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東京大学 新領域創成科学研究科 メディカル情報生命専攻 2015年8月実施 問題8

Author

zephyr

Description

Answer the following questions about linear algebra.

(1)

Compute the inverse matrix of the following matrix,

\[ \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}. \]

(2)

Consider data points \((x_i, y_i), i = 1, \dots, n\) in a two-dimensional space. Variance with respect to the x-axis, variance with respect to the y-axis, and covariance are respectively defined as

\[ \sigma_x = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2, \quad \sigma_y = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2, \quad \sigma_{xy} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \]

where \(\bar{x}, \bar{y}\) denote the averages with respect to the x and y axes, respectively.

A: Compute the variance-covariance matrix

\[ \begin{pmatrix} \sigma_x & \sigma_{xy} \\ \sigma_{xy} & \sigma_y \end{pmatrix} \]

for the following data points, \((-2, -2), (2, 2), (1, -1), (-1, 1)\).

B: Compute all eigenvalues and eigenvectors of the variance-covariance matrix.

(3)

Prove that, if the eigenvalues of a regular matrix A are \(\lambda_1, \dots, \lambda_n\), those of the inverse matrix A\(^{-1}\) are \(1/\lambda_1, \dots, 1/\lambda_n\).

回答以下关于线性代数的问题。

(1)

计算以下矩阵的逆矩阵，

\[ \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}. \]

(2)

考虑数据点 \((x_i, y_i), i = 1, \dots, n\) 在二维空间中。相对于 x 轴的方差、相对于 y 轴的方差和协方差分别定义为

\[ \sigma_x = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2, \quad \sigma_y = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2, \quad \sigma_{xy} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \]

其中 \(\bar{x}, \bar{y}\) 分别表示相对于 x 和 y 轴的平均值。

A: 计算方差-协方差矩阵

\[ \begin{pmatrix} \sigma_x & \sigma_{xy} \\ \sigma_{xy} & \sigma_y \end{pmatrix} \]

对于以下数据点，\((-2, -2), (2, 2), (1, 1), (-1, 1)\)。

B: 计算方差-协方差矩阵的所有特征值和特征向量。

(3)

证明，如果一个正规矩阵 A 的特征值是 \(\lambda_1, \dots, \lambda_n\)，那么其逆矩阵 A\(^{-1}\) 的特征值是 \(1/\lambda_1, \dots, 1/\lambda_n\)。

Kai

(1)

To find the inverse of the matrix

\[ \mathbf{A} = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix}, \]

we use the formula for the inverse of a \(2 \times 2\) matrix:

\[ \mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \]

where \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(\det(\mathbf{A}) = ad - bc\).

For our matrix,

\[ a = 1, \quad b = 2, \quad c = 2, \quad d = 5. \]

First, compute the determinant:

\[ \det(\mathbf{A}) = (1 \cdot 5) - (2 \cdot 2) = 5 - 4 = 1. \]

Then, the inverse is

\[ \mathbf{A}^{-1} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix}. \]

(2)

A: Variance-Covariance Matrix

Given data points \((-2, -2), (2, 2), (1, -1), (-1, 1)\), we first compute the mean values:

\[ \bar{x} = \frac{-2 + 2 + 1 - 1}{4} = 0, \quad \bar{y} = \frac{-2 + 2 - 1 + 1}{4} = 0. \]

Next, we compute the variances and covariances:

\[ \sigma_{xx} = \frac{1}{4} \sum_{i=1}^4 (x_i - \bar{x})^2 = \frac{1}{4} [(2)^2 + (2)^2 + (1)^2 + (1)^2] = \frac{10}{4} = 2.5, \]
\[ \sigma_{yy} = \frac{1}{4} \sum_{i=1}^4 (y_i - \bar{y})^2 = \frac{1}{4} [(2)^2 + (2)^2 + (1)^2 + (1)^2] = \frac{10}{4} = 2.5, \]
\[ \sigma_{xy} = \frac{1}{4} \sum_{i=1}^4 (x_i - \bar{x})(y_i - \bar{y}) = \frac{1}{4} [(-2)(-2) + (2)(2) + (1)(-1) + (-1)(1)] = \frac{6}{4} = 1.5. \]

Thus, the variance-covariance matrix is:

\[ \mathbf{C} = \begin{pmatrix} 2.5 & 1.5 \\ 1.5 & 2.5 \end{pmatrix}. \]

