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Units and Nilpotents

Given $A^{2012}=0$ prove that $A+I$ is invertible and find an expression for $(A+I)^{-1}$ in terms of $A$. ($I$ is the identity matrix).

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marked as duplicate by Bill Dubuque, Benjamin Lim, Bruno Joyal, AD., t.b. May 4 '12 at 9:58

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On the other hand, if I + A not invertible there is a v so that $0 = (I + A) v$, so $Av = -v, A^{2012}v = v, $ A not nilpotent. – mike May 3 '12 at 13:36
@ugoolm If you are asking for help with your MATH 1115 assignment due this monday, you should provide a bit more context and where you are stuck. – user38268 May 3 '12 at 14:44
I think it would be sufficient with a hint like "Do you remember the proof of the geometric formula?". – AD. May 3 '12 at 21:51
Please don't do this. There's nothing wrong with a closed question while an impenetrable non-question has no value whatsoever. – t.b. May 4 '12 at 10:37

4 Answers 4

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 $A$.

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Maybe you should add that $n=2012$, otherwise it seems that $A^n=0$ for every $n>1$. – Siminore May 3 '12 at 11:34
@Siminore - fixed, thanks! – nbubis May 3 '12 at 11:41

More generally: A (square) matrix $A$ is invertible if and only if $\lambda=0$ is not an eigenvalue.

Independently of this, we have that if $\lambda$ is an eigenvalue of $A$, then $\lambda+\mu$ is an eigenvalue of $A+\mu I$: if $\mathbf{x}$ is an eigenvalue of $A$ corresponding $\lambda$, then $(A+\mu I)\mathbf{x} = A\mathbf{x}+\mu I\mathbf{x} = \lambda\mathbf{x}+\mu\mathbf{x} = (\lambda+\mu)\mathbf{x}$.

Also independently of all of this, if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. (Careful, though, this one is not reversible: a rotation of $90^{\circ}$ of the plane has no eigenvalues over $\mathbb{R}$, but its square has eigenvalue $-1$. The square of the identity has $(-1)^2$ as an eigenvalue, but the identity does not have $-1$ as an eigenvalue).

So: $$\begin{align*} A+I\text{ is invertible}&\iff 0\text{ is not an eigenvalue of }A+I\\ &\iff -1\text{ is not an eigenvalue of }A. \end{align*}$$ And if $A^n=0$ for some $n\gt 0$, then $-1$ is not an eigenvalue of $A$.

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Hint: the series $\sum_{j=0}^{+\infty}(-1)^jA^j$ has only finitely many terms. In fact, it works it there is a positive integer $p$ such that $A^p=0$.

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A matrix $A$ is nilpotent if and only if all its eigenvalues are zero. It is not hard also to see that the eigenvalues of $A+I$ will all be equal to $1$ (when we add $I$ to any matrix, we just shift its spectrum by 1). Thus $A+I$ is invertible, since all its eigenvalues are non-zero.

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You want to be careful: you can have a matrix for which all eigenvalues are $0$, but the matrix is not nilpotent (because it doesn't have "enough" eigenvalues); e.g., the $3\times 3$ matrix over $\mathbb{R}$ given by $$\left(\begin{array}{rrr}0 & -1 & 0\\1 & 0 & 0\\0 & 0 & 0\end{array}\right).$$ The characteristic polynomial is $t(t^2+1)$, so the only (real) eigenvalue is $0$, but the matrix is not nilpotent. So it depends on how you interpret "all its eigenvalues" (true if that means "in an algebraic closure of the ground field"). – Arturo Magidin May 3 '12 at 21:20

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