Todai Entrance Exam: Math 2019 – Problem 1

(1)

    \[ AB(AB)^{*} = ABB^{*}A^{*} = AA^{*} = I \]

(2)

Prove that if F is unitary then G is orgthogonal:

As F^{*} = C^{T} - iD^{T}, if F is unitary:

    \[ FF^{*} = (C + iD)(C^{T} - iD^{T}) = CC^{T} + DD^{T} + i(DC^{T} - CD^{T}) = I \]

Then, CC^{T} + DD^{T} = I (because they are the real part) and DC^{T} - CD^{T} = 0 (because they are the imaginary part).

Therefore:

    \[ GG^{T} = \begin{pmatrix} C & -D \\  D & C \end{pmatrix}\begin{pmatrix} C^{T} & D^{T} \\  -D^{T} & C^{T} \end{pmatrix} = \begin{pmatrix} CC^{T} + DD^{T} & CD^{T} - DC^{T} \\  DC^{T} - CD^{T} & DD^{T} + CC^{T} \end{pmatrix} = \begin{pmatrix} I & 0 \\  0 & I \end{pmatrix} = I \]

So the statement follows.

Prove that if G is orthogonal then F is unitary:

If GG^{T} = I then as shown above, it must be the case that CC^{T} + DD^{T} = I and DC^{T} - CD^{T} = 0, which later leads to FF^{*} = I.

So the statement follows.

Therefore, G is orthogonal if and only if F is unitary.

(3) Thanks to this:

Let the matrix be A, we have:

    \[ \det(A - \lambda I) = 0 \Leftrightarrow \lambda(\lambda - 2)(\lambda + 2i) = 0 \Leftrightarrow \lambda \in \{ 0, 2, -2i \} \]

So the eigenvalues of the matrix are 0, 2, and -2i.

(4)

Let b_{jk} be the (j, k)-th element of the matrix B = QQ^{*}, q^{*}_{jk} = \exp \left ( -\frac{2\pi i(j-1)(k-1)}{n} \right ) be the (k, j)-th element of Q^{*}.

According to Euler’s formula: e^{ix} = \cos(x) + i\sin(x), let x = \frac{2\pi (j-1)(k-1)}{n}, so:

    \[ b_{jj} = \sum_{k = 1}^{n} q_{jk}q^{*}_{jk} = \sum{k = 1}^{n}\frac{1}{\sqrt{n}}(\cos(x) + i\sin(x))\frac{1}{\sqrt{n}}(\cos(x) - i\sin(x)) = \frac{1}{n}\sum_{k = 1}^{n}\cos^{2}(x) + \sin^{2}(x) = 1 \]

So the diagonal elements of B equals to 1. Now we prove that other elements b_{jm} (with j \neq m) of B equals 0:

    \[ b_{jm} = \sum_{k = 1}^{n} q_{jk}q^{*}{(j + m)k} = \frac{1}{n}\sum_{k = 1}^{n}\exp \left (\frac{2\pi i(j-1)(k-1)}{n} \right )\exp \left (-\frac{2\pi i(j+m-1)(k-1)}{n} \right )  \]

    \[ b_{jm} = \frac{1}{n}\sum_{k = 1}^{n}\exp \left (-m\frac{2\pi i (k-1)}{n} \right ) = \frac{1}{n} \sum_{k = 1}^{n} \cos \left (-m\frac{2\pi (k-1)}{n} \right ) + i\sin \left (-m\frac{2\pi (k-1)}{n} \right ) \]

    \[ b_{jm} = \frac{1}{n} \sum_{k = 1}^{n} \cos \left (m\frac{2\pi (k-1)}{n} \right ) - i\sin \left (m\frac{2\pi (k-1)}{n} \right ) \]

We observe that:

    \[ \sin(x) = -\sin(2m\pi - x) \]


    \[ \Leftrightarrow \sin \left(m\frac{2\pi (k-1)}{n} \right ) = -\sin \left( 2m\pi - m\frac{2\pi (k-1)}{n} \right ) \]


    \[ \Leftrightarrow \sin \left(m\frac{2\pi (k-1)}{n} \right ) = -\sin \left( m\frac{2\pi (n-k+ 1)}{n} \right ) \]

For k = 1, \sin(0) = 0. For 1 < k \leq n, there exists a k_{2} = n - k + 2 with \sin \left(m\frac{2\pi (k-1)}{n} \right ) = -\sin \left(m\frac{2\pi (k_{2}-1)}{n} \right ).

We also observe that:

    \[ \cos(x) = -\cos(m\pi + x) \]


    \[ \Leftrightarrow \cos \left(m\frac{2\pi (k-1)}{n} \right ) = -\cos \left( m\pi + m\frac{2\pi (k-1)}{n} \right ) \]


    \[ \Leftrightarrow \cos \left(m\frac{2\pi (k-1)}{n} \right ) = -\cos \left( m\frac{2\pi (\frac{n}{2}+k+ 1)}{n} \right ) \]

Working… A note here: What if n is odd? Then \frac{n}{2} will not be an integer number. And what if m is even?

