1

I've two doubts:

1) If $A,B$ are square matrices and $AB=I_n$, but not necessary $BA=I_n$, is true that $A$ is invertible and $A^{-1}=B$?

2) If $AB=B$, then $A=I_n$.

Well, I know that the second afirmation if false, but I don't know why. Look:

Consider $A = \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & { - 1} \\ \end{array}} \right),$ $B = \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 0 \\ \end{array}} \right)$, here $AB=B$ but $A\neq I_n$.

Someone can help me?

Michael Hardy
  • 1
DGs
  • 11
  • 1
  • For 1 : For a finite dimensional linear map, injectivity implies surjectivity. – Prahlad Vaidyanathan Sep 01 '13 at 18:55
  • The first is true, perhaps most easily seen by looking at the linear maps the matrices induce. For the second, if $AB = B$, then $(A-I)B = 0$ (and vice versa). If $B$ has not full rank, there are matrices $C \neq 0$ with $CB=0$. – Daniel Fischer Sep 01 '13 at 18:56

1 Answers1

2

Let $$f_B:\mathcal M_n(\mathbb R)\rightarrow \mathcal M_n(\mathbb R),\quad X\mapsto BX $$ then $f_B$ is an endomorphism and we have $$X\in \ker f_B\iff BX=0\Rightarrow ABX=X=0$$ so $f_B$ is injective and then bijective since we are in the finite dimensional space hence there's $C$ such that $$BC=I\Rightarrow ABC=A\iff C=A$$

  • I have proved that if $AB=I$ then $BA=I$ and that means if$B$ has left inverse then it's also a right inverse and then $A=B^{-1}$. –  Sep 01 '13 at 19:09
  • You are very smart! And how about the second problem? – DGs Sep 01 '13 at 19:10
  • We have this general result $AB=0\iff \mathrm{Im}(B)\subset \ker(A)$ so can you construct a matrices $A$ and $B$ such that $AB=0$? –  Sep 01 '13 at 19:12