B: Eigenvalues and Eigenvectors

To find the eigenvalues \(\lambda\) of \(\mathbf{C}\), solve the characteristic equation:

\[ \det(\mathbf{C} - \lambda \mathbf{I}) = 0. \]

For our matrix \(\mathbf{C}\),

\[ \mathbf{C} - \lambda \mathbf{I} = \begin{pmatrix} 2.5 - \lambda & 1.5 \\ 1.5 & 2.5 - \lambda \end{pmatrix}, \]

the determinant is:

\[ \det(\mathbf{C} - \lambda \mathbf{I}) = (2.5 - \lambda)(2.5 - \lambda) - (1.5)(1.5) = \lambda^2 - 5\lambda + 4 = 0. \]

Solving for \(\lambda\), we get:

\[ \lambda^2 - 5\lambda + 4 = 0 \implies (\lambda - 4)(\lambda - 1) = 0 \implies \lambda_1 = 4, \quad \lambda_2 = 1. \]

To find the eigenvectors corresponding to the eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 1\), we solve the equation \((\mathbf{C} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}\).

For \(\lambda_1 = 4\)
\[ \mathbf{C} - 4\mathbf{I} = \begin{pmatrix} 2.5 - 4 & 1.5 \\ 1.5 & 2.5 - 4 \end{pmatrix} = \begin{pmatrix} -1.5 & 1.5 \\ 1.5 & -1.5 \end{pmatrix}. \]

The equation \((\mathbf{C} - 4\mathbf{I})\mathbf{v} = \mathbf{0}\) becomes:

\[ \begin{pmatrix} -1.5 & 1.5 \\ 1.5 & -1.5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \]

This gives us the system of equations:

\[ -1.5v_1 + 1.5v_2 = 0, \\ 1.5v_1 - 1.5v_2 = 0. \]

From the first equation, we obtain \(v_1 = v_2\). Therefore, an eigenvector corresponding to \(\lambda_1 = 4\) is:

\[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. \]
For \(\lambda_2 = 1\)
\[ \mathbf{C} - 1\mathbf{I} = \begin{pmatrix} 2.5 - 1 & 1.5 \\ 1.5 & 2.5 - 1 \end{pmatrix} = \begin{pmatrix} 1.5 & 1.5 \\ 1.5 & 1.5 \end{pmatrix}. \]

The equation \((\mathbf{C} - 1\mathbf{I})\mathbf{v} = \mathbf{0}\) becomes:

\[ \begin{pmatrix} 1.5 & 1.5 \\ 1.5 & 1.5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \]

This gives us the system of equations:

\[ 1.5v_1 + 1.5v_2 = 0, \\ 1.5v_1 + 1.5v_2 = 0. \]

From the first equation, we obtain \(v_1 = -v_2\). Therefore, an eigenvector corresponding to \(\lambda_2 = 1\) is:

\[ \mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. \]

Thus, the eigenvectors corresponding to the eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 1\) are \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\), respectively.

(3)

Let \(\mathbf{A}\) be a regular matrix with eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\) and corresponding eigenvectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\). By definition, we have:

\[ \mathbf{A} \mathbf{v}_i = \lambda_i \mathbf{v}_i \implies \mathbf{A}^{-1} \mathbf{A} \mathbf{v}_i = \mathbf{A}^{-1} (\lambda_i \mathbf{v}_i) \implies \mathbf{v}_i = \lambda_i \mathbf{A}^{-1} \mathbf{v}_i \implies \mathbf{A}^{-1} \mathbf{v}_i = \frac{1}{\lambda_i} \mathbf{v}_i. \]

Thus, the eigenvalues of \(\mathbf{A}^{-1}\) are \(\frac{1}{\lambda_i}\) for \(i = 1, 2, \ldots, n\).

Knowledge

矩阵逆 方差协方差矩阵 特征值和特征向量

解题技巧和信息

  1. 计算逆矩阵时，确保熟记 \(2 \times 2\) 矩阵的逆矩阵公式。
  2. 计算方差-协方差矩阵时，需准确计算均值、方差和协方差。
  3. 找特征值和特征向量时，熟悉特征值方程和特征向量的计算方法。
  4. 证明部分注意利用特征值和特征向量的定义和性质。

重点词汇

  • Inverse matrix: 逆矩阵
  • Variance-Covariance matrix: 方差-协方差矩阵
  • Eigenvalue: 特征值
  • Eigenvector: 特征向量

参考资料

  1. Gilbert Strang, Linear Algebra and Its Applications, Chapter 3.
  2. Axler, Sheldon, Linear Algebra Done Right, Chapter 5.