Show that if $P$ is an invertible $m$ x $m$ matrix, then$rank(PA)=rank(A)$, then show if $Q$ is invertible, then rank $AQ$= rank $A$.

linear algebraproof-writingsolution-verification

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Show that if $P$ is an invertible $m$ x $m$ matrix, then$rank(PA)=rank(A)$.

solution:

$col(PA)=PA$x

$col(PA)=P(A$x$)$

$col(A)⊆col(PA)$

$rank(A)≤rank(PA)$

$nul(PA)=$x such that $PA$x$=0$

$nul(PA)=$x such that $P(A$x$)=0$

$nul(A)⊆nul(PA)$

$n-rank(A)≤n-rank(PA)$

$rank(PA)≤rank(A)$

$rank(PA)=rank(A)$

Part two: Now, use the above exercise to show that: if $Q$ is invertible, then rank $AQ$= rank $A$

From the above, we know that $rank(Q^TA^T)=rank(A^T)$.

$rank(Q^TA^T)=rank((AQ)^T)=rank(AQ)$

$rank(A^T)=rank(A)$

so, rank $AQ$= rank $A$.

Best Answer

By the rank theorem it suffices to show $\dim(\ker(\mathbf A))=\dim(\ker(\mathbf{PA}))=\dim(\ker(\mathbf {AP}))$. $$\ker(\mathbf{PA})=\{\mathbf x:\mathbf 0=\mathbf{PA}(\mathbf x)=\mathbf P(\mathbf {Ax})\}=\{\mathbf x:\mathbf{A}(\mathbf x)=\mathbf 0\}=\ker(\mathbf A)$$ $$\therefore \dim(\ker(\mathbf {PA}))=\dim(\ker(\mathbf A))$$ Finally, because $\mathbf P^T$ is invertible and the row and column space of $\mathbf{AP}$ have the same dimension then $$\dim(\ker(\mathbf{AP}))=\dim(\ker(\mathbf{P}^T\mathbf{A}^T))=\dim(\ker(\mathbf A^T))=\dim(\ker(\mathbf A))$$ where $^T$ denotes matrix transposition.

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