Diagonalisation of symmetric matricesEdexcel International A Level Further Maths: Revision notes

A square matrix is symmetric if $\mathbf{A}^T=\mathbf{A}$, so the entry in row $i$, column $j$ equals the entry in row $j$, column $i$. Symmetric matrices have two properties that make them special:

• all their eigenvalues are real;
• eigenvectors corresponding to different eigenvalues are perpendicular, i.e. their scalar product is $0$. Example: $\mathbf{S}=\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ has $\lambda=4$ with $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\lambda=2$ with $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, and $(1)(1)+(1)(-1)=0$.

A square matrix $\mathbf{P}$ is orthogonal if $\mathbf{P}^T\mathbf{P}=\mathbf{I},\quad\text{so}\quad\mathbf{P}^{-1}=\mathbf{P}^T.$ Equivalently, the columns of $\mathbf{P}$ are orthonormal: each column has magnitude $1$ and every pair of columns has scalar product $0$ (the same is true of the rows). Then $\det\mathbf{P}=\pm1$. For example $\frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ is orthogonal: both columns have magnitude $\frac{1}{\sqrt2}\sqrt2=1$ and their scalar product is $\frac12(1-1)=0$.

For a symmetric matrix $\mathbf{A}$, the method is:

1. Find the eigenvalues from $\det(\mathbf{A}-\lambda\mathbf{I})=0$.
2. Find an eigenvector for each eigenvalue by solving $(\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}$.
3. Normalise each eigenvector (divide by its magnitude).
4. Put the normalised eigenvectors in the columns of $\mathbf{P}$, and the matching eigenvalues, in the same order, on the diagonal of $\mathbf{D}$. Then $\mathbf{P}$ is orthogonal and $\mathbf{P}^T\mathbf{A}\mathbf{P}=\mathbf{D}.$ Equivalently $\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^T$, because $\mathbf{P}^{-1}=\mathbf{P}^T$.

Diagonalise $\mathbf{S}=\begin{pmatrix} 1 & 2 \\ 2 & -2 \end{pmatrix}$. $\det(\mathbf{S}-\lambda\mathbf{I})=(1-\lambda)(-2-\lambda)-4=\lambda^2+\lambda-6$, so $\lambda=2$ or $-3$. $\lambda=2$: $-x+2y=0$ gives $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. $\lambda=-3$: $4x+2y=0$ gives $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$. Each has magnitude $\sqrt5$. $\mathbf{P}=\frac{1}{\sqrt5}\begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix},\qquad\mathbf{D}=\begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix}.$ Check: the eigenvectors are perpendicular since $2\times1+1\times(-2)=0$.

For $\mathbf{A}=\begin{pmatrix} 2 & 0 & -1 \\ 0 & 2 & -1 \\ -1 & -1 & 3 \end{pmatrix}$ the eigenvalues are $1$, $2$, $4$ with eigenvectors $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$ of magnitudes $\sqrt3$, $\sqrt2$, $\sqrt6$. So $\mathbf{P}$ has columns $\frac{1}{\sqrt3}(1,1,1)^{T}$, $\frac{1}{\sqrt2}(1,-1,0)^{T}$, $\frac{1}{\sqrt6}(1,1,-2)^{T}$ and $\mathbf{D}=\mathrm{diag}(1,2,4)$. Checks: the three eigenvectors are mutually perpendicular; the trace of $\mathbf{A}$ equals the sum of the diagonal of $\mathbf{D}$; $\det\mathbf{A}=\det\mathbf{D}$ (since $\det\mathbf{P}\det\mathbf{P}^T=1$).