東京大学 情報理工学系研究科 コンピュータ科学専攻 2015年2月実施 問題1

Author

kainoj

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(1) Suppose that \(A\) is a real symmetric matrix. Let \(\boldsymbol{x}\) and \(\lambda\) be an eigenvector and the corresponding eigenvalue of \(A\), respectively. Prove that \(\lambda\) is a real number.

(2) Suppose that \(A\) is a real symmetric matrix. Let \(\boldsymbol{x}_1\) and \(\boldsymbol{x}_2\) be eigenvectors of \(A\); and \(\lambda_1\) and \(\lambda_2\) be the eigenvalues of \(A\) that correspond to \(\boldsymbol{x}_1\) and \(\boldsymbol{x}_2\), respectively. Furthermore, assume that \(\lambda_1 \neq \lambda_2\).

Prove that \(\boldsymbol{x}_1 \cdot \boldsymbol{x}_2 = 0\). Here \(\boldsymbol{x}_1 \cdot \boldsymbol{x}_2\) denotes the inner product of the two vectors \(\boldsymbol{x}_1\) and \(\boldsymbol{x}_2\).

(3) Let \(A\) be a matrix

\[ A = \begin{pmatrix} x & a_1 & b_1 & c_1 \\ y & a_2 & b_2 & c_2 \\ z & a_3 & b_3 & c_3 \\ 1 & 1 & 1 & 1 \end{pmatrix} \]

where \(a_i\), \(b_i\), and \(c_i\) ( \(i \in \{1, 2, 3\}\)) are real numbers such that the two vectors \((b_1 - a_1, b_2 - a_2, b_3 - a_3)\) and \((c_1 - a_1, c_2 - a_2, c_3 - a_3)\) are linearly independent. Let us define a function \(f: \mathbb{R}^3 \to \mathbb{R}\) by

\[ f(x, y, z) = \text{det}(A) \]

Prove that the equation \(f(x, y, z) = 0\) determines a plane in the \(xyz\) space.

(4) In the setting of Question (3), give three points that lie on the plane determined by \(f(x, y, z) = 0\).

Kai

(1)

Let \(A\in \mathbb{R}^{n\times n}\) such that \(A = A^T\) and \(Ax = \lambda x\) for some eigenvector \(x\) and corresponding eigenvalue \(\lambda\). Suppose that \(\lambda \in \mathbb{C}\), i.e. \(\lambda = a + ib\) for some \(a,b\in\mathbb{R}\). Let's start with \(Ax = \lambda x\):

\[ \begin{aligned} Ax &= \lambda x && \text{take conjugate transpose} \\ x^H A^H &= \lambda^H x^H && \text{$A$ is real, so $A=A^H$} \\ x^H A &= \lambda^H x^H && \text{multiply by $x$} \\ x^H A x &= \lambda^H x^H x && \text{now, $Ax = \lambda x$} \\ \lambda x^H x &= \lambda^H x^H x \\ \lambda &= \lambda^H \\ a + bi &= a - bi \end{aligned} \]

Hence, \(\text{Im}(\lambda)\) = 0, which means \(\lambda \in \mathbb{R}\).

(2)

Start with the first eigenpair:

\[ \begin{aligned} Ax_1 &= \lambda_1 x_1 && \text{take transpose. $A = A^T$} \\ x_1^T A &= \lambda_1 x_1^T && \text{multiply by $x_2$} \\ x_1^T A x_2 &= \lambda_1 x_1^T x_2 && \\ \end{aligned} \]

Now fiddle with the second eigenpair:

\[ \begin{aligned} Ax_2 &= \lambda_2 x_2 && \text{left-multiply by $x_1^T$} \\ x_1^T Ax_2 &= \lambda_2 x_1^T x_2 \end{aligned} \]

If we subtract both result, we get:

\[ \begin{aligned} 0 = (\lambda_1 - \lambda_2) x_1^T x_2 \end{aligned} \]

Since \(\lambda_1 \neq \lambda_2\), then \(x_1^T x_2 = 0\). In other words, \(x_1 \cdot x_2 =0\).

(3) - send help

\[ \begin{aligned} A = \begin{pmatrix} x & a_1 & b_1 & c_1 \\ x & a_2 & b_2 & c_2 \\ x & a_3 & b_3 & c_3 \\ 1 & 1 & 1 & 1 \end{pmatrix} \end{aligned} \]

\[ \begin{aligned} f(x,y,z) = \text{det}(A) \end{aligned} \]

This seems to be difficult. I succeeded in computing determinant using Laplace expansion wrt first column, but it gets hardcore later. Any ideas?

(4)

Give 3 points that lie on plane \(f(x,y,z) = 0\). When is determinant \(=0\) ? For example, when some columns are dependent. Let's make some columns dependent. Zum Beispiel, let's take:

\[ \begin{aligned} x = a_1 && y = a_2 && z = a_3 \end{aligned} \]

Now first and second column are dependent, so \(\text{det}(A) = f(x,y,z) = 0\). To get two others solutions, take \(x = b_1, y = b_2, z = b_3\) and \(x = c_1, y = c_2, z = c_3\) or simply multiply the first solution by some constants.