Matrices (Further Pure 2)
Matrix equations & eigenvalues
- Three linear equations in three unknowns → a matrix equation $A\mathbf{x} = \mathbf{b}$. A non-singular $A$ gives one solution.
- The characteristic equation $\det(A - \lambda I) = 0$ gives the eigenvalues $\lambda$.
- For each $\lambda$, the eigenvector $\mathbf{v}$ satisfies $A\mathbf{v} = \lambda\mathbf{v}$.
Practice
The matrix [[2,1],[1,2]] has characteristic equation (2−λ)² − 1 = 0, giving λ = 1 or λ = 3. What is the larger eigenvalue?
(2−λ)² = 1 → 2−λ = ±1 → λ = 1 or 3; the larger is 3.
Practice
The eigenvalues of a matrix A are found by solving:
The characteristic equation det(A − λI) = 0 gives the eigenvalues.
Diagonalisation
- You can write $A = QDQ^{-1}$, where $D$ is a diagonal matrix of eigenvalues and the columns of $Q$ are the eigenvectors.
- Example: $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ has $(2-\lambda)^2 - 1 = 0$, so $\lambda = 1$ or $3$.
Practice
Multiplying an eigenvector by A only stretches it (Av = λv).
An eigenvector keeps its direction; A just scales it by the eigenvalue λ.
You've got it
Key idea
- eigenvalues solve $\det(A - \lambda I) = 0$; eigenvectors satisfy $A\mathbf{v} = \lambda\mathbf{v}$
- multiplying an eigenvector by $A$ only stretches it
- diagonalise: $A = QDQ^{-1}$ (D = eigenvalues, Q's columns = eigenvectors)