# Eigen values of $I+A$ if $A^2=0$

Let $A\in M_{n\times n}(\mathbb{C})$. Suppose that $A^2=0$. Show that $\lambda$ is an eigenvalue of $A+I$ if and only if $\lambda=1$

As $A^2=0$, $A^2-t^2I=-t^2I$ i.e., $(A+tI)(A-tI)=-t^2I$ this imply $A-tI$ is invertible iff $t\neq 0$.

So, $(A+I)-\lambda I=A-(\lambda-1)I$ is invertible iff $\lambda-1=0$ i.e., $\lambda=1$. So, $1$ is the only eigen value of $I+A$.

After posting this question i got this idea.. Let me know if this is fine..

• $(A+I)-I$, not $(A+I)+I$. – Christopher Carl Heckman Mar 1 '16 at 7:01
• (After the above fix was made:) $A^2=0$ means more than "0 is one of the eigenvalues of $A$" ... If $\vec v$ is a nonzero vector, $A^2\vec v = 0_{n\times n}\vec v = 0\vec v$, so every vector is an eigenvector belonging to the number 0; 0 is the ONLY eigenvalue of $A$. – Christopher Carl Heckman Mar 1 '16 at 7:04
• @CarlHeckman That makes very little sense. You only proved that $0$ is the only eigenvalue of $A^2$, you did nothing with $A$. – 5xum Mar 1 '16 at 7:10
• @CarlHeckman : PLease see the edited version – user312648 Mar 1 '16 at 8:50
• @5xum : PLease see the edited version – user312648 Mar 1 '16 at 8:56

Since $A^2=0$, we have $$(I+A)^n=I+nA$$ . Let $v$ be an eigenvector of $I+A$, then \begin{align} (I+A)^nv &= (I+nA)v \\ \lambda^nv &= v+nA v \\ \frac{(\lambda^n-1)}{n}v &= Av \end{align} for all $n\in \Bbb N$. Since $v\ne 0$, the term on the left implies that $\lambda=1$ or else we'd have a contradiction ($Av$ not well defined).

• Can be done more directly. – Christopher Carl Heckman Mar 3 '16 at 6:56
• @CarlHeckman There's no doubt about that. I've already acknowledged that in the comment to S. W. Cheung's answer. – BigbearZzz Mar 3 '16 at 7:04

Suppose $\lambda$ is an eigenvalue of $A+I$ and $v \neq 0$ is an eigenvector of $A + I$ with respect to $\lambda$. Then $$Av = (\lambda - 1)v \implies 0 = A^2v = (\lambda - 1)^2 v \implies (\lambda - 1)^2 = 0 \implies \lambda = 1.$$ This shows that if $\lambda$ is an eigenvalue of $A+I$, $\lambda$ has to be $1$.

On the other hand, we have $$A^2 = 0 \implies \det(A)^2 = \det(A^2) = 0 \implies \det(A) = 0,$$ which shows $\lambda = 1$ is indeed an eigenvalue of $A+I$.

• This looks much simpler than my answer, +1 to this. – BigbearZzz Mar 1 '16 at 7:35

$\lambda$ is an eigenvalue of $A$ if and only if $\lambda+a$ is an eigenvalue of $A+aI$.

Sketch of proof: $$Av=\lambda v\iff Av+av=\lambda v+av$$

If $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$.

Sketch of proof: if $Av=\lambda v$, then $$A^{n+1}v=A^n(Av)=A^n(\lambda v)=\lambda(A^nv)=\lambda(\lambda^nv)= \lambda^{n+1}v$$

So the only eigenvalue of $A$ is $0$, and the result follows.

If $\vec v$ is a nonzero vector, then $$A^2 \vec v = 0_{n\times n}\cdot \vec v = 0\cdot \vec v,$$ where $0_{n\times n}$ is the $n\times n$ zero matrix. Hence every (nonzero) vector is an eigenvector belonging to the eigenvalue $0$.

The eigenvalues of $A^2$ are the eigenvalues of $A$, squared. Hence. The eigenvalues of $A$ are all $0$.

The eigenvalues of $A+I$ are the eigenvalues of $A$ plus one; hence there is only one eigenvalue of $A+I$, namely $0+1=1$.