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I have got a question and I would appreciate if one could help.

I start with an example to explain what I am looking for. Assume a scaled unitary matrix like

$U_2 = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $.

For this $2 \times 2$ matrix, I know $U_2 U_2^H = 2 I$ where $I$ is identity matrix. Moreover, $|U_2(i,j)|=1$.

I am looking for scaled unitary matrices with the same properties for arbitrary $n$; i.e., assuming arbitrary $n$, is there any method to define a $U_n$ such that $U_n U_n^H = n I$ and $|U_n(i,j)|=1$.

I am stuck with this problem for sometime now. Any hint/help would greatly be appreciated.

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  • $\begingroup$ Whats H hermitian ... $\endgroup$ – Archis Welankar Jan 25 '16 at 14:08
  • $\begingroup$ that is hermitian or conjugate transpose $\endgroup$ – M.X Jan 25 '16 at 14:09
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Let the entry in row $j$, column $k$ be

$$ u_{jk} = e^{2\pi i j k/n}$$

This is related to the properties of the finite Fourier transform.

If you want your $u_{jk}$ to be real, you're looking for Hadamard matrices. These can only exist if $n$ is $1$, $2$ or a multiple of $4$. The existence of such matrices for all multiples of $4$ is the Hadamard conjecture.

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  • $\begingroup$ Thank you very much @Robert $\endgroup$ – M.X Jan 25 '16 at 18:50
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Unitary matrices can be created from anti-Hermitian matrices via exponentiation, just like Rotations can be generated from anti-symmetric matrices via exponentialation (this is just the complex extension).

So if you want to generate a unitary matrix then write down some Hermitian matrixand do it's matrix exp,

$U(H,t) \equiv \exp(Ht)$

If you want to make the determinant non-zero then you add a multiple of the identity matrix to your exponent,

$U(H,t,\lambda) \equiv \exp(\lambda \mathbb{I} + Ht) = \exp(\lambda \mathbb{I})\exp(Ht) = \exp(\lambda)U(H,t)$

[edit I've noticed your columns are the other way around. What follows applies to the case where the -1 is on the off diagonal. You've just an extra factor of H overall]

As a simple example you can work with the SO(2) example you've given (SO(2) being equivalent to U(1)). If you consider the 2x2 matrix $H = \pmatrix{ 0 & 1 \\ -1 & 0 }$ you'll find $H^{2} = -\mathbb{I}$ and it follows that $\exp(Ht) = \pmatrix{\cos t & \sin t \\ -\sin t & \cos t}$. Put in $t = \frac{\pi}{4}$ and you have the unit determinat version of your matrix U. To scale it up to have det $\sqrt{2}$ you need to to consider $U = \exp(\log(\sqrt{2})\mathbb{I} + \frac{\pi}{4}H)$. That'll give you your matrix U.

The extension to larger dimensions works in exactly this manner. Scaling by an overall factor is than "adding a constant" to the exponential. It's a straightforward thing since the Identity commutes with everything, unlike general matrices.

FYI this is the basis of Lie theory - you can generate rotations, unitary groups, any "smooth" group, by exponentiating up a linear combination of appropriate matrices. For rotations they are anti-symmetric and for unitary they are anti-hermitian. Some definitions pull out a factor of i so that $iH$ is anti-Hermitian (so $H$ is Hermitian) but that's a matter of preference.

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  • $\begingroup$ Thank you very much @AlphaNumeric . I am just going to try it for larger dimensions. $\endgroup$ – M.X Jan 25 '16 at 14:36
  • $\begingroup$ Just one point. By $|U_n(i,j)|=1$, I mean the absolute values of the entries of $U_n$ should be equal to $1$. I tried this for n=3 but I can't fulfil this requirement. I would appreciate if you could help me with $n=3$ case too. $\endgroup$ – M.X Jan 25 '16 at 18:25
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    $\begingroup$ Ah, okay, you want all the elements to have unit modulus. You'll have to complexify things because what you're asking for cannot be done for arbitrary n using only Reals. As the answer above mine mentions, it CAN be done for n being a power of 2, see the Hadamard Matrices : en.wikipedia.org/wiki/Hadamard_matrix and their iterative construction. A specific complex matrix is the Fourier transform, as all its elements are complex phases and it is overall a unitary operator on L2. Is the Fourier Transform matrix enough or are you wanting to generate familes of such matrices? $\endgroup$ – AlphaNumeric Jan 26 '16 at 13:58
  • $\begingroup$ I can continue my work with Fourier transform matrix but is there a general way to generate such a matrix? a general form of a matrix would definitely be better $\endgroup$ – M.X Jan 27 '16 at 11:50

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