京都大学 情報学研究科 知能情報学専攻 2021年8月実施 情報学基礎 F1-1

Author

Isidore

Description

Kai

設問1

(1)

\[ A = \begin{bmatrix} -8 & 0 \\ 0 & -8 \\ \end{bmatrix} \]

(2)

\[ A^{-1} = \frac{1}{4} \begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \\ \end{bmatrix} \]

(3)

\[ A^{15}=(A^3)^5=(-8E)^5=-2^{15}E \]

(4)

Denote the required the value as \(I\), and we have

\[ I = \lim_{n \rightarrow \infty}\|(AB)^nx\| = \lim_{n \rightarrow \infty}\|(-8r^3E)^{\frac{n}{3}}x\| \]

Obviously, the limit converges if and only if

\[ -8r^3 \in [-1, 1] \]

Hence, when \(r \in (-\frac{1}{2}, \frac{1}{2})\), we have

\[ I = 0 \]

Particularly, when \(r=\pm \frac{1}{2}\), we have

\[ I = \lim_{n \rightarrow \infty}\|(-1)^{\frac{n}{3}}x\| = \|x\| = \sqrt{x_1^2+x_2^2} \]

設問2

(1)

Obviously, as \(A\) is a real matrix, \(B\) and \(C\) are both real matrices. Consider the symmetry, we have

\[ B^T = (A^TA)^T = A^TA=B \]

So as \(C\).

(2)

Consider B's eigenvalue, we have

\[ A^TAx=\lambda x \]

Perform the transpose and right-multiply by the eigenvector \(x\), we have

\[ x^TA^TAx = \lambda x^Tx \]

Hence, we have

\[ \lambda = \frac{x^TA^TAx}{x^Tx} = \frac{\|Ax\|}{\|x\|} \geq 0 \]

(3)

Consider the eigenvalue \(\lambda_i\) with its eigenvector \(p_i\), we have

\[ A^TAp_i=\lambda_ip_i \]

Left-multiply by \(\frac{1}{\sqrt{\lambda_i}}A\), we have

\[ \frac{1}{\sqrt{\lambda_i}}AA^TAp_i = \sqrt{\lambda_i}Ap_i \\ AA^T(\frac{1}{\sqrt{\lambda_i}}Ap_i) = \lambda_i(\frac{1}{\sqrt{\lambda_i}}Ap_i) \]

Hence, insert \(q_i = \frac{1}{\sqrt{\lambda_i}}Ap_i\) and we have

\[ Cq_i = \lambda_i q_i \]

Similarly, we have \(Cq_j = \lambda_j q_j\) for a different eigenvalue \(\lambda_j\). Therefore, \(q_i\) and \(q_j\) are eigenvectors of C corresponding to eigenvalues \(\lambda_i\) and \(\lambda_j\).

Now we consider the inner product of \(q_i\) and \(q_j\)

\[ q_i^Tq_j = \frac{1}{\sqrt{\lambda_i \lambda_j}}(Ap_i)^T(Ap_j) = \frac{\sqrt{\lambda_j}}{\sqrt{\lambda_i}}p_i^Tp_j \]

Obviously, if \(i \neq j\), as \(B\) is a real symmetric matrix, \(p_i^Tp_j = 0\). If \(i = j\), \(p_i^Tp_j = 1\) Therefore, we have

\[ q_i^Tq_j = \begin{cases} 0 & \text{if } i \neq j \\ 1 & \text{if } i = j \end{cases} \]

Finally, insert \(q_i = \frac{1}{\sqrt{\lambda_i}}Ap_i\) to \(\frac{1}{\sqrt{\lambda_i}}A^Tq_i\), we immediately have

\[ \frac{1}{\sqrt{\lambda_i}}A^Tq_i = \frac{1}{\sqrt{\lambda_i}}A^T\frac{1}{\sqrt{\lambda_i}}Ap_i = \frac{1}{\lambda_i}A^TAp_i = p_i \]

Q.E.D