Suppose A is nilpotent, i.e. [tex]A^k=0[/tex], [tex]k>0[/tex]. Show that [tex]A+I[/tex] is invertible.(adsbygoogle = window.adsbygoogle || []).push({});

Here's my go at it... although I'm unsure about step marked with asterisks... is it ok? I'm not sure how to show that it's true for sure.

Contrapositive: Suppose [tex]A+I[/tex] isn't invertible. Then [tex](A+I)X =0 [/tex] has a non-zero solution, say [tex]X = X_0[/tex].

Then

[tex](A+I)X_0=0[/tex]

[tex]AX_0+X_0=0[/tex]

[tex]A^kX_0 +A^{k-1}X_0=0[/tex]

If [tex]A^k=0[/tex] then [tex]A^{k-1}X_0=0[/tex]... but [tex]X_0[/tex] is not a solution of [tex]A^{k-1}X=0[/tex].************

Hence, [tex]A[/tex] must not nilpotent, QED

Thanks

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# Proving that for nilpotent A, A+I is invertible

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