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Is there a quick and easy way to show that a $n\times m$ Matrix (with $n<m$ ) has rank n ?

Is the determinant method in general the best way ? (I mean, if you find that the determinants of 2 $n\times n$ submatrizes, cannot be simultaneously 0.)

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    $\begingroup$ It normally is. Otherwise you could sometime (if lucky) spot some easy linear combination or a zero column or row. $\endgroup$ – user88595 Feb 5 '14 at 12:33
  • $\begingroup$ I'm not sure if I underastand the question correctly: Do you suppose your matrix to have full rank and want to show that it has rank $n$? Do you want to show that the matrix in general has at most rank $n$? Or do you want to test a given matrix on its rank? $\endgroup$ – Roland Feb 5 '14 at 12:48
  • $\begingroup$ It is easier to add columns one by one and see, whether the new column is in the span of previous columns. $\endgroup$ – user68061 Feb 5 '14 at 12:49
  • $\begingroup$ @Roland to show: $n\times m$-Matrix has rank min{n,m} $\endgroup$ – derivative Feb 5 '14 at 12:51
  • $\begingroup$ The most effective systematic way is the Gaussian elimination procedure. When the matrix is in reduced form you can compute the rank by counting pivots. $\endgroup$ – Giuseppe Negro Feb 5 '14 at 13:04
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If $X$ is $n \times m$ with $n<m$ , then consider $Y = X X ^t$ which is $n \times n$. See that $Y$ is nonsingular iff $X$ has rank $n$. So, you only need to compute the determinant of $Y$.

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You can use $${\rm rank\,}A = {\rm rank\,}A^t,$$

which means that the row rank of $A$ coincides with the row rank of $A^t$, i.e. the column rank of $A$. Thus, the rank of $A$ is the maximal number of linear independent rows of $A$, but also the maximal number of linear independent columns of $A$, which implies $${\rm rank\,}A \leq \min(n,m).$$

If $A$ is singular, equality will not be attained - consider the $n\times m$ matrix which is filled with zeros, which has rank $0$.

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