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This question is connected with the previous one .

Suppose we know distinct $n$ eigenvalues $\lambda_1,\lambda_2,..\lambda_n$ for an unknown matrix $A_{{n}\times{n}}$ and dot products ${v_i}^T {v_j}$ for any pair of unit length eigenvectors ${v_i} , {v_j}$(they represent cosines of angles between these unit vectors. We can assume - if needed - that they are all non-negative).

Question:

  • how to reconstruct from these data any matrix $A$ with given properties?

Of course, there are plenty of such matrices and all are probably similar to each other, so we can choose a basis for a searched representation of the matrix - for example - the eigenvector $v_1$ might be equal $ [ 1 \ 0 \ 0 \ ... \ 0]^T$, other vectors should be calculated taking into account this starting point. As in the previous question it's relatively easy to calculate it for the dimension $n=2$. For higher dimensions problem seems to be more complicated and hard to deal with... but maybe some method exists..

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3 Answers 3

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Every nonsingular matrix has a unique polar decomposition. It follows that if $P$ is the unique positive definite matrix square root of $V^TV$, then $A=U(P\Lambda P^{-1})U^T$ for some real orthogonal matrix $U$. Since inner products are preserved under changes of orthonormal bases, there is not enough information to determine $U$ and you can only determine $A$ up to unitary equivalence.

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  • $\begingroup$ Up the unitary equivalence ? What does it mean ? What method would be for that? $\endgroup$
    – Widawensen
    Nov 22, 2016 at 10:56
  • $\begingroup$ @Widawensen It means you can find $P$ (from the inner products, form the Gramian matrix $V^TV$; then take its square root $P=\sqrt{V^TV}$) but not $U$. You see, if $v_1,\ldots,v_n$ are a set of feasible vectors and you rotate the whole system, the rotated vectors still give the same inner products. Therefore, unless further conditions are imposed, there is no way to solve $V$ uniquely. All you know is that if the columns of $A=P\Lambda P^{-1}$ is a solution, all other solutions are of the form $UAU^T$ for some unitary (i.e. real orthogonal) matrix $U$. $\endgroup$
    – user1551
    Nov 22, 2016 at 11:04
  • $\begingroup$ What additional conditions could be appropriate for unique solution? $\endgroup$
    – Widawensen
    Nov 22, 2016 at 11:10
  • $\begingroup$ @Widawensen I don't know. Are you sure you need uniqueness? $\endgroup$
    – user1551
    Nov 22, 2016 at 11:33
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    $\begingroup$ @Widawensen Then just take $A=(V^TV)^{1/2}\Lambda (V^TV)^{-1/2}$. $\endgroup$
    – user1551
    Nov 22, 2016 at 12:34
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Here's an alternative view: think of the endpoints of the unit eigenvectors as points on a sphere. Then the inverse cosines of the known dot-products determine their geodesic distances on the sphere. So suppose you have three points on a sphere with certain given distances --- one red, one blue, one green.

Rotate the sphere any way you like: the pairwise distances remain the same.

Now hold the sphere so that the red point is up. Rotate about the up-down axis...and the blue and grene points move, but the pairwise distances remain the same.

One possible constraint to make this unique:

Insist that $V_1$ lies in the $e_1$ direction, $V_2$ in the span of $e_1, e_2$, $V_3$ in $span\{e_1, e_2, e_3\}$, and so on. Then you've got uniqueness up to sign. If you say that $V_1$ is a positive multiple of $e_1$, and $$ V_2 = c_{21}e_1 + c_{22} e_2 $$ with $c_{22} > 0$, and similarly for later ones, then you have uniqueness, although there's no guarantee that $c_{nn}$ will actually turn out positive: the location of $V_n$ in general will be completely determined once you've constrained $V_1 \ldots V_{n-1}$ as described.

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  • $\begingroup$ Excellent. It seems to be a proper method. So we need a span and it is systematic approach. Thank you. Your both answers (as the answer of user1551) greatly extend my understanding of the problem. $\endgroup$
    – Widawensen
    Nov 22, 2016 at 11:44
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Let $c_{ij}$ denote $v_i^T v_j$.

Suppose you somehow found $A$. Then from that, you found the eigenvectors $v_i$, and made them the columns of a matrix $V$. Then you'd have

$$ AV = V \Lambda $$ where $\Lambda$ is a diagonal matrix with the $v_i$ on the diagonal. And $$ V^T V = C, $$ where $C$ is the matrix of the $c_{ij}$s.

Let $R$ be any rotation matrix, and let $W = R^TV$, so that $V = RW$. Then you'd have

$$ ARW = RW \Lambda $$ and $$ (RW)^T (RW) = C, $$ which simplifies to $$ (W^T R^T) (RW) = W^T R^T R W = W^T W = C. $$ In short: rotating the eigenvectors doesn't change their pairwise dot products.

And the first equation becomes $$ (R^TAR) W = W \Lambda $$

In short: if $A$ is a solution to your problem, so is $R^t A R$ for any rotation matrix $R$. So there's no way to find $A$ just from the data given.

If you fix $V_1$...then you can conjugate $A$ by any rotation of the plane orthogonal to $V_1$ to get a different solution. If you fix $V_1$ and $V_2$, you can conjugate by a rotation in the plane orthogonal to both, and so on. So only once you fix $V_1...V_{n-1}$ is the final $V$ determined.

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  • $\begingroup$ Assigning $V_1$ is not sufficient to start the procedure ? Let's take dimension $n=3$ for example. We starting from $v_1=[1 0 0]^T$ and we know that ${v_1}^T{v_2}= c_{12}$ and ${v_1}^T{v_3}= c_{13}$. Additionally we know $c_{23}$ . It's not enough data to determine $v_2$ and $v_3$ ? We can visualize the process of the unit sphere.. $\endgroup$
    – Widawensen
    Nov 22, 2016 at 11:07
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    $\begingroup$ Nope. For if $A$ is one solution, so is $R^t A R$, where $R$ is any rotation in the $yz$-plane, such as $y \to z, z \to -y$. $\endgroup$ Nov 22, 2016 at 11:11
  • $\begingroup$ Hmm, so we need some additional constraint.. but what could it be? $\endgroup$
    – Widawensen
    Nov 22, 2016 at 11:12
  • $\begingroup$ I'm trying to visualize process on the unit sphere. We are starting from the known $v_1= [ 1 0 0]^T$ Maybe additional constraint might be to assign 0 for the second or third component of $v_2$ vector ? The third vector would be calculated (almost) uniquely in this case... how to develop procedure into higher dimension it's hard to imagine because it is hard to imagine 4-d unit sphere. What do you think about this idea, John? $\endgroup$
    – Widawensen
    Nov 22, 2016 at 11:29
  • $\begingroup$ See my additional answer. $\endgroup$ Nov 22, 2016 at 11:38

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