Matrices (Further Pure 2)
Introduced| English | Meaning |
|---|---|
| stretch/stretʃ/ | A transformation that multiplies distances or coordinates in a specified direction by a scale factor. |
| eigenvector/ˈaɪdʒənvektə/ | A nonzero vector that a linear transformation maps to a scalar multiple of itself. |
| eigenvalue/ˈaɪdʒənvæljuː/ | eigenvalue |
| diagonalisation/daɪˌæɡənəlaɪˈzeɪʃn/ | diagonalisation |
| matrix equation/ˈmeɪtrɪks ɪˈkweɪʒn/ | An equation involving matrices and their operations, such as a matrix representation of simultaneous linear equations. |
| characteristic equation/ˌkærɪktəˈrɪstɪk ɪˈkweɪʒn/ | For a square matrix A, the equation det(A minus lambda I) equals zero, whose solutions are the matrix's eigenvalues. |
| scalar/ˈskeɪlə/ | A quantity described by magnitude alone, with no direction. |
| diagonal matrix/daɪˈæɡənl ˈmeɪtrɪks/ | A square matrix whose entries outside the main diagonal are all zero. |
The hidden directions in a matrix
- Every matrix transforms space. Some directions are special: the matrix only stretches 拉伸 them, without rotating. These are the eigenvectors 特征向量, and the stretch factors are the eigenvalues 特征值.
- Eigenvalues unlock the structure of a matrix — they're the key to diagonalisation 对角化, stability analysis, and Google's PageRank algorithm.
Matrix equations 矩阵方程
- Three linear equations in three unknowns → a matrix equation $A\mathbf{x} = \mathbf{b}$. A non-singular $A$ gives one unique solution: $\mathbf{x} = A^{-1}\mathbf{b}$.
Worked example. $\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \end{pmatrix}$. Solving: $2x + y = 5$ and $x + 3y = 7$. Solution: $x = \dfrac{8}{5}$, $y = \dfrac{9}{5}$.
An eigenvector keeps its direction under the matrix; a general vector is turned
Matrices (Further Pure 2)
(x', y') = M(x, y)
Rotations, reflections and shears are all matrices — read them off the transformed basis vectors.
Eigenvalues and eigenvectors
- The characteristic equation 特征方程 $\det(A - \lambda I) = 0$ gives the eigenvalues $\lambda$.
- For each $\lambda$, the eigenvector $\mathbf{v}$ satisfies $A\mathbf{v} = \lambda\mathbf{v}$.
- Multiplying an eigenvector by $A$ only stretches it (by factor $\lambda$), without changing its direction.
Eigenvectors are not unique. If $\mathbf{v}$ is an eigenvector, so is $k\mathbf{v}$ for any non-zero scalar 标量 $k$. We usually choose the simplest form (smallest integers).
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?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
(2−λ)² = 1 → 2−λ = ±1 → λ = 1 or 3; the larger is 3.
The eigenvalues of a matrix A are found by solving:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
The characteristic equation det(A − λI) = 0 gives the eigenvalues.
Multiplying an eigenvector by A only stretches it (Av = λv).
Difficulty 1/5 · 100 coins + a small bonus on first completion today
An eigenvector keeps its direction; A just scales it by the eigenvalue λ.
If v is an eigenvector of A, then 2v is also an eigenvector of A.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
Eigenvectors are not unique — any non-zero scalar multiple of an eigenvector is also an eigenvector.
Diagonalisation
- You can write $A = QDQ^{-1}$, where $D$ is a diagonal matrix 对角矩阵 of eigenvalues and the columns of $Q$ are the eigenvectors.
- This makes computing $A^n$ easy: $A^n = QD^nQ^{-1}$, and $D^n$ is just the diagonal entries raised to the $n$th power.
In the diagonalisation A = QDQ⁻¹, the matrix D contains:
Difficulty 1/5 · 100 coins + a small bonus on first completion today
D is a diagonal matrix with the eigenvalues of A on its main diagonal.
If A = QDQ⁻¹, then A³ = QD³Q⁻¹. If D has diagonal entries 1 and 3, D³ has diagonal entries 1 and ?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
D³ has diagonal entries 1³ = 1 and 3³ = 27.
Worked example
- $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. Characteristic equation: $(2-\lambda)^2 - 1 = 0 \Rightarrow \lambda^2 - 4\lambda + 3 = 0 \Rightarrow \lambda = 1$ or $3$.
- For $\lambda = 1$: eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. For $\lambda = 3$: eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
- A matrix with zero determinant has a singularity (it is singular and has no inverse).
You've got it
- 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)