Eigenvalues and eigenvectorsEdexcel A-Level Further Maths: Revision notes

A non-zero vector $\mathbf{v}$ is an eigenvector of a square matrix $\mathbf{A}$ if $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ for some scalar $\lambda$, the eigenvalue. Geometrically, the transformation $\mathbf{A}$ maps the line through the origin in the direction of $\mathbf{v}$ onto itself, scaling it by $\lambda$ (and reversing it if $\lambda<0$). Any non-zero multiple of an eigenvector is also an eigenvector.

$\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ means $(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$, which has a non-zero solution only if $\det(\mathbf{A}-\lambda\mathbf{I})=0$. This is the characteristic equation. For a $2\times2$ matrix: $\lambda^2-(\text{trace }\mathbf{A})\lambda+\det\mathbf{A}=0.$ Example: $\mathbf{A}=\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ has trace $5$, determinant $4$, so $\lambda^2-5\lambda+4=0$ and $\lambda=1$ or $4$. The sum of the eigenvalues is the trace and their product is the determinant, which gives a quick check. A matrix is singular exactly when $0$ is an eigenvalue.

For each eigenvalue solve $(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$. The two equations are multiples of each other (that is why $\lambda$ is an eigenvalue), so you get one equation in $x$ and $y$ and choose a convenient non-zero multiple. For $\lambda=4$ with the matrix above: $-x+y=0$, so $\mathbf{v}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda=1$: $2x+y=0$, so $\mathbf{v}=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Always check by multiplying: $\mathbf{A}\mathbf{v}$ must equal $\lambda\mathbf{v}$. A normalised eigenvector has magnitude $1$: divide by $|\mathbf{v}|$. For $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ this gives $\frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

If the characteristic equation has a repeated root, $\lambda$ has multiplicity $2$. For $\mathbf{C}=\begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}$: $\lambda^2-4\lambda+4=(\lambda-2)^2=0$, so $\lambda=2$ twice. Solving $(\mathbf{C}-2\mathbf{I})\mathbf{v}=\mathbf{0}$ gives $x=y$, a single direction $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (normalised: $\frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$). A repeated eigenvalue may give one eigenvector direction (as here) or, for a multiple of the identity such as $3\mathbf{I}$, every vector is an eigenvector.

If the discriminant of the characteristic equation is negative, the eigenvalues are complex. For a real matrix they come as conjugate pairs, and the eigenvectors are complex conjugates too. For $\mathbf{B}=\begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix}$: $(1-\lambda)^2+4=0$, so $\lambda=1\pm2i$. For $\lambda=1+2i$: $-2ix-2y=0$, so $y=-ix$ and $\mathbf{v}=\begin{pmatrix} 1 \\ -i \end{pmatrix}$. For $\lambda=1-2i$ the eigenvector is the conjugate, $\begin{pmatrix} 1 \\ i \end{pmatrix}$. Geometrically $\mathbf{B}$ is a rotation with an enlargement, so no real line is invariant.

If $\mathbf{A}\mathbf{v}=\lambda\mathbf{v}$ then $\mathbf{A}^n\mathbf{v}=\lambda^n\mathbf{v}$. To find $\mathbf{A}^n\mathbf{w}$ for any vector $\mathbf{w}$, write $\mathbf{w}$ as a combination of eigenvectors and scale each part by its $\lambda^n$. For a $3\times3$ matrix the characteristic equation is a cubic, solved by factorising; each eigenvector is found by solving $(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}$ with three equations (a $2\times2$ matrix is the AS requirement). Example: $\mathbf{M}=\begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ has eigenvalues $7$ and $2$ with eigenvectors $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$.