1. ## nilpotent

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

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

Is this correct?

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

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

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

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

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

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

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

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