ($l$ by $m$) of rank $m$, then $rank(CA) = rank(A)$, but that doesn't seem to help here.







share|cite|improve this question



















  • Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
    – LinAlg
    Jul 25 at 14:33












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $widetildeZ = ZA$ where $Z$ is a $N$ by $K$ matrix and $A$ is a $K$ by $M$ matrix with full column rank (with $K>M$). Also let $X$ be a $N$ by $M$ matrix. Can we say anything about the rank of $widetildeZ$? Is the product $widetildeZ'X$ invertible?



I know that in general, if we have matrices $A$ ($m$ by $n$), $C$ ($l$ by $m$) of rank $m$, then $rank(CA) = rank(A)$, but that doesn't seem to help here.







share|cite|improve this question











Let $widetildeZ = ZA$ where $Z$ is a $N$ by $K$ matrix and $A$ is a $K$ by $M$ matrix with full column rank (with $K>M$). Also let $X$ be a $N$ by $M$ matrix. Can we say anything about the rank of $widetildeZ$? Is the product $widetildeZ'X$ invertible?



I know that in general, if we have matrices $A$ ($m$ by $n$), $C$ ($l$ by $m$) of rank $m$, then $rank(CA) = rank(A)$, but that doesn't seem to help here.









share|cite|improve this question










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asked Jul 25 at 14:23









elbarto

1,519523




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  • Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
    – LinAlg
    Jul 25 at 14:33
















  • Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
    – LinAlg
    Jul 25 at 14:33















Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
– LinAlg
Jul 25 at 14:33




Any relation between $M$ and $N$? The rank of $tildeZ$ is at most $M$ and at most $N$.
– LinAlg
Jul 25 at 14:33















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