(5)

    \[ H = \begin{pmatrix} e^{ia_{1}}a_{2} & e^{ib_{1}}b_{2}\\  e^{ic_{1}}c_{2} & e^{id_{1}}d_{2} \end{pmatrix} \]

    \[ HH^{*} = \begin{pmatrix}  e^{ia_{1}}a_{2} & e^{ib_{1}}b_{2}\\   e^{ic_{1}}c_{2} & e^{id_{1}}d_{2}  \end{pmatrix}\begin{pmatrix}  e^{-ia_{1}}a_{2} & e^{-ic_{1}}c_{2}\\  e^{-ib_{1}}b_{2} & e^{-id_{1}}d_{2}  \end{pmatrix} \]

If H is unitary then HH^{*} = I:

    \[ HH^{*} = \begin{pmatrix}  a^{2}_{2} + b^{2}_{2} & e^{i(a_{1} - c_{1})}a_{2}c_{2} + e^{i(b_{1} - d_{1})}b_{2}d_{2}\\  e^{i(c_{1} - a_{1})}a_{2}c_{2} + e^{i(d_{1} - b_{1})}b_{2}d_{2} & c^{2}_{2} + d^{2}_{2}  \end{pmatrix} =  \begin{pmatrix} 1 & 0\\   0 & 1  \end{pmatrix} \]

    \[ \Leftrightarrow \begin{Bmatrix} a^{2}_{2} + b^{2}_{2} = 1 \Leftrightarrow a_{2} = \sin(x), b_{2} = \cos(x) & \vee  & a_{2} = \cos(x), b_{2} = \sin(x) & (5.1) \\  c^{2}_{2} + d^{2}_{2} = 1 \Leftrightarrow c_{2} = \sin(y), d_{2} = \cos(y) & \vee  & c_{2} = \cos(y), d_{2} = \sin(y) & (5.2) \\  e^{i(a_{1} - c_{1})}a_{2}c_{2} + e^{i(b_{1} - d_{1})}b_{2}d_{2} = 0 \Leftrightarrow e^{i(a_{1} - c_{1} - b_{1} + d_{1})} = -\frac{b_{2}d_{2}}{a_{2}c_{2}} & & & (5.3) \\  e^{i(c_{1} - a_{1})}a_{2}c_{2} + e^{i(d_{1} - b_{1})}b_{2}d_{2}  = 0 \Leftrightarrow e^{i(c_{1} - a_{1} - d_{1} + b_{1})} = -\frac{b_{2}d_{2}}{a_{2}c_{2}} & & & (5.4) \end{matrix} \]

    \[ (5.3), (5.4) \Rightarrow e^{i(a_{1} - c_{1} - b_{1} + d_{1})} = e^{i(c_{1} - a_{1} - d_{1} + b_{1})} \Leftrightarrow a_{1} - c_{1} - b_{1} + d_{1} = c_{1} - a_{1} - d_{1} + b_{1} \Leftrightarrow a_{1} + d_{1} = c_{1} + b_{1} \text{ } (5.5) \]

We also employ the fact that \det(H) = 1:

    \[ \det(H) = e^{i(a_{1} + d_{1})}a_{2}d_{2} - e^{i(c_{1} + b_{1})}c_{2}b_{2} = 1 \text{ } (5.6) \]

    \[ (5.5), (5.6) \Rightarrow e^{i(a_{1} + d_{1})}(a_{2}d_{2} - c_{2}b_{2}) = 1 \Leftrightarrow e^{i(a_{1} + d_{1})} = \frac{1}{a_{2}d_{2} - c_{2}b_{2}} \Leftrightarrow \cos(a_{1} + d_{1}) + i\sin(a_{1} + d_{1}) = \frac{1}{a_{2}d_{2} - c_{2}b_{2}} \text{ } (5.7) \]

    \[ \Rightarrow  a_{1} + d_{1} = k\pi \]

Without loss of generality, it is either the case a_{1} + d_{1} = 0 or a_{1} + d_{1} = \pi. (5.1), (5.2) and (5.7) suggests that a_{2} = d_{2}, c_{2} = - b_{2} in case a_{1} + d_{1} = 0 or a_{2} = -d_{2}, c_{2} = b_{2} in case a_{1} + d_{1} = \pi.

In case a_{1} + d_{1} = \pi:

    \[ H = \begin{pmatrix}  e^{ix}a_{2} & e^{ix}b_{2}\\   e^{i(\pi-x)}c_{2} & -e^{i(\pi-x)}d_{2}  \end{pmatrix} = \begin{pmatrix}  e^{ix}a_{2} & e^{ix}b_{2}\\   e^{i\pi}e^{-ix}c_{2} & -e^{i\pi}e^{-ix}d_{2}  \end{pmatrix}  \]

Remember: e^{i\pi} = \cos(\pi) + i\sin(\pi) = - 1. So:

    \[ H = \begin{pmatrix}  e^{ix}a_{2} & e^{ix}b_{2}\\   -e^{-ix}c_{2} & e^{-ix}d_{2}  \end{pmatrix}  \]

Therefore a unitary matrix of size 2 with determinant 1 has a form of:

    \[ H = \begin{pmatrix}  e^{i\varphi}\cos\theta & e^{i\varphi}\sin\theta\\   -e^{-i\varphi}\sin\theta & e^{-i\varphi}\cos\theta  \end{pmatrix} \]

So the statement follows.

(6)

From (5.1), (5.2), (5.5), we can derive the general form of a unitary matrix of size 2:

    \[ H = \begin{pmatrix}  e^{i\alpha}\cos\theta & e^{i\gamma}\sin\theta \\   e^{i(\alpha + \beta - \gamma)}\sin\varphi & e^{i\beta}\cos\varphi  \end{pmatrix} \]


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