京都大学情報学研究科知能情報学専攻 2024年2月実施基礎科目 F1-1

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Author

itsuitsuki, 祭音Myyura (assisted by ChatGPT 5.4 Thinking)

Description

Q.1

Consider simultaneous linear equations given by

\begin{cases} \lambda x_1 + 3x_3 = 0, \\ x_1 + (1 - \lambda)x_2 + 2x_3 = 0, \\ 2x_1 + (5 - \lambda)x_3 = 0. \end{cases}

Find all values of $\lambda$ such that there is a solution except for $x_1 = x_2 = x_3 = 0$ .

Q.2

Consider a quadratic equation given by

5x^2 - 4xy + 8y^2 = 1.

Draw the ellipse represented by this equation in the $xy$ -plane. The semi-major and semi-minor axes of the ellipse must be specified.

Q.3

Consider an inner product space $\mathbb{R}^3$ , where we have two vectors given by

\mathbf{x}_1 = \begin{bmatrix} 3 \\ 0 \\ 4 \end{bmatrix} \quad \text{and} \quad \mathbf{x}_2 = \begin{bmatrix} 4 \\ 5 \\ -3 \end{bmatrix}.

Let $W$ be a subspace of $\mathbb{R}^3$ spanned by $\mathbf{x}_1$ and $\mathbf{x}_2$ .

(1) Compute an orthonormal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ of $W$ .

(2) Compute $\mathbf{v}_3 \in \mathbb{R}^3$ such that $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ forms an orthonormal basis of $\mathbb{R}^3$ .

题目描述

考虑齐次线性方程组
$\begin{cases} \lambda x_1+3x_3=0,\\ x_1+(1-\lambda)x_2+2x_3=0,\\ 2x_1+(5-\lambda)x_3=0. \end{cases}$
求使其存在非零解的全部 $\lambda$ 。
对二次曲线
$5x^2-4xy+8y^2=1,$
在 $xy$ 平面画出椭圆，并标明半长轴、半短轴。
在 $\mathbb R^3$ $R^{3}$ 中给定
$x_1=(3,0,4)^\top,\qquad x_2=(4,5,-3)^\top,$
令 $W=\operatorname{span}\{x_1,x_2\}$ $W = span {x_{1}, x_{2}}$ 。
1. 求 $W$ 的一组标准正交基 $\{v_1,v_2\}$ ；
2. 求 $v_3$ ，使 $\{v_1,v_2,v_3\}$ 成为 $\mathbb R^3$ 的标准正交基。

考点

齐次线性方程组：令系数矩阵行列式为零确定存在非零解的参数。
二次型正交对角化：求对称矩阵特征值、旋转主轴方向及椭圆半轴长度。
Gram–Schmidt 正交化：把给定生成向量标准正交化，再求正交补中的单位向量。

Kai

Q.1

This is a homogeneous linear system. A non-zero solution exists if and only if the coefficient matrix is singular.

The coefficient matrix is

A(λ) = [[λ, 0, 3],
        [1, 1-λ, 2],
        [2, 0, 5-λ]].

Its determinant is

det A(λ)
= λ((1-λ)(5-λ)) + 3(2(1-λ))
= (λ - 1)(λ - 2)(λ - 3).

Therefore, a non-trivial solution exists exactly when

λ = 1, 2, 3.

Q.2

Write the quadratic form as

[x y] [[5, -2], [-2, 8]] [x y]^T = 1.

The symmetric matrix

M = [[5, -2], [-2, 8]]

has eigenvalues 4 and 9.

Eigenvalue 4 has eigenvector (2, 1)^T.
Eigenvalue 9 has eigenvector (-1, 2)^T.

So if we rotate coordinates to the orthonormal axes

u = (2x + y) / sqrt(5),
v = (-x + 2y) / sqrt(5),

then the equation becomes

4u^2 + 9v^2 = 1.

Hence:

the center is the origin;
the semi-major axis length is 1/2, along the direction (2, 1);
the semi-minor axis length is 1/3, along the direction (-1, 2).

Q.3

(1) Orthonormal basis of `W`

First,

||x1|| = sqrt(3^2 + 0^2 + 4^2) = 5,

so we may take

v1 = x1 / ||x1|| = (3/5, 0, 4/5)^T.

Next, check orthogonality:

x1 · x2 = 3·4 + 0·5 + 4·(-3) = 12 - 12 = 0.

Thus x2 is already orthogonal to x1. Also,

||x2|| = sqrt(4^2 + 5^2 + (-3)^2) = sqrt(50) = 5sqrt(2).

Therefore

v2 = x2 / ||x2|| = (4, 5, -3)^T / (5sqrt(2)).

So an orthonormal basis of W is

{
  (3/5, 0, 4/5)^T,
  (4, 5, -3)^T / (5sqrt(2))
}.

(2) A vector `v3` completing an orthonormal basis of `R^3`

Take a unit vector orthogonal to both v1 and v2. Using the cross product:

x1 × x2 = (-20, 25, 15)^T = 5(-4, 5, 3)^T.

The norm of (-4, 5, 3)^T is sqrt(16 + 25 + 9) = 5sqrt(2), so we can choose

v3 = (-4, 5, 3)^T / (5sqrt(2)).

Then {v1, v2, v3} is an orthonormal basis of R^3.

(Any choice differing by sign is also correct.)

Author​

Description​

Q.1​

Q.2​

Q.3​

题目描述​

考点​

Kai​