
We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!
Definitions
Let $A$ be an $n \times n$ matrix. The number $\lambda$ is an
eigenvalue of $A$ if there exists a nonzero vector ${\bf v}$ such
that
$$
A{\bf v} = \lambda {\bf v}.
$$
In this case, vector ${\bf v}$ is called an eigenvector of $A$
corresponding to $\lambda$.
Computing Eigenvalues and Eigenvectors
ExampleLet $A = \left[ \begin{array}{rr} 2 & 4\\ 1 & 1 \end{array}\right] $. Then $$\begin{array}{rcl} p(\lambda) & = & \det \left[\begin{array}{cc} 2\lambda & 4 \\ 1 & 1\lambda \end{array}\right]\\ & = & (2\lambda)(1\lambda)(4)(1)\\ & = & \lambda^{2} \lambda 6\\ & = & (\lambda 3)(\lambda +2). \end{array}$$ Thus, $\lambda_1 =3$ and $\lambda_2 = 2$ are the eigenvalues of $A$. To find eigenvectors ${\bf v} = \left[ \begin{array}{c} v_1\\ v_2\\ \vdots\\ v_n \end{array}\right] $ corresponding to an eigenvalue $\lambda$, we simply solve the system of linear equations given by $$ (A\lambda I) {\bf v} = {\bf 0}. $$
ExampleThe matrix $A = \left[ \begin{array}{rr} 2 & 4\\ 1 & 1 \end{array}\right] $ of the previous example has eigenvalues $\lambda_1 =3$ and $\lambda_2 = 2$. Let's find the eigenvectors corresponding to $\lambda_1 =3$. Let ${\bf v}= \left[{v_1 \atop v_2}\right]$. Then $(A3I){\bf v}={\bf 0}$ gives us $$ \left[\begin{array}{cc} 23 & 4\\ 1 & 13 \end{array}\right]\left[\begin{array}{c} v_1\\ v_2 \end{array}\right] = \left[\begin{array}{c} 0\\ 0 \end{array}\right], $$ from which we obtain the duplicate equations \begin{eqnarray*} v_14v_2 & = & 0 \\ v_14v_2 & = & 0. \end{eqnarray*} If we let $v_2=t$, then $v_1=4t$. All eigenvectors corresponding to $\lambda_1 =3$ are multiples of $\left[{4 \atop 1}\right] $ and thus the eigenspace corresponding to $\lambda_1 =3$ is given by the span of $\left[{4 \atop 1}\right] $. That is, $\left\{\left[{4 \atop 1}\right]\right\}$ is a basis of the eigenspace corresponding to $\lambda_1 =3$. Repeating this process with $\lambda_2 = 2$, we find that \begin{eqnarray*} 4v_1 4V_2 & = & 0 \\ v_1 + v_2 & = & 0 \end{eqnarray*} If we let $v_2=t$ then $v_1=t$ as well. Thus, an eigenvector corresponding to $\lambda_2 = 2$ is $\left[{1 \atop 1}\right]$ and the eigenspace corresponding to $\lambda_2 = 2$ is given by the span of $\left[{1 \atop 1}\right]$. $\left\{\left[{1 \atop 1}\right]\right\}$ is a basis for the eigenspace corresponding to $\lambda_2 = 2$. In the following example, we see a twodimensional eigenspace.
ExampleLet $A=\left[\begin{array}{ccc} 5 & 8 & 16\\ 4 & 1 & 8\\ 4 & 4 & 11 \end{array}\right]$. Then $p(\lambda ) = \det\left[\begin{array}{ccc} 5\lambda & 8 & 16\\ 4 & 1\lambda & 8\\ 4 & 4 & 11\lambda \end{array}\right] = (\lambda1)(\lambda+3)^{2}$ after some algebra! Thus, $\lambda_1 = 1$ and $\lambda_2=3$ are the eigenvalues of $A$. Eigenvectors ${\bf v} = \left[\begin{array}{c} v_1\\ v_2\\ v_3 \end{array}\right]$ corresponding to $\lambda_1=1$ must satisfy Letting $v_3=t$, we find from the second equation that $v_1=2t$, and then $v_2=t$. All eigenvectors corresponding to $\lambda_1=1$ are multiples of $\left[\begin{array}{c} 2\\ 1\\ 1 \end{array}\right]$, and so the eigenspace corresponding to $\lambda_1=1$ is given by the span of $\left[\begin{array}{c} 2\\ 1\\ 1 \end{array}\right]$. $\left\{\left[\begin{array}{c} 2\\ 1\\ 1 \end{array}\right]\right\}$ is a basis for the eigenspace corresponding to $\lambda_1=1$. Eigenvectors corresponding to $\lambda_2=3$ must satisfy The equations here are just multiples of each other! If we let $v_3 = t$ and $v_2 = s$, then $v_1 = s 2t$. Eigenvectors corresponding to $\lambda_2=3$ have the form $$ \left[\begin{array}{c} 1\\ 1\\ 0 \end{array}\right]s+\left[\begin{array}{c} 2\\ 0\\ 1 \end{array}\right]t. $$ Thus, the eigenspace corresponding to $\lambda_2=3$ is twodimensional and is spanned by $\left[\begin{array}{c} 1\\ 1\\ 0 \end{array}\right]$ and $\left[\begin{array}{c} 2\\ 0\\ 1 \end{array}\right]$. $\left\{\left[\begin{array}{c} 1\\ 1\\ 0 \end{array}\right],\left[\begin{array}{c} 2\\ 0\\ 1 \end{array}\right]\right\}$ is a basis for the eigenspace corresponding to $\lambda_2=3$.
Key Concepts 