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For any matrix the column rank and row rank are equal. As I understand it rank means the number of linearly independent vectors, where vectors is either the rows or columns of the matrix.

This seems to mean that the number of linearly independent rows in a matrix is equal to the number of linearly independent columns? But my gut tells me this shouldn't be the case, at least intuitively I can't see why it would be the case.

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It is true.

There are many ways to show this.

One of the most constructive ones is to transform the matrix to its "echelon form", using elementary transformations which do not change the number of linearly independent rows or columns.

What you obtain in the end is a diagonal matrix, with ones followed by zero in the diagonal. Clearly, in such a matrix the number of linearly independent rows is the same with the number of linearly independent columns.

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  • $\begingroup$ so this also means only square matrices can be full rank? (or at least, a non sure matrix must have at least one linearly dependant row or column) $\endgroup$ – Jonathan. Sep 6 '15 at 15:45
  • $\begingroup$ If $A\in\mathbb R^{m\times n}$, then Rank$(A)\le\,$ min$\{m,n\}$. $\endgroup$ – Yiorgos S. Smyrlis Sep 7 '15 at 8:23
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For instance the rank of the matrix is the largest dimension of an invertible square submatrix. This criterion is independenty of whether you work with rows or with columns.

You also can say it is the size of the largest non-zero minor of the associated determinant.

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