Take a $3\times5$ matrix $A$ then we have that $rank(A)\leq3$. More specifically we have that when we think about the rows and columns of $A$ as vectors then any collection of more than three $3$‐vectors is automatically dependent.
Why is this true?
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$\begingroup$ row rank $A$ = column rank $A$ $\endgroup$– hardmathMar 12, 2015 at 23:25
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$\begingroup$ Take four 3-vectors and form the matrix, corresponding to a system of equations for a linear combination of them to be 0. There aren't enough rows for each column to have a pivot, so there must be a free variable. (Write out an example and you'll probably see it more clearly.) $\endgroup$– John BrevikMar 12, 2015 at 23:26
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It's true because the rank is the dimention of the image of the linear function represented by the matrix A. As the image is contained in a space of dimension 3, the rank of A cannot be bigger than 3
