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My professor was hinting this was going to be on the exam, but wasn't telling us if this is true.

I do believe in fact it is true though, because both the rowspace and column space are determined by the number of leading 1's in the row-reduced matrix. Therefore, if there is only 1 row with a leading 1, then both the rowspace and columnspace will contain only 1 vector each, thus making each of their dimensions 1.

Is this logic and the original statement correct? Thanks

Johnathon Svenkat
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  • Yes it does. It's called the "rank" of the matrix. Do you have a textbook? Look in there. – Stefan Smith May 06 '13 at 01:38
  • Relevant: https://en.wikipedia.org/wiki/Rank_(linear_algebra)#Main_definitions – vadim123 May 06 '13 at 01:39
  • Yes, my textbook lists rank as "the number of nonzero rows in any row-echelon form of a matrix A". But what does this have to do with the rowspace and columnspace, as they depend on the number of rows with leading 1's, not just nonzero rows – Johnathon Svenkat May 06 '13 at 02:18
  • I'm slightly confused as to what you are trying to say, but it seems you may be confusing Gaussian elimination and Gauss-Jordan elimination? The rank of a matrix is equal to the number of non-zero rows when it is reduced to a triangular matrix. Not just the rows with leading 1's. – Upside May 06 '13 at 02:22

1 Answers1

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Yes, given a matrix $A$, the dimension of the row space of $A$ is equal to the dimension of the column space of $A$.

These are always equal to the rank of the matrix: which can also be defined as the number of nonzero rows of $A$ when $A$ is in row-echelon form.

See the equivalent definitions for the rank of a matrix.

amWhy
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