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I'm having a bit of trouble with the following proof, and I was hoping someone could help. I started with some things already but I'm not sure where to go from there.

Suppose $A\in M_{n\times n}(\mathbb{R})$ for $n\geq 2$ and let $I_n$ be the $n\times n$ identity matrix. Prove, or give a counter example, that if $A^3=0$, then $I_n+A$ is invertible.

Proof:

We know that $A^3=0$ and $det(0)=0$ such that $det(A^3)=det(A)\times det(A)\times det(A)=0$. From that it follows that $det(A)=0$.

We want to prove that $I_n+A$ is invertible, so $det(I_n+A)$ must be unequal to $0$.

We know that $det(I_n+A)=\sum_{j=1}^n\Big[(-1)^{j+1}(I_n+A)_{1j}\times det\widetilde{(I_n+A)}_{1j}\Big]$ with the tilde determinant representing the minor, deleting row $i$ and $j$... I'm not sure how to go from here, anyone?

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Because just $$(I+A)(I-A+A^2)=I+A^3=I$$

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  • $\begingroup$ Do matrices have a distributive property...? I didn't expect that given they don't commute. $\endgroup$ – Marc Jan 16 '18 at 17:59
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    $\begingroup$ Yes of course. $A(B+C)=AB+AC.$ If $X=A^n$ and $Y=A^m$ then $XY=YX$. The last follows from $A(BC)=(AB)C$. $\endgroup$ – Michael Rozenberg Jan 16 '18 at 18:00
  • $\begingroup$ $I$ commutes with any matrix, that's why it works. $\endgroup$ – Bernard Jan 16 '18 at 18:38
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If $A^3=0$, its only eigenvalue is $0$, therefore the only eigenvalue of $A+I$ is $1$. The determinant of a matrix is equal to the product of its eigenvalues, so $\det(A+I)\ne0$.

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