1. ## [SOLVED] Nilpotence

A nilpotent matrix is any $\displaystyle n\times n$ matrix $\displaystyle A$ where $\displaystyle \exists~k$ such that $\displaystyle A^k=0$. I need to prove that if $\displaystyle A$ is nilpotent, then $\displaystyle A+I$ is invertible (where $\displaystyle I$ 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 $\displaystyle A+I$ is not invertible; that is, $\displaystyle \forall~B$, $\displaystyle B(A+I)\neq I$.

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

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

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

2. Originally Posted by redsoxfan325
A nilpotent matrix is any $\displaystyle n\times n$ matrix $\displaystyle A$ where $\displaystyle \exists~k$ such that $\displaystyle A^k=0$. I need to prove that if $\displaystyle A$ is nilpotent, then $\displaystyle A+I$ is invertible (where $\displaystyle I$ 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 $\displaystyle A+I$ is not invertible; that is, $\displaystyle \forall~B$, $\displaystyle B(A+I)\neq I$.

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

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

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

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

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