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Assume we have the matrix equation:

$(U^{\top} A) (Q \Lambda Q^{\top}) (A^{\top} U) = \Sigma$

such that all matrices in this equation have real values and ($d > n$):

$U$ is $d \times d$ and is orthonormal

$A$ is $d \times n$

$Q$ is $n \times n$ and is orthonormal

$\Lambda$ is $n \times n$ and is diagonal (with only non-negative values)

$\Sigma$ is $d \times d$ and is diagonal (with only non-negative values)

Is there a way to extract the matrix $U^{\top} A$ if we know: $\Sigma$, $\Lambda$ and $Q$ but don't know $A$, $U$? (I think this can be phrased more "mathematically" as a question about a uniqueness of a solution, but not sure how.)

I wouldn't mind getting $U^{\top} A$ up to an invertible linear transformation, as long as this transformation is a function of $U$ only. What if $A$ contains $n$ orthonormal vectors of length $d$? I think this has an easier solution.

Here is a SOLUTION which I think is correct:

Here is my conclusion, I am not sure if it is completely correct.

Set $F = U^{\top} A Q$. Then we have an equation of the form:

$F \Lambda F^{\top} = \Sigma$ where $\Lambda$ and $\Sigma$ are diagonal. To solve for $F$, we note that a solution would be a diagonal matrix that equals:

$F = \Sigma^{1/2} \Lambda^{-1/2}$. (I am assuming $d = n$ for simplicity, this can be easily generalized to any $d > n$).

So we have:

$U^{\top} A Q = \Sigma^{1/2} \Lambda^{-1/2}$ and therefore,

$U^{\top} A = \Sigma^{1/2} \Lambda^{-1/2} Q^{\top}$.

The solution, I am guessing is very much not unique (for $U^{\top} A$).

I just found a specific $F$ for which it works, there could be others.

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Sketch of an answer: If you define $U' = U^TAQ$ and $V'=V^TAQ$ then your equation becomes $\Sigma=U'\Lambda V'^T$ which describes the singular value decomposition of $\Sigma$. So you perform the SVD to get $U'$, $\Lambda'$ and $V'$ and then you have $U^TA = U'Q^{-1}$ and similarly for $V^TA$. Unfortunately the svd is only defined up to row and column permutations, so there's one additional step - you should permute the rows and columns of $\Lambda'$ to match those entries in $\Lambda$, and then do the equivalent permuations on $U'$ and $V'$ before hitting them with $Q^{-1}$ on the left. –  Chris Taylor Feb 1 '12 at 12:43
    
@Chris, I am afraid this is not true - $U'$ and $V'$ don't have to be unitary, so $\Sigma = U' \Lambda V'^{\top}$ is not an SVD of $\Sigma$. In addition, $\Sigma$ is diagonal, so its SVD is trivial ($I\Sigma I$). I edited the question a bit, it is actually simpler now (no $V$). –  harmonic Feb 1 '12 at 13:02
    
Oops, didn't notice that $\Sigma$ was diagonal! –  Chris Taylor Feb 1 '12 at 13:12

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