# Find a matrix such that $XX'= I$

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

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

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

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 $YX' = I$.