nilpotent

**manjohn12** · May 24th 2009, 10:40 AM

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?

**manjohn12** · May 24th 2009, 10:43 AM

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

**NonCommAlg** · May 24th 2009, 10:44 AM

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 $(I+A)(I-A+A^2- \cdots + (-1)^{n-1}A^{n-1})=I+(-1)^{n-1}A^n=I.$

**Isomorphism** · May 24th 2009, 10:49 AM

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.

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