Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The problem is given all the inner products of N N-dimensional vector, can we work out the vector set?

For example given a N-by-N matrix $K$

$\{K\}_{ij} = v_i^Tv_j$ for all $i,j \in \{1,...,N\}$

Is it possible to find all $v_i$?

And will it be easier if we add norm constrain to the problem such that $||v_i||_2=1$?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The answer in general is negative. Essentially, if $v_i^T v_j=K_{ij}$, then for any real orthogonal matrix $Q$, we also have $w_i^Tw_j=K_{ij}$, where $w_i=Qv_i$. Therefore, if you have found a set of vectors $\{v_i\}$ that produces matrix $K$, any set of vectors $\{w_i\}$ obtained by applying rotations and/or reflections on $\{v_i\}$ is also able to generate $K$. In other words, $\{v_i\}$ is not unique. Since $\|w_i\|=\|Qv_i\|=\|v_i\|$, your additional constraint that each $v_i$ is a unit vector will not make the problem conceptually easier.

If you just want pick a set of vectors that can generate $K$, you may simply perform a Cholesky decomposition $K=LL^T$. Then $V=L^T$ will be a solution and the vectors $v_i$s are the columns of $V$. Cholesky decomposition has been implemented in many software libraries and computer algebra systems.

share|improve this answer

No. You are essentially asking for a kind of matrix "square root": given a matrix $K$, find a matrix $V$ (whose columns are your $v_i$) such that $V^TV=K$. This $V$ is generally not unique; multiplying $V$ on the left by any orthogonal matrix is still a solution. In fact, the only matrix $K$ with a unique $V$ is the zero matrix.

Requiring $v_i$ to be unit does not help; for $K=I$, any orthogonal matrix $V$ has this property.

If you only care about finding a $V$, even if it isn't unique, then a solution exists if and only if $K$ is positive semi-definite, in which case one solution is given by the Cholesky decomposition.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.