---
title: "東京大学大学院　情報理工入試過去問（数学）解答・２０１９年　第一問"
tags: u-tokyo, exam
---

# 東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０１９年　第一問</span>

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+ こちらの解答は非公式で個人記録用に作成されており、大学院側に認められたものではありません。正確性についての保証は致しかねます。
+ 問題本文は研究科のウェブサイトから見ることができます。
+ 満足出来た解答でしたらコメントしていただけると幸いです。

## ２０１９年　第一問
Let $A^*$ is the conjugate transpose matrix of a complex square matrix $A$.
If $A^*A = I$, then $A$ is **unitary**.

### (1) Prove that if $A, B$ are unitary, $AB$ is also unitary.
If $A, B$ are unitary, then:
$$
(AB)^*(AB) = B^*A^*AB = B^*IB = B^*B = I
$$

So $AB$ is also unitary.

### (2) Let for some real square matrix $C, D$ with size $n$, $$F = C + iD\;\;\;\;\;\;\;\;G = \bmat{C & -D \\ D & C}$$ <br>&emsp;&ensp;Prove that $G$ is orthogonal if and only if $F$ is unitary.
If $G$ is orthogonal, then:
$$
\begin{align}
\trps{G}G &= \bmat{\trps{C} & \trps{D} \\ -\trps{D} & \trps{C}}\bmat{C & -D \\ D & C} \\
&= \bmat{
\trps{C}C + \trps{D}D & C\trps{D} - \trps{C}D \\
\trps{C}D - C\trps{D} & \trps{C}C + \trps{D}D
} = I
\end{align}
$$

Inducing that,
+ $C\trps{D} - \trps{C}D = \trps{C}D - C\trps{D} = 0 \Rightarrow \trps{C}D = C\trps{D}$
+ $\trps{C}C + \trps{D}D = I$

If $F$ is unitary, then:
$$
\begin{align}
(C + iD)^*(C + iD) &= \trps{C - iD}(C + iD) \\
&= (\trps{C} - i\trps{D})(C + iD) \\
&= \trps{C}C + i\trps{C}D + i\left( C\trps{D} - \trps{D}D \right) = I
\end{align}
$$

Inducing that:
+ $\trps{C}C + i\trps{C}D = I$
+ $C\trps{D} - \trps{D}D = 0 \Rightarrow C\trps{D} = \trps{D}D$
which are the same as the properties of $G$ when its orthogonal.

Thus, $G$ is orthogonal if and only if $F$ is unitary.

### (3) Find the eigenvalues of $$\displaystyle{1 \over 2}\bmat{1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\1 & -1 & 1 & -1 \\ 1 & -i & -1 & i}$$
For some scalar $C$ scaling matrix $M$ having eigenvalue $\lambda$:
$$
(cM)\vec{v} = c(M\vec{v}) = c(\lambda\vec{v}) = (c\lambda)\vec{v}
$$

So we can instead find eigenvalues for $\bmat{1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\1 & -1 & 1 & -1 \\ 1 & -i & -1 & i}$ then scale it down by $\displaystyle {1 \over 2}$.

To find the eigenvalues, we let $\det{(A - \lambda I)} = 0$, then:
$$
\vmat{
1 - \lambda & 1 & 1 & 1 \\
1 & i - \lambda & -1 & -i \\
1 & -1 & 1 - \lambda & -1 \\
1 & -i & -1 & i - \lambda
} = \vmat{
1 - \lambda & 1 & 1 & 1 \\
1 & i - \lambda & -1 & -i \\
2 - \lambda & 0 & 2 - \lambda & 0 \\
1 & -i & -1 & i - \lambda
} = \vmat{
1 - \lambda & 1 & \lambda & 1 \\
1 & i - \lambda & -2 & -i \\
2 - \lambda & 0 & 0 & 0 \\
1 & -i & -2 & i - \lambda
} \\
\Rightarrow (2 - \lambda)\vmat{
1 & \lambda & 1 \\
i - \lambda & -2 & -i \\
-i & -2 & i - \lambda
} = 0 \\
\Rightarrow (2 - \lambda)(-\lambda^3 - 2i\lambda^2 + 4\lambda - 16i) = 0 \\
\Rightarrow (2 - \lambda)^2(1 - 2i)(2 + \lambda) = 0
$$

Multiplying by $1 \over 2$, thus the four eigenvalues are:
$$
\begin{align}
\lambda_1 &= i\\
\lambda_2 &= -1\\
\lambda_3 &= \lambda_4 = 1
\end{align}
$$

### (4) For some integer $n$ and a matrix $Q$ whose entry is defined as: $$q_{jk} = \frac{1}{\sqrt{n}}e^{\frac{2i\pi(j - 1)(k - 1)}{n}}$$&emsp;, Prove that $Q$ is unitary.
By Euler's formula, we know that:
$$
q_{jk} = \frac{1}{\sqrt{n}}\left[ \cos\frac{2(j-1)(k-1)}{n}\pi + i\sin\frac{2(j-1)(k-1)}{n}\pi \right] = \frac{1}{\sqrt{n}}e^{\frac{2(j-1)(k-1)}{n}\pi i}
$$

