Diagonalisation and the Cayley-Hamilton theoremEdexcel A-Level Further Maths: Revision notes

If $\mathbf{A}$ ($2\times2$) has eigenvalues $\lambda_1,\lambda_2$ with independent eigenvectors $\mathbf{v}_1,\mathbf{v}_2$, form $\mathbf{P}$ with the eigenvectors as columns. Then $\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{D}=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}.$ The order of the eigenvalues in $\mathbf{D}$ must match the order of the eigenvectors in $\mathbf{P}$. Example: $\mathbf{A}=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ has $\lambda=2$ with $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\lambda=5$ with $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, so $\mathbf{P}=\begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$. Swapping the columns of $\mathbf{P}$ swaps the diagonal entries of $\mathbf{D}$.

Since $\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, the inner factors cancel in powers: $\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1},\qquad \mathbf{D}^n=\begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}.$ Example: $\mathbf{M}=\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ has $\mathbf{P}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\mathbf{D}=\mathrm{diag}(2,3)$, giving $\mathbf{M}^n=\begin{pmatrix} 2^n & 3^n-2^n \\ 0 & 3^n \end{pmatrix}$. Check $n=1$ returns $\mathbf{M}$.

A real symmetric matrix ($\mathbf{S}=\mathbf{S}^{\mathrm{T}}$) always has real eigenvalues, and eigenvectors for distinct eigenvalues are perpendicular. If the eigenvectors are normalised, $\mathbf{Q}$ (with them as columns) is an orthogonal matrix: $\mathbf{Q}^{\mathrm{T}}\mathbf{Q}=\mathbf{I}$, so $\mathbf{Q}^{-1}=\mathbf{Q}^{\mathrm{T}}$. Then $\mathbf{Q}^{\mathrm{T}}\mathbf{S}\mathbf{Q}=\mathbf{D}.$ Example: $\mathbf{S}=\begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}$ has $\lambda=6$ with $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\lambda=1$ with $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Their scalar product is $0$. Normalised: $\mathbf{Q}=\frac{1}{\sqrt5}\begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix}$.

A $2\times2$ matrix with a repeated eigenvalue that has only one eigenvector direction cannot be diagonalised, since $\mathbf{P}$ would not be invertible. For example $\begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}$ has $\lambda=2$ twice and only the eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. With complex eigenvalues a real $\mathbf{P}$ does not exist, though a complex one does. Always check that the two eigenvectors are independent (not multiples) before forming $\mathbf{P}$.

Every square matrix satisfies its own characteristic equation. For a $2\times2$ matrix with $\lambda^2-t\lambda+d=0$ (trace $t$, determinant $d$): $\mathbf{A}^2-t\mathbf{A}+d\mathbf{I}=\mathbf{0}.$ The constant term becomes $d\mathbf{I}$, not just $d$. For $\mathbf{C}=\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$: $\lambda^2-6\lambda+5=0$, so $\mathbf{C}^2=6\mathbf{C}-5\mathbf{I}=\begin{pmatrix} 7 & 6 \\ 18 & 19 \end{pmatrix}$. This reduces any power of $\mathbf{A}$ to a combination of $\mathbf{A}$ and $\mathbf{I}$. The same holds for $3\times3$ matrices with a cubic (the AS course uses $2\times2$ only).

Rearrange $\mathbf{C}^2-6\mathbf{C}+5\mathbf{I}=\mathbf{0}$ as $\mathbf{C}(6\mathbf{I}-\mathbf{C})=5\mathbf{I}$. Then $\mathbf{C}^{-1}=\frac15(6\mathbf{I}-\mathbf{C})=\frac15\begin{pmatrix} 4 & -1 \\ -3 & 2 \end{pmatrix}$. For higher powers multiply the relation by $\mathbf{C}$ and substitute repeatedly: $\mathbf{C}^3=6\mathbf{C}^2-5\mathbf{C}=31\mathbf{C}-30\mathbf{I}$. These give the same results as diagonalisation, so use whichever the question hints at.