# What is wrong with this "proof"? $1$ is always an eigenvalue for $I + A$ ($A$ is nilpotent)?

Consider the nilpotent matrix $A$ ($A^k = 0$ for some positive $k$). It is well known that the only eigenvalue of $A$ is $0$.

Then suppose $\lambda$ is any eigenvalue of $I + A$ such that $(I + A) \mathbf{v} = \lambda \mathbf{v}$ where ($\mathbf{v} \neq \mathbf{0}$).

Then $I \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies \mathbf{v} + A \mathbf{v} = \lambda \mathbf{v} \implies A \mathbf{v} = (\lambda - 1) \mathbf{v}$

We know that $\lambda - 1 = 0$ because the only eigenvalue of a nilpotent matrix is $0$. Therefore $\lambda = 1$

This "proof" seems to indicate that $1$ is always an eigenvalue for the sum of the identity matrix with any nilpotent matrix, but I believe I have a counterexample that disproves this.

I believe my error was in assuming that $I + A$ has eigenvalues -- but I do not know how I could prove/disprove this.

If someone could help me see where I've gone wrong I would greatly appreciate it!

• Whoops! Sorry, I'll edit the question. But the question still stands. Is this correct? May 10, 2017 at 12:10
• No, you start out with $\lambda$ is an eigenvalue for $I+A$, then after the block, you change $\lambda$ to be an eigenvalue for $A$. May 10, 2017 at 12:11
• What you have found is an eigenvalue of $I+A$. smh
– user384138
May 10, 2017 at 12:11
• Note also that all square matrices have eigenvalues, they come from roots of the characteristic polynomial. May 10, 2017 at 12:28
• @MichaelBurr Thanks for the tips. You're right, my counterexample was completely bogus on further inspection. By the way, doesn't the existence of eigenvalues depend on the closure of the field? May 10, 2017 at 12:30

You have proven that if $A$ is nilpotent, then the eigenvalue of $A+I$ (NOT the eigenvalues of $A$) is equal to $1$. There's nothing wrong with the proof.

• You are absolutely right. I believe I somehow forgot how eigenvectors work. My "counterexample" was poorly constructed. May 10, 2017 at 12:28

Here's an alternate argument (assuming the base field is algebraically closed):

• For a nilpotent matrix with $A^k=0$, all eigenvalues are $0$.

Proof: Let $\lambda$ be an eigenvalue for $A$ and $v\not=0$ an eigenvector for $\lambda$. Then $$0=0v=A^kv=\lambda^kv$$ Since $v\not=0$, it must be that $\lambda^k=0$ so that $\lambda=0$

• $1$ is an eigenvalue for $I+A$. Let $v$ be an eigenvector for $A$.

Proof: We know that all eigenvalues for $A$ are $0$, so $Av=0$. Therefore, $$(I+A)v=Iv+Av=v+0=v.$$ Hence, $1$ is the eigenvalue for $v$ in $I+A$.

• You can prove more, every eigenvalue of $I+A$ is $1$.

Proof: Suppose that $\mu$ is an eigenvalue for $I+A$ and $w$ the corresponding eigenvector. Then $$Aw=((I+A)-I)w=(I+A)w-Iw=\mu w-w=(\mu-1)w.$$ Therefore, $w$ is an eigenvector for $A$ with eigenvalue $\mu-1$. Since the only eigenvalues for $A$ are $0$, $\mu=1$.

If $\lambda$ is an eigenvalue of a matrix $A$ then $p(\lambda)$ is eigenvalue of a polynomial $p(A)$.
In this case we have polynomial $p(A)=A+I$ so in this case the eigenvalues of $A$ (all zeros) should be shifted by $1$.

If $\lambda$ is an eigenvalue for the matrix $A$, then, for every scalar $\alpha$, $\lambda+\alpha$ is an eigenvalue for $A+\alpha I$.
Indeed, if $v\ne0$ and $Av=\lambda v$, then $$(A+\alpha I)v=Av+\alpha v=\lambda v+\alpha v=(\lambda+\alpha)v$$
Now you can apply this to show that if $A$ has only the eigenvalue $0$, then $A+I$ has only the eigenvalue $1$. Conversely, if $A+I$ has only the eigenvalue $1$, then $A=(A+I)-I$ has only the eigenvalue $1-1=0$.
This doesn't depend on the field of scalars. However a real matrix is nilpotent if and only if its only eigenvalue over the complex numbers is $0$. As an example, the matrix $$\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ has $0$ as its only real eigenvalue, but is not nilpotent.