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I need you to do just what any math genuis in a shallow Hollywood movie does: looking at big tables of numbers and seeing exact structure!

These $3 \times 3$ matrices are solutions to a well-posed optimization problem in three-dimensional projective geometry (see below). So I expect the solutions to be rational expressions of integers, $\sqrt{2}$, $\sqrt{3}$, and the likes, but I cannot exclude crazier stuff like transcendental numbers.

I am particularly (but not only) interested in a closed-form expression of the value $\color{red}{0.9595767}$, which occurs six times in $\mathbf{A}$. One can see that the numbers are accurate to the sixth or seventh digit.

A = -0.959576712     0.959576694    -0.959576722
     0.959576703     0.484117298     0.959576704
     0.959576691    -0.080846583    -0.484117295

U =  0.766455562    -0.598872401    -0.232158820
     0.233804715    -0.076515828     0.969268117
    -0.598231749    -0.797180766     0.081373209

V = -0.766455553     0.232158826    -0.598872410
    -0.632310726    -0.108948904     0.767015828
     0.112823001     0.966556990     0.230301009

Background: Matrices $\mathbf{U}$ and $\mathbf{V}$ represent bases, so they are orthogonal, i.e. $\mathbf{U}^\text{T}\mathbf{U} = \mathbf{V}^\text{T}\mathbf{V} = \mathbf{I}$. The matrices are related by $$\mathbf{A} = \mathbf{U}^\text{T} \left[ \begin{array}{ccc} 2 && \\ & -1 & \\ && -1 \end{array} \right] \mathbf{V} \ .$$ The given numerical values are the solutions to the minimization problem $$\text{argmin}_{\mathbf{U},\mathbf{V}} \left(\max_{i,j} |A_{i,j}| \right) \ .$$

Thereby, $\max_{i,j} |A_{i,j}| \in [0.9595767, \ 2]$. If you think you can solve this riddle directly instead of interpreting the numerical values, then by all means go ahead :-)

(Simple tests for $\sqrt{2}$ and $\sqrt{3}$ didn't yield anything useful. I looked at the azimuth and polar angles of the vectors in $\mathbf{U}$ and $\mathbf{V}$ and didn't find anything meaningful either.)


EDIT:

Here is what I worked out from the answers: Proper order and signs of the base vectors give

$$A = \left[\begin{array}{rrr}a&a&a \\a&a&b \\a&b&c \end{array}\right]$$.

  • The eigenvalues of $\mathbf{A}$ are $2$, $1$, $-1$. Since the trace is the sum of eigenvalues, we get $$2a+c = 2 \ .$$
  • $\text{det}(\mathbf{A})$ is the product of eigenvalues, so $\text{det}(\mathbf{A}) = -2$. Expanding the determinant gives $$a(a-b)^2 = 2 \ .$$
  • The eigenvalues of $\mathbf{A}^2$ are $4$, $1$, $1$. Again, the trace is the sum of eigenvectors, giving $$6a^2+2b^2+c^2 = 6 \ .$$

This can be rewritten to finding the root $$\left( 6a^3 - 4a^2 - a + 2 \right)^2 - 8a^3 = 0 \ .$$

The largest real solution is $a=0.9595767$, and in consequence $b=-0.4841173$ and $c=0.0808466$.


Here is a nicer version of the values, with the unnecessary degree of freedom fixed and proper ordering:

