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You can easily prove that if $A^n=0$: $$(A+I)(I-A+A^2-...+(-1)^n A^{n-1}) = I +(-1)^{n-1} A^n = I$$$$\left(A+I\right)\left(I-A+A^2-...+(-1)^n A^{n-1}\right) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent $A$.
You can easily prove that if $A^n=0$: $$(A+I)(I-A+A^2-...+(-1)^n A^{n-1}) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent $A$.
You can easily prove that if $A^n=0$: $$\left(A+I\right)\left(I-A+A^2-...+(-1)^n A^{n-1}\right) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent $A$.
You can easily prove that if $A^n=0$: $$(A+I)(I-A+A^2-...+(-1)^n A^{n-1}) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any positivenilpotent$n$$A$.
You can easily prove that: $$(A+I)(I-A+A^2-...+(-1)^n A^{n-1}) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any positive$n$.
You can easily prove that if $A^n=0$: $$(A+I)(I-A+A^2-...+(-1)^n A^{n-1}) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent$A$.
