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!

  • $\begingroup$ What do you denote $A^*$? $\endgroup$
    – Bernard
    Nov 2, 2016 at 2:58
  • $\begingroup$ 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? $\endgroup$
    – user137731
    Nov 2, 2016 at 3:02
  • $\begingroup$ A* is denoting the adjoint $\endgroup$ Nov 2, 2016 at 3:03
  • $\begingroup$ 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? $\endgroup$ Nov 2, 2016 at 3:05
  • $\begingroup$ Given your edit, you can use the rank-nullity theorem for $(b)$ as well. $\endgroup$
    – user137731
    Nov 2, 2016 at 3:05


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