Transposing Row Vectors For Solutions

I am learning about linear algebra and am solving span problems

the problems in the book give row vectors, and I usually place each row vector on top of each other to make a matrix, and then RREF to see if there exists solutions for a given span.

however, does it matter if I transpose the row vectors and make them as columns and find the solution with respect to the column space?

I think they should have the same solutions even if they are not symmetric matrices.

The problems in the book and in here show that given a row vector, they form the matrix by making them into columns. is this always necessary?

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If I understand the question correctly, the answer is no: row operations preserve the row space of a matrix, but can change the column space.

For example, if we have the matrix $$\begin{bmatrix} 1 & 1 \\ 2 & 2 \\ \end{bmatrix}$$ and apply the row operation $R_2 \gets R_2-2R_1$ we obtain $$\begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix}.$$ These two matrices have different column spaces.

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yes this is true. the row space is preserved, but for column space we have to revert back to the original matrix. given row vectors, how do I set them up to solve homogenous equation? Do i place them as rows, or place them as columns? –  arcolombo Feb 2 '14 at 4:19
for instance I have three vectors in R^{4}. do I create a a matrix 3 X 4, or do I transpose them into 4 X 3, augment it with a zero vector and reduce? I see examples where folks transpose the row vector into columns. not sure hwy. –  arcolombo Feb 2 '14 at 4:23
and do they have the same solution? –  arcolombo Feb 2 '14 at 4:24
Well, it depends on what you want to do with them: if you want to know if a vector $\mathbf{b}$ is in the span of vectors $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$ say, then we need to find a linear combination $a_1\mathbf{v}_1+a_2\mathbf{v}_2+a_3\mathbf{v}_3=\mathbf{b}$, in which case we should use them as columns $[\mathbf{v}_1\ \mathbf{v}_2\ \mathbf{v}_3]\mathbf{a}=\mathbf{b}$ and solve for $\mathbf{a}=(a_1,a_2,a_3)^T$. If we want to find a basis for the span, we might instead use them as rows of a matrix, and use the rows of the RREF as the basis. –  Rebecca J. Stones Feb 2 '14 at 4:28
So if I have $\alpha_1$ = (1,1,-2,1), $\alpha_2$ = (3,0,4,-1) and $\alpha_3$ = (1, 2, 3,4). My matrix would be the transpose of these vectors? –  arcolombo Feb 2 '14 at 5:01