A =
   0.959576   0.959576   0.959576
   0.959576   0.959576  -0.484117
   0.959576  -0.484117   0.080846

U =
   0.766455   0.598872   0.232158
   0          0.361450  -0.932391
   0.642297  -0.714636  -0.277035

V =
   0.766455   0.598872   0.232158
   0.547861  -0.798226   0.250365
   0.335252  -0.064702  -0.939904
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    $\begingroup$ Your stated relation between $\mathbf{U}$ and $\mathbf{V}$ mentions $\mathbf{A}$, $\mathbf{U}$ and $\mathbf{R}$ but not $\mathbf{V}$ $\endgroup$ – Henry Feb 10 '16 at 12:24
  • $\begingroup$ @Henry Thanks, I intermingle different notations from my notes. Fixed. $\endgroup$ – GDumphart Feb 10 '16 at 12:29
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    $\begingroup$ @GDumphart: Check out the closed forms using WA - make sure to click More to see them all. Also, do you know even more digits - which would be better? $\endgroup$ – Moo Feb 10 '16 at 13:26
  • $\begingroup$ Side note: by "rational expressions of integers" you probably mean algebraic (which is the opposite of transcendental). $\endgroup$ – barak manos Feb 10 '16 at 13:36
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    $\begingroup$ The three numbers in $A$ satisfy $6a^2+2b^2+c^2=6=2^2+(-1)^2+(-1)^2$ $\endgroup$ – Empy2 Feb 10 '16 at 13:37
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You can tweak $U$ and $V$ so that $A$ becomes $\left[\begin{array}{ccc}a&a&a\\a&a&b\\a&b&c\end{array}\right]$. $(a=0.9595767,b=-0.4841173,c=0.0808466)$
Then $A^TA=A^2$ has eigenvalues $1,1,4$, so $A$ has eigenvalues $\pm1,\pm1,\pm2$
The choice $1,-1,2$ gives $2a+c=2,a(a-b)^2=2$ and I think: $2ac-b^2-a^2=-1$.
I think it turns into a degree-6 polynomial in $a$.

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  • $\begingroup$ Great stuff, I'll look at the polynomials in detail later. Thank you so far. $\endgroup$ – GDumphart Feb 10 '16 at 17:54
  • $\begingroup$ Alright, I worked out your solution and appended that to my question. Thanks! $\endgroup$ – GDumphart Feb 11 '16 at 13:29
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This is a svd decomposition, and it's not unique, because two singular values are equal. That means that U and V are not unique, and probably not in the most symmetric form. The same degeneracy may be present in the solution itself, although the absolute value in the condition on component may then push this into an awkward angle. The fact that the same number is seen everywhere is quite obvious actually, as you are "squashing" the system until it reaches the limits in as many components as possible, even if there's some wiggle room left in the degenerate components.

When I do the SVD transform myself (rounded to 6 places on purpose, so that we reinforce the degeneracy), I get a bit different result for the U and V matrices, as the degeneracy makes the result arbitrary. U and V that I get are almost the same to each other, just with different signs and exchanged rows. Here I have $A=USV^T$ (matlab uses this convention, as do most textbooks).

U =
  -0.76646   0.31402  -0.56030
   0.59887   0.66469  -0.44669
   0.23216  -0.67792  -0.69751

V =
   0.76646  -0.31402  -0.56030
  -0.23216   0.67792  -0.69751
   0.59887   0.66469   0.44669

In this case, S=diag(2,1,1) (singular values are positive and the signs go to V).

We can probably assume, that $P=UV=\begin{bmatrix}-1&0&0\\0&0&-1\\0&-1&0\end{bmatrix}$ in the exact solution. Unless the tiny difference is what pushes the components of A from 1 to the given number - but we'll see about that.

This makes it all quite a bit easier. Using these empirically obtained constraints (which make sense due to symmetry), you only have 3 variables left - for instance, the euler angles of the matrix U (or V). You may not even have to look for an actual minimum, because two constraints can be guessed from A: all three components must be equal by magnitude in the first row.

Using the above condition, we now have: $$A=USV^T=U(SP)(PV^T)=U(SP)U$$ where $SP=\begin{bmatrix}-2&0&0\\0&0&-1\\0&-1&0\end{bmatrix}$ and $U$ is the orthogonal matrix to look for. I'm surprised at this form: it's mixing dimensions, the last U isn't transposed, so it's hard to see geometric meaning of this.

Now, more speculation - I don't know if this is justified. Using symmetry, $U$ must be a rotational matrix that rotates about an axis with last two components in equal measures (rotation about $z$ will maximize, not minimize). Let's try to rotate about (0,1,±1). This gives (I think, It's quite ugly) $$U=\begin{bmatrix}\cos\alpha & \sin\alpha/\sqrt{2}&-\sin\alpha/\sqrt{2}\\ -\sin\alpha/\sqrt{2}&(1+\cos\alpha)/2 & \pm (1-\cos\alpha)/2 \\ \sin\alpha/\sqrt{2} & \pm (1-\cos\alpha)/2 & (1+\cos\alpha)/2 \end{bmatrix}$$ This could lead to equations for $\alpha$.

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  • $\begingroup$ Great stuff! I was aware of the SVD and equal value aspects, I fooled around with these matrices a lot for other reasons and for much easier problems. I'll study your answer thoroughly and calmly later or tomorrow, but thank you very much so far. $\endgroup$ – GDumphart Feb 10 '16 at 17:58

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