Matrices (Further Pure 2)
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| stretch/stretʃ/ | kéo dãn |
| eigenvector/ˈaɪdʒənvektə/ | véc-tơ riêng |
| eigenvalue/ˈaɪdʒənvæljuː/ | giá trị riêng |
| diagonalisation/daɪˌæɡənəlaɪˈzeɪʃn/ | khử chéo hóa |
| matrix equation/ˈmeɪtrɪks ɪˈkweɪʒn/ | phương trình ma trận |
| characteristic equation/ˌkærɪktəˈrɪstɪk ɪˈkweɪʒn/ | phương trình đặc trưng |
| scalar/ˈskeɪlə/ | thực |
| diagonal matrix/daɪˈæɡənl ˈmeɪtrɪks/ | ma trận chéo |
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?
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(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:
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The characteristic equation det(A − λI) = 0 gives the eigenvalues.
Multiplying an eigenvector by A always rotates it by 90°.
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False. An eigenvector keeps its line: Av = λv, so A only scales it by the eigenvalue λ (reversing it if λ < 0).
If v is an eigenvector of A, then 2v is also an eigenvector of A.
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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:
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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 ?
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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)
Write from a larger eigenvalue of 5
- The matrix has rows (4, 1) and (1, 4).
- Find the larger eigenvalue. An eigenvector can be scaled by any non-zero number.
The matrix has rows (4, 1) and (1, 4). What is its larger eigenvalue, and why is an eigenvector not a single vector?
The eigenvalues solve (4 minus lambda) squared minus 1 = 0. Scaling an eigenvector does not change the eigenvalue equation.
What is the larger eigenvalue?
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The eigenvalues are 3 and 5, so the larger one is 5. Other options — 3 (the smaller eigenvalue, and also the lesson larger eigenvalue); 125 (the cube of the larger eigenvalue).