What does "$n$-fold eigenvalue" mean?

Question

I am trying to find the eigenvalues of the matrix $A = I + N$ where $N^2 = 0$. The answer I have says

$1$ is the $n$-fold eigenvalue.

I am not sure what is meant by "$n$-fold" but my guess is that $1$ is the only eigenvalue and is repeated $n$ times. Indeed, when I have the $2$ by $2$ matrices

$$I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$$ and $$N = \begin{bmatrix}0&1\\0&0\end{bmatrix}$$

the only eigenvalue of $A$ is $1$ and it is repeated $2$ times.

Is there a way to prove this for $n$ by $n$ matrix?

Answer 1

You have interpreted "$n$-fold" correctly.

We have $(A - I)^2 = 0$. However, if $\lambda$ is an eigenvalue of $A$ and $x$ is an associated eigenvector, then we must have $$ (A - I)x = Ax - x = \lambda x - x = (\lambda - 1)x, $$ and similarly $$ (A - I)^2x = (A - I)[(A - I)x] = (A - I)[(\lambda - 1)x] = (\lambda-1)[(A - I)x] = (\lambda - 1)^2 x. $$ It follows that $$ (\lambda - 1)^2 x = (A - I)^2 x = 0 \cdot x = 0. $$

Answer 2

Since the polynomial $p(z) = (z-1)^2$ annihilates $A$, the minimal polynomial of $A$ divides $p$. So the minimal polynomial of $A$ is either $p(z)$ or $q(z) = z - 1$. We know the minimal polynomial vanishes at every distinct eigenvalue of $A$, but from the above, the minimal polynomial of $A$ only vanishes at $1$. So $1$ is the only eigenvalue of $A$ and so it must be repeated $n$ times.