# Thread: nilpotent

1. ## nilpotent

Suppose $\displaystyle A$ is nilpotent. Prove that $\displaystyle I+A$ is invertible.

So $\displaystyle \det A = 0$. Then $\displaystyle \det (A+I) = 1$ which implies that $\displaystyle A+I$ is invertible.

Is this correct?

2. Ok let $\displaystyle B = \sum_{n=0}^{\infty} (-1)^{n} A^{n}$ which converges. So $\displaystyle B(I+A) = I$. And so $\displaystyle B = (I+A)^{-1}$.

3. Originally Posted by manjohn12
Suppose $\displaystyle A$ is nilpotent. Prove that $\displaystyle I+A$ is invertible.

So $\displaystyle \det A = 0$. Then $\displaystyle \det (A+I) = 1$ which implies that $\displaystyle A+I$ is invertible.

Is this correct?
if $\displaystyle A^n=0,$ then $\displaystyle (I+A)(I-A+A^2- \cdots + (-1)^{n-1}A^{n-1})=I+(-1)^{n-1}A^n=I.$

4. Originally Posted by manjohn12
Suppose $\displaystyle A$ is nilpotent. Prove that $\displaystyle I+A$ is invertible.

So $\displaystyle \det A = 0$. Then $\displaystyle \det (A+I) = 1$ which implies that $\displaystyle A+I$ is invertible.

Is this correct?
What is the reason for"So $\displaystyle \det A = 0$. Then $\displaystyle \det (A+I) = 1$" ... I mean what is the reason to assume det(A+I) = 1?

Since the matrix is nilpotent, its characteristic polynomial is $\displaystyle t^n$. So $\displaystyle \det(tI - A) = t^n$, now put $\displaystyle t = -1$, we get $\displaystyle \det(-I - A) = (-1)^n\det(I + A) = (-1)^n \implies \det(A+I) = 1$, thus A+I is invertible.

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