Eigenvalues and eigenvectorsEdexcel International A Level Further Maths: Revision notes

For a square matrix $\mathbf{A}$, a non-zero vector $\mathbf{x}$ is an eigenvector of $\mathbf{A}$ with eigenvalue $\lambda$ if $\mathbf{A}\mathbf{x}=\lambda\mathbf{x}.$ The transformation $\mathbf{A}$ maps $\mathbf{x}$ to a vector in the same (or exactly opposite) direction, scaled by $\lambda$. The zero vector is never counted as an eigenvector. Any non-zero multiple of an eigenvector is also an eigenvector for the same $\lambda$, so eigenvectors are only defined up to a scalar multiple.

Rewrite $\mathbf{A}\mathbf{x}=\lambda\mathbf{x}$ as $(\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}$. A non-zero solution exists only if $\mathbf{A}-\lambda\mathbf{I}$ is singular, so $\det(\mathbf{A}-\lambda\mathbf{I})=0.$ This is the characteristic equation. For a $2\times2$ matrix it is $\lambda^2-(\text{trace})\lambda+\det\mathbf{A}=0$; for a $3\times3$ matrix it is a cubic. Example: $\mathbf{A}=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ gives $(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=0$, so $\lambda=2$ or $5$. Checks: the eigenvalues sum to the trace ($2+5=7=4+3$) and multiply to the determinant ($2\times5=10=12-2$).

For each eigenvalue $\lambda$, solve $(\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}$. Because $\mathbf{A}-\lambda\mathbf{I}$ is singular, the equations are dependent, so there is a line (or plane) of solutions; choose any convenient non-zero one. Example ($\lambda=5$ for the matrix above): $-x+y=0$ and $2x-2y=0$ are the same equation, so $y=x$ and an eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For a $3\times3$ matrix the three equations reduce to two independent ones: eliminate to find the ratios $x:y:z$. Check by multiplying: $\mathbf{A}\mathbf{x}$ must equal $\lambda\mathbf{x}$.

A normalised (unit) eigenvector has magnitude $1$. Divide any eigenvector by its magnitude: $\hat{\mathbf{e}}=\frac{\mathbf{x}}{|\mathbf{x}|}.$ For $\mathbf{x}=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $|\mathbf{x}|=\sqrt3$, so $\hat{\mathbf{e}}=\frac{1}{\sqrt3}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. For $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ the magnitude is $\sqrt5$. A normalised vector is still determined only up to sign, so either sign is acceptable unless the question fixes a direction.

Find the eigenvalues of $\mathbf{C}=\begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 5 \end{pmatrix}$. Expanding $\det(\mathbf{C}-\lambda\mathbf{I})$ along the bottom row gives $(5-\lambda)\big[(2-\lambda)(3-\lambda)-2\big]=(5-\lambda)(\lambda-1)(\lambda-4)$, so $\lambda=1,4,5$. For $\lambda=4$: $-2x+2y+z=0$, $x-y+z=0$, $z=0$. Hence $z=0$ and $x=y$, giving $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$. Look for rows or columns with many zeros before expanding. A triangular or block matrix gives its eigenvalues from the diagonal blocks.