Suppose that $X$ and $Y$ are both $n \times k$ matrices with $n > k$.

Are there any matrices that satisfy $YX' = I_n$?

  • $\begingroup$ Why did you suppose there is a $X$ matrix? $\endgroup$ – Hugocito Jun 13 '18 at 11:45
  • 3
    $\begingroup$ Sadly, no there are not. The rank of product $YX'$ is at most the rank of $X'$, which is at most $k\lt n$, while $I_n$ has rank $n$. This has been brought out in many previous Questions. $\endgroup$ – hardmath Jun 13 '18 at 11:45
  • 2
    $\begingroup$ Title and question do not match. $\endgroup$ – Rodrigo de Azevedo Jun 13 '18 at 11:45
  • $\begingroup$ If $X'$ means transpose a matrix having orthonormal rows will have the property that $XX'=I$ $\endgroup$ – N8tron Jun 13 '18 at 11:46
  • $\begingroup$ See for example Inverse of non-square matrix and invertible matrix is a square matrix, as well as this Question about rank of a product. $\endgroup$ – hardmath Jun 13 '18 at 11:59

The rank of $YX'$ is less than the rank of $Y$ and the rank of $X$, both of which are at most $k$. The identity has rank $n >k$ which would violate $YX' = I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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