TITLE: Linear Algebra: Identity matrix and its relation to eigenvalues and eigenvectors
QUESTION [8 upvotes]: Kindly help me understand this statement made by my prof.
"The identity matrix I has the property that any non zero vector V is an eigenvector of eigenvalue 1."
My assumption of this statement is that the column vector (1,1) multiplied by the identity matrix is equal to the identity matrix. But the confusing part is when he says "...any non zero..". This is implying we can use other values that don't equal one. I believe the eigenvalue would change in light of the different non-1 values. 
Thank you in advance.

REPLY [3 votes]: Eigenvectors & Eigenvalues
3Blue1Brown's video on eigenvectors and eigenvalues explains the eigenvectors and eigenvalues visually.
In general, matrix-vector multiplication $Av = b$ maps the vector $v$ to the vector $b$.
Accordingly, the matrix multiplication with the identity matrix $I$ maps the vector $v$ to itself $Iv = v$.
Here is my attempt to visualize an example for $Av = b$ in 2D space:

By multiplying $A$ and $v$ we get $b$.
As you can see, to get $b$ we need to stretch and rotate $v$.
The rotation and scaling factor are given by $A$.
Given that $A$ is an arbitrary matrix, there can exist nonzero vectors such that when multiplied with $A$ they do not get rotated (only scaled).
These vectors are eigenvectors.
Therefore, we can write $\lambda v = b$ for every eigenvector $v$, where $\lambda$ is a scalar factor.
We call $\lambda$ the eigenvalue.
In the following example, $A$ describes a reflection along the diagonal from bottom left to top right:

Note that every vector on one of the diagonals does not get rotated, only scaled by a factor $\pm \lambda$.
This means that all vectors on one of the diagonals are eigenvectors.
To answer the question
When you look at the matrix-vector multiplication with the identity matrix $Iv = v$, all vectors stay the same.
In particular:

There is no rotation, which means all vectors are eigenvectors (except $\vec{0}$).
All vectors get scaled by a factor of $1$, which means the eigenvalue for every eigenvector is $1$.

And therefore,

The identity matrix I has the property that any non-zero vector V is an eigenvector of eigenvalue 1.