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  1. #1
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    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?
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  2. #2
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    Ok let  B = \sum_{n=0}^{\infty} (-1)^{n} A^{n} which converges. So  B(I+A) = I . And so  B = (I+A)^{-1} .
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  3. #3
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    Quote Originally Posted by manjohn12 View Post
    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.
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    Quote Originally Posted by manjohn12 View Post
    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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