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I'm confused in Linear Algebra when finding the basis.

In my textbook there are two methods: Row space and Casting out

In the Row Space algorithm I form the Matrix whose rows are the given vectors, then I reduce it to echelon and my basis are the non zero rows:

In the Casting out method, I basically form the matrix whose columns are the given vectors. Reduce to echelon and my entries with pivots form the basis.

I would really appreciate if someone could tell me the difference between using the column and row interpretation (this part really blocks me in the subject), is it basically the same (the difference being how to interpret the echeloned matrix? If yes what is the point of having two distinct methods. And when do I know which one I need to use.

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  • $\begingroup$ When you row reduce a matrix to reduced row echelon form, the non-zero rows forms a basis for the rowspace and the pivot columns correspond to a basis for the columnspace. Your two methods serve two different purposes: the first finds a basis for the rowspace and the second finds a basis for the columnspace. In the first method, you are constructing a matrix with a rowspace spanned by the vectors and finding a basis for the rowspace. In the second method, you are constructing a method with a columnspace spanned by the vectors and finding a basis for the columnspace. $\endgroup$ – EuYu Dec 15 '13 at 13:01
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    $\begingroup$ thanks, but what is the difference between the row space and the column space? $\endgroup$ – user115919 Dec 15 '13 at 15:12
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Given a subspace $V$ of $\Bbb R^n$ spanned by $m$ vectors $v_1,\dots,v_m$, it is your choice whether to put the vectors in as the rows of an $m\times n$ matrix $A$ or as the columns of an $n\times m$ matrix $B$. You will have $V = \text{Row space}(A)$ and $V=\text{Column space}(B)$. Whichever algorithm you prefer to find a basis is up to you. (Of course, $B = A^\top$.)

Here is an effective difference. When you use the row space approach, the basis you give will consist of vectors perhaps not obviously related to your original $v_1,\dots,v_m$, whereas when you use the column space approach, you will be providing a subset of the original set of vectors as your basis. So, specifically, if you want to obtain a basis by discarding some of your original vectors and using the remaining ones, you will want the column space approach.

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  • $\begingroup$ Another difference (which Ted mentions but I wanted to emphasize more) is that in the row space approach, the pivot rows in the reduced row echelon form give a basis for the span of the rows of the original matrix, whereas this is not true in the column space approach: you have to go back and pick the columns in the corresponding positions as the pivot columns in the reduced row echelon form. $\endgroup$ – Pete L. Clark Dec 15 '13 at 22:05

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