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 $A$.

Question

Did I show this properly? Any mistakes, any steps left out? Thank you. $$$$

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 $A$.

Does this answer your question? If $A \in \mathbb{C}^{m\times n}$ is full-column rank matrix, then is rank($AB$) = rank ($BA$) = rank($B$)? — PinkyWay, Nov 01 '20 at 13:18
Is $X\in\Bbb F^n$? Second, $\operatorname{col(PA)}={v\in\Bbb F^n\mid v=APX}$, as you wrote, $APX=\operatorname{col}(AP)$ is meaningless unless $X=0$, because a non-trivial subspace doesn't have one single element. — PinkyWay, Nov 01 '20 at 13:39
x is an arbitrary vector in R^n representing all possible weights that can be used to form a linear combination of the columns of PA, which is exactly what the column space is. — bfff, Nov 01 '20 at 13:46
are you saying that my failure to use the set notation makes the statement meaningless? — bfff, Nov 01 '20 at 13:48
Yes, that part is meaningless because you haven't even mentioned what $x$ is. I just assumed what it meant and then saw what you wrote about the kernel and realized what you probably wanted to say. Keep in mind you have to write everything down on the exam. Good luck! (: — PinkyWay, Nov 01 '20 at 16:22

Oliver Kayende · Accepted Answer · 2020-11-03T07:56:24.033

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.

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 $A$.

1 Answers1