Rank Question- with adjoints

Asked Nov 02 '16 at 02:52

Active Nov 02 '16 at 03:04

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I was hoping for help with a math practice problem for an exam from my book.

We are wanted to use the property $\operatorname{Ker}(A)=\operatorname{Ker}(A^*A)$ to show:

a) Rank(A)=rank(A*A)

b) IF Ax=0 has only the trivial solution, A is left invertible.

I am very unsure as to how to start this problem so I was hoping someone could point me in the right direction!

edited Nov 02 '16 at 03:04

asked Nov 02 '16 at 02:52

readyforthebettys

What do you denote $A^*$? – Bernard Nov 02 '16 at 02:58
For $(a)$ you just need the rank-nullity theorem. Is $(b)$ supposed to be an if ... then statement or are you saying $\operatorname{Ker}(A)=\operatorname{Ker}(A^*A)$ implies both of those properties? – Nov 02 '16 at 03:02
A* is denoting the adjoint – readyforthebettys Nov 02 '16 at 03:03
To use the rank nullity theorem can you say the dim(Ker(A))=dim(Ker(A* A)) and since they are both in V we use this to say dim(Im(A))=dim(Im(A* A)) since dim(ker)+dim(im)=dim(V) so then we know they have the same dimension of the image, which is the same as rank? – readyforthebettys Nov 02 '16 at 03:05
Given your edit, you can use the rank-nullity theorem for $(b)$ as well. – Nov 02 '16 at 03:05
One little change, $A$ and $A^A$ both have domain $V$* -- they're not both in $V$. Other than that, yes. That's the argument. – Nov 02 '16 at 03:07
how would you use rank-nullity for b)? – readyforthebettys Nov 02 '16 at 03:22
$Ax=0$ has only trivial solution $\implies$ nullity = 0 $\implies$ rank = $\dim(V)$ $\implies$ what? – Nov 02 '16 at 03:26
so that would make it invertible, ah I see that makes sense. Thanks! – readyforthebettys Nov 02 '16 at 03:27
Well, there are a couple of extra steps. But yeah, it eventually implies that $A$ is left invertible (hint: if you can't immediately see where to go from the link, try multiplying the equation $Ax=0$ by $A^*$). – Nov 02 '16 at 03:31

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