# Eigenvalues of Square Matrices

Let $A$ be an $n \times n$ matrix and let $\lambda \in \mathbb{R}$. Consider the matrix equation:

(1)This matrix equation has the trivial solution $x = 0$. The equation in general has either $1$ solution or infinitely many solutions.

Now recall from linear algebra that a matrix equation of the form $Ax = 0$ has infinitely many solutions if and only if $\det A = 0$. For the matrix equation above, this means that $(A - \lambda I)x = 0$ has infinitely many solutions provided that $\det (A - \lambda I) = 0$.

Definition: Let $A$ be an $n \times n$ matrix and consider the matrix equation $Ax = \lambda x$. The Characteristic Polynomial for the matrix $A$ is the polynomial $\det (A - \lambda I) = 0$. The roots/solutions of the characteristic polynomial are called the Eigenvalues of $A$. |

For example, consider the following $2 \times 2$ matrix:

(2)From this we can obtain the matrix $A - \lambda I$:

(3)The characteristic polynomial for $A$ is:

(4)There are two eigenvalues for $A$. Namely $\lambda_1 = 1$ and $\lambda_2 = -5$.

For another example, consider the following matrix:

(5)From this we can obtain the matrix $A - \lambda I$:

(6)To find the characteristic polynomial for $A$ we need to find the determinant of $A$. We can do this by cofactor expansion along the first row of $A$. We have that:

(7)There are two distinct eigenvalues of $A$. $\lambda_1 = 3$ is an eigenvalue with multiplicity $2$, and $\lambda_2 = 5$ is an eigenvalue with multiplicity $1$.