0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – sophie-germain Feb 2 '14 at 4:19
  • $\begingroup$ 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. $\endgroup$ – sophie-germain Feb 2 '14 at 4:23
  • $\begingroup$ and do they have the same solution? $\endgroup$ – sophie-germain Feb 2 '14 at 4:24
  • $\begingroup$ 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. $\endgroup$ – Rebecca J. Stones Feb 2 '14 at 4:28
  • $\begingroup$ 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? $\endgroup$ – sophie-germain Feb 2 '14 at 5:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.