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

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

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  • $\begingroup$ Why did you suppose there is a $X$ matrix? $\endgroup$ – Hugocito Jun 13 '18 at 11:45
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $YX' = I$.

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