Finding a basis for a spanning list by columns vs. by rows
From this answer, I see that, given a list of vectors, we can put them as the columns of a matrix, then reduce it, and the pivots will tell us which of the original vectors form a basis for the span of those vectors.
However, I'm wondering, can't we just put the list of vectors as the rows of a matrix, then reduce it, and find a basis this way? Since row operations don't change the row space, this method should also work. We just look at the rows that have a leading variable.
Though, I can see that the first method of putting the vectors in a column might be useful if we want to find out which of the original vectors form a basis.
linear-algebra matrices
twosigma
twosigmatwosigma
[I'm sure that I've answered this before, but can't find anything relevant at the moment.]
Yes, both methods will get you a basis for the span of the vectors, and your observation is correct: writing them as columns of a matrix will let you find a linearly-independent subset of the original vectors that has the same span. On the other hand, writing them as rows generally gives you a "nicer" basis: The first part of each vector in the basis will consists of a bunch of zeros with a single one somewhere. Which one is preferable depends entirely on what you're then going to do with that basis.
amdamd
$\begingroup$ Your answer was the one I linked :) $\endgroup$
– twosigma
Consider the matrix $A$ whose columns are the vectors $v_1, \dots, v_n$ from the $m$-dimensional vector space $V.$ We have that $A^t$ (the transpose of $A$) is the matrix whose rows are the vectors $v_1, \dots, v_n.$
Like you mentioned, we can determine which of the vectors $v_1, \dots, v_n$ are linearly independent by putting the matrix $A$ in reduced row-echelon form. Explicitly, the vectors $v_j$ for each column $j$ with a pivot are linearly independent, and the rest of the vectors can be written as a linear combination of the $v_j.$ (In fact, the coefficients of the relation of linear dependence are determined by the entries of the column of the corresponding vector. Check that for $v_1 = (1, 2, 5)$ and $v_2 = (-3, -5, -13),$ we have that $v_4 = (2, 1, 4) = -7v_1 - 3v_2.$ Compare this to the problem you mention.)
Like you have seen, there are other advantages of writing the vectors as the columns; however, it is true that the column rank and the row rank of any matrix are equal, hence the number of pivots of the matrix $A^t$ is the same as the number of pivots of the matrix $A,$ and one can just as legitimately row-reduce $A^t.$ Unfortunately, the interpretation of the coefficients of the matrix $A^t$ is not as clear as in the other method. (Check that the transpose of the matrix in the problem you mention is $$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix};$$ however, the coefficients of this matrix have no immediate meaning.)
Dylan C. BeckDylan C. Beck
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