1. ## [SOLVED] Nilpotence

A nilpotent matrix is any $n\times n$ matrix $A$ where $\exists~k$ such that $A^k=0$. I need to prove that if $A$ is nilpotent, then $A+I$ is invertible (where $I$ is the identity matrix). I am not allowed to use any theorems regarding determinants, trace, linear (in)dependence, eigenvalues, etc. I am only supposed to use basic matrix operations and properties of invertible matrices.

I wrote this proof, but it doesn't feel right, though I can't place my finger on the error.

Proof: Assume for a contradiction that $A+I$ is not invertible; that is, $\forall~B$, $B(A+I)\neq I$.

Thus, $BA+B\neq I\implies BA\neq I-B$. Multiplying through (on the right) by $A^{k-1}$ gives us $BA\cdot A^{k-1}\neq (I-B)A^{k-1}\implies (I-B)A^{k-1}\neq0$.

However, choosing $B=I$ implies that $(I-B)A^{k-1}=0$, giving us a contradiction. Therefore $A+I$ is invertible. $\square$

Is that a valid proof, or did I screw up somewhere?

2. Originally Posted by redsoxfan325
A nilpotent matrix is any $n\times n$ matrix $A$ where $\exists~k$ such that $A^k=0$. I need to prove that if $A$ is nilpotent, then $A+I$ is invertible (where $I$ is the identity matrix). I am not allowed to use any theorems regarding determinants, trace, linear (in)dependence, eigenvalues, etc. I am only supposed to use basic matrix operations and properties of invertible matrices.

I wrote this proof, but it doesn't feel right, though I can't place my finger on the error.

Proof: Assume for a contradiction that $A+I$ is not invertible; that is, $\forall~B$, $B(A+I)\neq I$.

Thus, $BA+B\neq I\implies BA\neq I-B$. Multiplying through (on the right) by $A^{k-1}$ gives us $BA\cdot A^{k-1}\neq (I-B)A^{k-1}\implies (I-B)A^{k-1}\neq0$.

However, choosing $B=I$ implies that $(I-B)A^{k-1}=0$, giving us a contradiction. Therefore $A+I$ is invertible. $\square$

Is that a valid proof, or did I screw up somewhere?
no, it's not a valid proof because $BA\neq I-B$ does not necessarily imply that $BA\cdot A^{k-1}\neq (I-B)A^{k-1}.$

the proof is quite easy: $A^k=0$ implies that $A^m=0,$ for all $m \geq k.$ so we may assume that $k$ is odd and thus: $I=A^k+I=(A+I)(A^{k-1}-A^{k-2} + \cdots -A + I).$

3. Thank you. It always seems so obvious once I see the solution...