# Matrix completion when the eigenvectors are a tensor product?

Suppose we have incomplete observations of the square matrix $X$. Most matrix completion algorithms assume the matrix is low-rank. What if instead we assume the matrix of eigenvectors is a tensor product: $V = A\otimes B$, $X=VDV^{-1}$? This still dramatically reduces the space of possible matrices, so it may be possible to exactly recover a matrix of this form. How would we solve this algorithmically?

• If $V$ is the tensor product of two vectors $A,B$ (I suppose this is what you mean), then it has rank 2 and $V^{-1}$ in your formula does not exist unless you are only considering $2\times 2$ matrices. – Wolfgang Bangerth Feb 8 '14 at 13:05
• No, $A$ and $B$ can be full matrices. For instance, the 2D Fourier basis is the tensor product of two 1D Fourier bases, both of which are invertible matrices. – David P Feb 9 '14 at 2:38
• Do you mean V is the kroneker product of matrices A and B ? – sebas Feb 9 '14 at 9:26