Three-dimensional matrices and eigenvectors

What you’ll learn

• Use determinants, inverses and transposes in three dimensions.
• Find eigenvalues and diagonalise symmetric matrices.

Understand the idea

Three-by-three matrices represent transformations of three-dimensional column vectors. Cofactor expansion gives a determinant, and the transposed cofactor matrix divided by a nonzero determinant gives the inverse. Reversing a product reverses the order of its inverses or transposes.

An eigenvector is mapped to a scalar multiple of itself, with the eigenvalue as the multiplier. A negative eigenvalue reverses its direction; a zero eigenvalue maps it to the zero vector. Solve the characteristic equation for eigenvalues, then find nonzero vectors in the corresponding null spaces. For a real symmetric matrix, choose orthonormal eigenvectors as columns of $P$; then $P^TAP$ is diagonal, with eigenvalues in the matching column order.

Key formulae

$$A\mathbf v=\lambda\mathbf v,\qquad\det(A-\lambda I)=0$$

$$A^{-1}=\frac{\operatorname{adj}A}{\det A}$$

$$(AB)^T=B^TA^T,\qquad(AB)^{-1}=B^{-1}A^{-1}$$

$$P^TAP=D,\qquad P^TP=I\quad\text{(orthogonal }P\text{)}$$

Worked example

Find the eigenvalues of $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ and one eigenvector for each.

1. The characteristic equation is $(3-\lambda)^2-1=0$.
2. This gives eigenvalues $4$ and $2$.
3. For $4$, solve $-x+y=0$; for $2$, solve $x+y=0$.

Answer: $\lambda=4$ has eigenvector $(1,1)^T$; $\lambda=2$ has eigenvector $(1,-1)^T$.