Eigenvalues and eigenvectorsAQA A-Level Further Maths: Revision notes

For a square matrix $M$, a non-zero vector $\mathbf{v}$ is an eigenvector of $M$ with eigenvalue $\lambda$ if $M\mathbf{v}=\lambda\mathbf{v}.$ Multiplying by $M$ only scales $\mathbf{v}$; it does not change its line. Any non-zero multiple of an eigenvector is also an eigenvector with the same eigenvalue, so an eigenvector is only ever found up to a scalar multiple. Eigenvalues are the scale factors, eigenvectors are the directions that do not turn.

Rearranging $M\mathbf{v}=\lambda\mathbf{v}$ gives $(M-\lambda I)\mathbf{v}=\mathbf{0}$. A non-zero solution exists only if $M-\lambda I$ is singular, so $\det(M-\lambda I)=0.$ This is the characteristic equation. For a $2\times2$ matrix it is a quadratic, $\lambda^2-(\text{trace})\lambda+\det M=0$. For a $3\times3$ matrix it is a cubic in $\lambda$. Example: $M=\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ gives $(3-\lambda)(2-\lambda)-2=\lambda^2-5\lambda+4=0$, so $\lambda=1$ or $\lambda=4$. Check: the eigenvalues add up to the trace ($1+4=5=3+2$) and multiply to the determinant ($1\times4=4$).

For each eigenvalue $\lambda$, solve $(M-\lambda I)\mathbf{v}=\mathbf{0}$. For $M=\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ and $\lambda=4$: $\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\mathbf{0}$ gives $y=x$, so an eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda=1$: $2x+y=0$, so an eigenvector is $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$. The equations are dependent (they give the same condition), because $M-\lambda I$ is singular. Choose one free value, such as $x=1$, and find the rest.

The method is the same, but the characteristic equation is a cubic and the eigenvector has three components. Example: $Q=\begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{pmatrix}$. Expanding $\det(Q-\lambda I)$ gives $(2-\lambda)\left((2-\lambda)^2-2\right)=0$, so $\lambda=2$ or $\lambda=2\pm\sqrt2$. For $\lambda=2$: $-y=0$ and $-x-z=0$, so an eigenvector is $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$. Once one root is known, factorise the cubic to find the rest. To find the eigenvector, set one component to 1, solve two of the three equations, and use the third as a check.

When $M$ represents a linear transformation:

• An eigenvector gives the direction of an invariant line through the origin: every point on it is mapped to a point on the same line.
• The eigenvalue $\lambda$ is the scale factor of the stretch along that line. If $\lambda>1$ points move away from the origin, if $0<\lambda<1$ they move towards it, and if $\lambda<0$ they also swap sides of the origin.
• If $\lambda=1$ every point on the line is invariant (it stays where it is). Example: $A=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ has eigenvalue $3$ with eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ (stretch factor 3 along $y=x$) and eigenvalue $-1$ with eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ (points on $y=-x$ are reflected through the origin). A rotation through $90^\circ$ has no real eigenvalues, because no line through the origin is mapped to itself.