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It is obvious that elementary row operations will not change the dimension of the row space of a matrix. But is there an easy way to understand that it also will not change the dimension of the column space? For example, I can understand that swapping two rows has no effect, because this is just to switch the order of two basis vectors in the column space. What about the other two operations?

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  • $\begingroup$ Any thoughts on my answer, Kyson? $\endgroup$ – Gerry Myerson Feb 15 '15 at 23:52
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All you need to see is that elementary row operations do not change linear dependence relations among the columns; pick a set of columns, they will be linearly dependent after the row operation if and only if they were dependent before the row operation. And the way to see this is to note that when you multiply the matrix by a vector, you get a linear combination of the columns of the matrix, the coefficients in the linear combination being the entries of the vector. So a linear dependence among the columns corresponds to an element of the nullspace, and row operations don't affect the nullspace (that's the idea behind using row operations to solve a system of equations).

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  • $\begingroup$ Dear Gerry Myerson. Thank you for the answer. But how to show that the rank of the Column space will not change by row operations? $\endgroup$ – velut luna Feb 16 '15 at 10:33
  • $\begingroup$ Those operations do not introduce any new linear dependencies among the columns, nor do they get rid of any linear dependencies among the columns, so they can't change the dimension of the column space. If you know all the linear dependence relations that hold among a collection of vectors, then you can calculate the dimension of the space spanned by those vectors. $\endgroup$ – Gerry Myerson Feb 16 '15 at 11:34
  • $\begingroup$ I need some time to digest the whole thing. I am still confused about the meaning of the rows vectors. For the column vectors, the meaning is clear. We have a linear transformation from $F^n$ to $F^m$. The column vectors are just the images of the standard basis vectors of $F^n$. Do the row vectors have simple meaning of this kind? $\endgroup$ – velut luna Feb 17 '15 at 14:03
  • $\begingroup$ The row vectors form a spanning set for the orthogonal complement of the kernel of the linear transformation. Row operations don't change the row space, so they don't change the dimension of the row space, so they don't change the dimension of the kernel, so they don't change the dimension of the image, so they don't change the dimension of the column space. $\endgroup$ – Gerry Myerson Feb 17 '15 at 22:44
  • $\begingroup$ How do you define orthogonal complement? And how to show that the row vectors span this space? $\endgroup$ – velut luna Feb 18 '15 at 1:24

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