Basis and rank of a matrix My professor assigned this question and I am a bit confused by the wording of it.
Find (a) a basis for the row space and (b) rank of the matrix:
\begin{bmatrix}2&5\\-2&-5\\-6&-15\end{bmatrix} 
There is only one pivot row, r1 so the rank is one. What I don't understand is finding a basis for the row space.
- There isn't a basis for the row spaces in the vector space of M3,2 matrices because there is only one pivot row.
- Since there is only one pivot row, then the row space can only be a basis for M1,2 right?  
 A: Before we begin, let's define something first:


*

*Let the matrix in your question denoted as $A$.

*Let the row space of $A$ denoted as $RS(A)$.

*RREF is the abbreviation of reduced row echelon form, which is the result of Gauss-Jordan elimination.

*Given any tuple $(a,b)\in\mathbb R^2$:


*

*It can be seen as
$\left[
  \begin{array}{c}
   a\\ b
  \end{array}
 \right]\in\mathbb R^{2\times1}$

*It can be seen as
$\left[
  \begin{array}{r}
   a, b
  \end{array}
 \right]\in\mathbb R^{1\times2}$

*So let's just use $(a,b)\in\mathbb R^2$ to mean one of them that makes sense in the context.




["] There isn't a basis for the row spaces in the vector space of M3,2 matrices because there is only one pivot row [."]


*

*By definition of row space, it's the span of the rows of $A$, which is a subspace of $\mathbb R^2$(or $\mathbb R^{1\times2}$ as explained above) so it must has a basis.

*$\dim \mathbb R^2=2,$ and $\dim RS(A)=1$, no contradiction.

*$(2,5)$ is not a basis of $\mathbb R^2$.

*$(2,5)$ is a basis of $RS(A)$. 


["] Since there is only one pivot row, then the row space can only be a basis for M1,2 right? [."]


*

*$\textrm{One pivot row in RREF}$ $\iff$ $\textrm{rank}(A)=1$.

*Basis is a set of vector(s) that spans a (sub)space.

*Row space is a subspace, which means it contains $\mathbf 0$, so it cannot be a basis.


From the comment, you might also consider some questions/issues:


*

*Why turning $A$ into RREF doesn't change the $RS(A)$?

*Why you said the first row in RREF is $(1,2.5,0)\in\mathbb R^3$? It should still be in $\mathbb R^2$.

*The difference is that $(1,2.5)\in\mathbb R^2$ and $(1,2.5,0)\in\mathbb R^3$.

