京都大学 情報学研究科 知能情報学専攻 2024年8月実施 情報学基礎 F1-1

Author

amongtrees, 空想性錯視, 祭音Myyura

Description

In the questions below, denotes the set of all real numbers, stands for the transpose of , is the inverse of a matrix , and denotes the identity matrix of size .

Q.1

Let be a nonzero column vector, and define a matrix as

Answer the following questions.

(1) Show that is a symmetric matrix.

(2) Show that is an orthogonal matrix.

(3) Compute all the eigenvalues of .

(4) Compute the determinant of .

(5) Define a column vector as

Given a column vector , which is not multiplied by any scalar. Determine so that becomes multiplied by some scalar, and express it using and .

Q.2

Answer the following questions.

(1) Let be an arbitrary real matrix of size . Assume that is non-singular. Show that the following equation holds.

(2) Let and be arbitrary real matrices of size and , respectively. Assume that is non-singular. Show that is non-singular and that the following equation holds.

Kai

Q.1

Background: Householder Transformation

(1)

Since is symmetric and is a scalar, we have:

Thus, is a symmetric matrix.

(2)

According to (1), we know , so we have:

Thus, is an orthogonal matrix.

(3)

The form is the projection matrix of vector , we can make the following classification discussion:

Case 1: is orthogonal to

Let be any non-zero vector such that , obviously the subspace of satisfied vectors in has demensions, and we have:

Thus, is an eigenvalue with multiplicity ;

Case 2: is parallel to

Let for some scalar , we have:

Thus, is an eigenvalue with multiplicity 1.

(4)

The determinant of a matrix is the product of its eigenvalues.

(5)

We want to find a vector such that for some scalar , where and is not a scalar multiple of .

Since is an orthogonal matrix, it preserves the norm of the multiplied vector. Thus, , and we have:

So we are actually looking for a such that . From the definition of , we have:

Rearranging this equation:

This shows that the vector must be parallel to . Let's choose to be just this vector and verify the result. Let , we have . Now compute the terms in the equation (1):

Thus, equation (1) stands true, which confirms our choice of is correct. And the different signs influence be of different scalar of .

Q.2

(1)

Left-multiplying both sides of the equation by , we get .

(2)

Firstly we need prove is non-singular. Assume it's singular, then there must exist a non-zero vector such that:

Left-multiplying by gives:

Since we are given that is non-singular, the only solution to this equation is . Substituting this back into the initial assumption:

This contradicts our assumption that is a non-zero vector. Thus, must be non-singular.

Next, we have:

Similarly left-multiplying both sides of the equation by , we get .