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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