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  1. #1
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    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?
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    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} $.
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    Quote Originally Posted by manjohn12 View Post
    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 $\displaystyle (I+A)(I-A+A^2- \cdots + (-1)^{n-1}A^{n-1})=I+(-1)^{n-1}A^n=I.$
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    Quote Originally Posted by manjohn12 View Post
    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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