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There's a linear algebra problem I'm having some trouble with:

Let $A$ and $B$ be square matrices with the dimensions $n\times n$.

Prove or disprove:

  1. If $A^2 + BA$ is invertible, then $A$ is also invertible.
  2. If $A^2 + BA$ is not invertible, then $A$ isn't invertible either.

Any help with this would be appreciated. I recognize that if $A^2 + BA$ is invertible then there is a matrix $C$ so that $(A^2 + BA)\cdot C = I$ but beyond that I'm a little lost.

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    $\begingroup$ The determinant of the product of two matrices is equal to the product of the determinants of each matrix $\endgroup$ – imranfat Nov 20 '15 at 16:25
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  1. $I=C(A^2 + BA)=C(A+B)A$ and so $C(A+B)$ is the inverse of $A$.

  2. This is false. Take $A=I$ (or any invertible matrix) and $B=-A$.

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  • $\begingroup$ Clear enough... $\endgroup$ – imranfat Nov 20 '15 at 16:37
  • $\begingroup$ Thanks! I understand your logic for the first argument but not for the second one... could you elaborate? $\endgroup$ – Or Bairey-Sehayek Nov 20 '15 at 16:53
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    $\begingroup$ @OrBairey-Sehayek, to prove that some claim is false, it is enough to give a counter example. $\endgroup$ – lhf Nov 20 '15 at 17:41
  • $\begingroup$ Oh I see thanks! $\endgroup$ – Or Bairey-Sehayek Nov 20 '15 at 17:46

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