1.15 Eigenvalues and eigenvectorsIB Maths: Applications and Interpretation HL: Revision notes

Most vectors change direction when multiplied by a matrix. An eigenvector $\mathbf{v}$ of a square matrix $M$ is a non-zero vector whose direction is unchanged (or reversed): $M\mathbf{v}=\lambda\mathbf{v},\qquad \mathbf{v}\neq\mathbf{0},$ where the scalar $\lambda$ is the eigenvalue. Example: for $M=\begin{pmatrix}4&1\\ 2&3\end{pmatrix}$, $M\begin{pmatrix}1\\ 1\end{pmatrix}=\begin{pmatrix}5\\ 5\end{pmatrix}=5\begin{pmatrix}1\\ 1\end{pmatrix}$, so $\begin{pmatrix}1\\ 1\end{pmatrix}$ is an eigenvector with eigenvalue $5$. Any non-zero multiple of an eigenvector is also an eigenvector.

$M\mathbf{v}=\lambda\mathbf{v}$ means $(M-\lambda I)\mathbf{v}=\mathbf{0}$. A non-zero solution exists only if $M-\lambda I$ has no inverse, so $\det(M-\lambda I)=0.$ For a $2\times2$ matrix this characteristic equation is $\lambda^2-(\text{trace})\lambda+\det M=0$, where the trace is $a+d$. Example: $M=\begin{pmatrix}4&1\\ 2&3\end{pmatrix}$ gives $\lambda^2-7\lambda+10=0$, so $\lambda=2$ or $5$. To find an eigenvector, substitute each $\lambda$ into $(M-\lambda I)\mathbf{v}=\mathbf{0}$. For $\lambda=2$: $2x+y=0$, so $\mathbf{v}=\begin{pmatrix}1\\ -2\end{pmatrix}$. For $\lambda=5$: $-x+y=0$, so $\mathbf{v}=\begin{pmatrix}1\\ 1\end{pmatrix}$. You can check with your GDC.

If a $2\times2$ matrix $M$ has two distinct real eigenvalues $\lambda_1,\lambda_2$ with eigenvectors $\mathbf{v}_1,\mathbf{v}_2$, form $P=\begin{pmatrix}\mathbf{v}_1&\mathbf{v}_2\end{pmatrix},\qquad D=\begin{pmatrix}\lambda_1&0\\ 0&\lambda_2\end{pmatrix},\qquad M=PDP^{-1}.$ The columns of $P$ and the entries of $D$ must be in the same order. Example: $M=\begin{pmatrix}1&2\\ 3&0\end{pmatrix}$ has $\lambda=3$ with $\begin{pmatrix}1\\ 1\end{pmatrix}$ and $\lambda=-2$ with $\begin{pmatrix}2\\ -3\end{pmatrix}$, so $P=\begin{pmatrix}1&2\\ 1&-3\end{pmatrix}$, $D=\begin{pmatrix}3&0\\ 0&-2\end{pmatrix}$ and $P^{-1}=\frac15\begin{pmatrix}3&2\\ 1&-1\end{pmatrix}$.

Because $M=PDP^{-1}$, the middle factors cancel in a product: $M^2=PD(P^{-1}P)DP^{-1}=PD^2P^{-1}$. In general $M^n=PD^nP^{-1},\qquad D^n=\begin{pmatrix}\lambda_1^n&0\\ 0&\lambda_2^n\end{pmatrix}.$ Only the diagonal entries are raised to the power. Check with $M=\begin{pmatrix}1&2\\ 3&0\end{pmatrix}$ and $n=2$: $PD^2P^{-1}=\frac15\begin{pmatrix}1&2\\ 1&-3\end{pmatrix}\begin{pmatrix}9&0\\ 0&4\end{pmatrix}\begin{pmatrix}3&2\\ 1&-1\end{pmatrix}=\begin{pmatrix}7&2\\ 3&6\end{pmatrix}$, which equals $M^2$. The eigenvalues of $M^n$ are $\lambda_1^n$ and $\lambda_2^n$.

A model $\mathbf{x}_{n+1}=M\mathbf{x}_n$ gives $\mathbf{x}_n=M^n\mathbf{x}_0$. Write $\mathbf{x}_0=c_1\mathbf{v}_1+c_2\mathbf{v}_2$; then $\mathbf{x}_n=c_1\lambda_1^n\mathbf{v}_1+c_2\lambda_2^n\mathbf{v}_2.$ For movement between two towns the eigenvalue $1$ gives a steady state and an eigenvalue of modulus below $1$ decays, so the populations settle. In predator-prey models an eigenvalue above $1$ gives growth, and the term with the largest $|\lambda|$ dominates in the long term, so the ratio of the populations tends to the ratio in its eigenvector. The same eigenvalue ideas solve coupled differential equations (AHL 5.17).