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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

Author​

itsuitsuki, 祭音Myyura (assisted by ChatGPT 5.4 Thinking)

Description (English)​

Q.1​

Consider simultaneous linear equations given by

Find all values of such that there is a solution except for .

Q.2​

Consider a quadratic equation given by

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

Q.3​

Consider an inner product space , where we have two vectors given by

Let be a subspace of spanned by and .

(1) Compute an orthonormal basis of .

(2) Compute such that forms an orthonormal basis of .

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.)

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