Eigenvalues and eigenvectors have numerous applications throughout mathematics and other fields such as physics and engineering. To understand eigenvectors, we return to the topic of matrices as linear transformations. If a non-zero vector 
is an eigenvector of the matrix 
, then the transformation 
changes the magnitude but not the direction of 
. The scale factor which changes the magnitude is denoted as 
and is called the corresponding eigenvalue of the eigenvector.
So how do we find eigenvalues and eigenvectors? To begin with, we write the definition stated above as
. It is obvious that this relationship says that the transformation 
is equivalent to scaling 
by a constant. As mentioned before, 
is a non-zero vector because for the zero vector this relationship will hold for any 
or 
and is thus a trivial solution. This relationship could equivalently be written as 
. Here 
is an isotropic (same in all directions) linear scale by 
. We can now solve this equation: 
Now, if we apply rules that we have learned earlier, we know that the system 
does not have a unique solution for a particular value of 
if and only if 
. We want this to be the case since the unique solution is the trivial case 
.
Thus, finding eigenvalues amounts to finding the roots of
. As a simple example let's start with a 2x2 matrix:
1) Find 
: 
2) Find 
: 
3) Find the roots of 
: 
Now that we have the eigenvalues, we can plug them back into the system and solve 
.
For
we must solve 
: 
Parameterizing the solution gives 
.
For
solve 
:
Parameterizing the solution gives 
.
is an eigenvector of the matrix , then the transformation changes the magnitude but not the direction of . The scale factor which changes the magnitude is denoted as and is called the corresponding eigenvalue of the eigenvector.So how do we find eigenvalues and eigenvectors? To begin with, we write the definition stated above as
. It is obvious that this relationship says that the transformation is equivalent to scaling by a constant. As mentioned before, is a non-zero vector because for the zero vector this relationship will hold for any or and is thus a trivial solution. This relationship could equivalently be written as . Here is an isotropic (same in all directions) linear scale by . We can now solve this equation: does not have a unique solution for a particular value of if and only if . We want this to be the case since the unique solution is the trivial case .Thus, finding eigenvalues amounts to finding the roots of
. As a simple example let's start with a 2x2 matrix:: : : .For
we must solve : .For
solve :.