Eigenvalues And Eigenvectors

# Eigenvalues and Eigenvectors

Consider the a system of linear equations in the form $Ax = \lambda x$ or written differently, $(\lambda I - A)x = 0$ where $\lambda$ is a scalar. Such systems arise frequently in linear algebra, and we will derive the definition of an eigenvalue and eigenvector from this.

 Definition: Given the system of linear equations $(\lambda I - A)x = 0$, we call any $\lambda$ that makes the system contain a non-trivial solution an eigenvalue of $A$, and we call the non-trivial solution an eigenvector of $A$. $\lambda$ is an eigenvalue of $A$ if and only if $\mathrm{det}(\lambda I - A) = 0$. We call the equation $p(x) = \mathrm{det}(\lambda I - A ) = 0$ the characteristic polynomial of $A$.

## Example 1

Find the eigenvalues of $A = \begin{bmatrix}1 & 2\\ 3 & 4 \end{bmatrix}$.

We first note that $\lambda I - A = \begin{bmatrix}\lambda & 0\\ 0 & \lambda\end{bmatrix} -\begin{bmatrix}1 & 2\\ 3 & 4 \end{bmatrix} = \begin{bmatrix} \lambda - 1 & -2\\ -3 & \lambda - 4 \end{bmatrix}$

We will now calculate the determinant of this matrix as

(1)
\begin{align} \begin{vmatrix} \lambda - 1 & -2\\ -3 & \lambda - 4 \end{vmatrix} = (\lambda - 1)(\lambda - 4) - (-2)(-3) \\ \begin{vmatrix} \lambda - 1 & -2\\ -3 & \lambda - 4 \end{vmatrix} = \lambda ^2 - 5\lambda + 4 - 6 = \lambda^2 - 5\lambda -2 = p(x) \end{align}

If we set our characteristic polynomial to zero, we can find values of $\lambda$ that make $p(x) = 0$. Using the quadratic formula, we obtain that $\lambda_1 = \frac{5 + \sqrt{33}}{2}$ and $\lambda_2 = \frac{5 - \sqrt{33}}{2}$.