Also, for the conjugate transpose $Q^*$, since $\begin{cases} -\sin\theta = \sin-\theta \\ \cos\theta = \cos-\theta \end{cases}$:
$$
\begin{align}
q^*_{jk} = \overline{q_{kj}} &= \frac{1}{\sqrt{n}}\left[ \cos\frac{2(j-1)(k-1)}{n}\pi - i\sin\frac{2(j-1)(k-1)}{n}\pi \right] \\
&= \frac{1}{\sqrt{n}}\left[ \cos\left(-\frac{2(j-1)(k-1)}{n}\pi\right) + i\sin\left(-\frac{2(j-1)(k-1)}{n}\pi\right) \right] \\
&= \frac{1}{\sqrt{n}}e^{-\frac{2(j-1)(k-1)}{n}\pi i}
\end{align}
$$

Then, for the entries of $Q' = QQ^*$:
$$
\begin{align}
q'_{jk} &= \displaystyle \sum_{s = 1}^n q_{js}q^*_{sk} \\
&= \displaystyle \sum_{s = 1}^n \frac{1}{\sqrt{n}}e^{\frac{2(j-1)(s-1)}{n}\pi i}\frac{1}{\sqrt{n}}e^{-\frac{2(s-1)(k-1)}{n}\pi i} \\
&= {1 \over n} \displaystyle \sum_{s = 1}^n e^{(s - 1)(j - k)\frac{2 \pi}{n}i} \\
&= {1 \over n} \displaystyle \sum_{s = 1}^n e^{s(j - k)\frac{2 \pi}{n}i}e^{-(j - k)\frac{2 \pi}{n}i} \\
&= {e^{-(j - k)\frac{2 \pi}{n}i} \over n} \displaystyle \sum_{s = 1}^n \left[e^{(j - k)\frac{2 \pi}{n}i}\right]^s
\end{align}
$$

+ If $j = k$, then the formula becomes:
$$
{e^0 \over n} \displaystyle \sum_{s = 1}^{n} e^0 = 1
$$

+ If $j \neq k$, since $0 < j - k \leq n$, $e^{(j - k)\frac{2 \pi}{n}i}$ is the $n^\text{th}$ root of unity, having the following property results in the result $0$
$$
\displaystyle \sum_{s = 1}^n \left[e^{(j - k)\frac{2 \pi}{n}i}\right]^s = 0
$$

In conclusion, the product matrix has $1$ in diagonol positions and $0$ in others, implies $QQ^* = I$, thus $Q$ is unitary.

### (5) Prove that the general form of a $2 \times 2$ unitary matrix with determinant $1$, where $\psi, \theta \in \mathbb{R}$, is:$$\bmat{e^{i\psi}\cos \theta & e^{i\psi}\sin \theta \\ -e^{-i\psi}\sin \theta & e^{-i\psi}\cos \theta}$$

### (6) Find the general form of a $2 \times 2$ unitary matrix.
A $2 \times 2$ complex matrix can represented as, where $a, b, c, d \in \mathbb{C}$:
$$
M = \bmat{
a & b \\ c & d
}
$$

If $M$ is unitary, it implies that:
$$
\inv{M} = M^* \Rightarrow \frac{1}{\det M}\,\bmat{d & -b \\ -c & a} = \bmat{\overline{a} & \overline{c} \\ \overline{b} & \overline{d}} \Rightarrow \begin{cases}
d = \overline{a} \det M \\
a = \overline{d} \det M \\[1ex]
-b = \overline{c} \det M \\
-c = \overline{b} \det M
\end{cases}
$$

The first two equations implies $d = d \cdot \overline{\det M} \cdot \det M = d \cdot \left\| \det M \right\|^2$;
+ If $d \neq 0$, then $\left\| \det M \right\|^2 = 1 \Rightarrow \det M = e^{i\theta}$ where $\theta \in \mathbb{R}$.
+ If $d = 0$, since $\inv{M}$ exists, proving $b, c \neq 0$, then $b = b \cdot \overline{\det M} \cdot \det M$ also implies $\det M = e^{i\theta}$ where $\theta \in \mathbb{R}$.

Substuting $\det M = e^{i\theta}$ back to the equation we get:
$$
\begin{cases}
d = e^{i \theta}\overline{a} \\
c = -e^{i \theta}\overline{b}
\end{cases}
$$

Hence the general form of a $2 \times 2$ unitary matrix is, where $a, b \in \mathbb{C}, \theta \in \mathbb{R}$:
$$
\bmat{
a & b \\
-e^{i \theta}\overline{b} & e^{i \theta}\overline{a}
}
$$