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This question is from Mathematics for Economists by Simon and Blume,Page 775, Proof of theorem 27.5.

I am asking just my question. I know that I should give complete detail including definitions and theorems already proved, but it is so time consuming that I omit it and just give the link for the e-book. I hope that this e-book covers the relevant chapters.e-book here.

By the definition of the row echelon form, the last $n-k$ entries in each column of a rank $k$ row echelon $n\times m$ matrix $A_r$ are all zeros. It follows that column space of $A_r$,call it $Col(A_r)$, cannot have a dimension strictly greater than $k$.

Now, I can't understand how it follows that dimension can't be strictly greater than $k$.

(Note that row spaces are defined in this book before column spaces and nullspaces are defined after column spaces. Theorem committing $dimension(Col(A))=dimension(Row(A))=rank\text{ }A$ is based on theorem 27.5 as well as some previous theorems.)

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The result you cite is about the column space of the row echelon form $A_r$, so I edited the title to reflect this narrowing. – hardmath Nov 16 '13 at 17:17
up vote 1 down vote accepted

Let's recall some setup leading to the row echelon form.

The row echelon form of an $n \times m$ matrix has leading ones in every row that is not entirely zeros, and these leading ones are in different columns (with the rows ordered by the appearance of the leading ones according to column).

If the matrix had rank $k$, then it will have $k$ leading ones. This is because the elementary row operations that yield its row echelon form do not change the row space of the matrix (even though individual rows are being changed).

Now consider the columns of the row echelon form of the matrix. If the bottom $n-k$ rows are all zeros, then in each column of the row echelon matrix the nonzero entries appear only in the upper $k$ places. Thus the column space of the matrix is contained in the subspace of $\mathbb{R}^n$, here considered as $n \times 1 $ column vectors, in which the last $n-k$ entries are zeros. This last space has a basis of "standard vectors" in which $e_i$ for $i=1,\ldots,k$ has $1$ in the $i$th entry and zeros elsewhere.

Thus the dimension of the column space of the row echelon form is at most $k$ (since it's contained in a vector space of dimension $k$). In fact with a little more work you can use the fact that the columns which have leading ones are linearly independent (because those leading ones are in distinct rows) to show that the dimension of the row echelon form's column space is exactly $k$.

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Many many thanks for so great answer. – Silent Nov 17 '13 at 2:32
Just have one doubt. "Standard vectors" $e_i$s are from $\mathbb{R}^n$, with $n$ elements in each vector. How then we know that $k$ standard vectors form at most $k$-dimension?$(k≤n)$ – Silent Nov 17 '13 at 6:52
The special case described has $k$ vectors $(1,0,0,\ldots)^T,(0,1,0,\ldots)^T,\ldots,(0,\ldots,0,1,0,\ldots)^T$, and these are clearly linearly independent. By definition they are a basis for their span, a $k$-dimensional subspace of $\mathbb{R}^n$. – hardmath Nov 17 '13 at 11:28

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