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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Why did you suppose there is a $X$ matrix? – Hugo Jun 13 '18 at 11:45
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3Sadly, 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. – hardmath Jun 13 '18 at 11:45
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2Title and question do not match. – Rodrigo de Azevedo Jun 13 '18 at 11:45
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If $X'$ means transpose a matrix having orthonormal rows will have the property that $XX'=I$ – N8tron Jun 13 '18 at 11:46
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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. – 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 $YX' = I$.
JohnKnoxV
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