Is there a matrix that is NOT I, but also has eigenvalue of 1? I am looking for a matrix that is not I, but also has eigenvalue of 1. Are there any? Can someone please show me an example and how it has eigenvalue 1.
 A: There are many, many matrices that have have an eigenvalue of 1, but aren't the identity matrix.
Consider the set of two-by-two matrices:
$${\bf M} = \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)$$
The eigenvalues are given by $\det({\bf M} - \lambda{\bf I})=0$ where ${\bf I}$ is the identity matrix and $\lambda$ is an eigenvalue.
$$\det({\bf M} - \lambda{\bf I}) \ = \ \left| 
\begin{array}{cc} a-\lambda & b \\ c & d-\lambda \end{array} \right|$$
The eigenvalues are given by $(a-\lambda)(d-\lambda)-bc=0$. Using the factor theorem: $\lambda=1$ is a root of a polynomial if, and only if, $\lambda = 1$ is a solution, i.e.
$$(a-1)(d-1)-bc=0$$
It follows that the set of all two-by-two matrices with an eigenvalue of 1 are given by
$$\left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array}\right) : ad-bc-a-d+1=0 \right\}$$
This can be expressed as $1+\det({\bf M}) = \mathbb{tr}({\bf M})$. This is not true for larger matrices.
If we think of the set of two-by-two matrices as a four dimensional space with coordinates $(a,b,c,d)$, then the set of matrices with eigenvalue 1 form an algebraic variety with equation $ad-bc-a-d+1=0$. Using the implicit function theorem, this variety is a smooth three-dimensional manifold away from the identity matrix, where there is a singularity. 
A: Here is a simple example:  set
$D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}; \tag 1$
we have
$D \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \tag 2$
and
$D \begin{pmatrix} -1 \\ 1 \end{pmatrix} =  \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} = -\begin{pmatrix} -1 \\ 1 \end{pmatrix}; \tag 3$
we see from (2) and (3) that the eigenvalues of $D$ are $\pm 1$.  We can also find the eigenvalues of $D$ by evaluating the characteristic polynomial $p_D(\mu)$:
$p_D(\mu) = \det(D - \mu I) = \det \left ( \begin{bmatrix} -\mu & 1 \\ 1 & -\mu \end{bmatrix} \right ) = \mu^2 -1; \tag 4$
the roots of
$p_D(\mu) = \mu^2 - 1 \tag 5$
are $\mu = \pm 1$, in agreement with (2)-(3).
