Eigenvalues of Square Matrices
 Table of Contents

# Eigenvalues of Square Matrices

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

(1)
\begin{align} \quad Ax = \lambda x \quad \Leftrightarrow \quad (A - \lambda I)x = 0 \end{align}

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)
\begin{align} \quad A = \begin{bmatrix} 2 & 7 \\ -1 & -6 \end{bmatrix} \end{align}

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

(3)
\begin{align} \quad A - \lambda I &= \begin{bmatrix} 2 & 7 \\ -1 & -6 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 - \lambda & 7 \\ -1 & -6 - \lambda \end{bmatrix} \end{align}

The characteristic polynomial for $A$ is:

(4)
\begin{align} \quad \det (A - \lambda I) &= 0 \\ (2 - \lambda)(-6 - \lambda) - (7)(-1) &= 0\\ -12 - 2\lambda + 6 \lambda + \lambda^2 + 7 &= 0\\ \lambda^2 + 4 \lambda -5 &= 0 \\ (\lambda - 1)(\lambda + 5) &= 0 \end{align}

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

For another example, consider the following matrix:

(5)
\begin{align} \quad A = \begin{bmatrix} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 1 & 0 & 4 \end{bmatrix} \end{align}

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

(6)
\begin{align} \quad A - \lambda I &= \begin{bmatrix} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 1 & 0 & 4 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 4 - \lambda & 0 & 1 \\ 2 & 3 -\lambda & 2 \\ 1 & 0 & 4 - \lambda\end{bmatrix} \end{align}

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)
\begin{align} \quad \det (A - \lambda) &= 0 \\ \quad (4 - \lambda) \begin{vmatrix} 3 - \lambda & 2 \\ 0 & 4 - \lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 - \lambda \\ 1 & 0 \end{vmatrix} &= 0 \\ (4 -\lambda)(3 - \lambda)(4 - \lambda) - (3 - \lambda) &= 0 \\ [3 - \lambda][(4 - \lambda)^2 - 1) &= 0 \\ [3 - \lambda](16 -8 \lambda + \lambda^2 - 1) &= 0 \\ [3 - \lambda](\lambda^2 - 8 \lambda + 15) &= 0 \\ [3 - \lambda](3 - \lambda)(5 - \lambda) &= 0 \end{align}

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 $1$.

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