Krdytkyu

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&-15end{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 M_3,2 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 M_1,2 right?

Before we begin, let's define something first:

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


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


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


4. 
   

   Given any tuple $(a,b)inmathbb R^2$:

   
   
   
   
   
   • It can be seen as
     $left[
     begin{array}{c}
     a\ b
     end{array}
     right]inmathbb R^{2times1}$
     
   
   
   • It can be seen as
     $left[
     begin{array}{r}
     a, b
     end{array}
     right]inmathbb R^{1times2}$
     
   
   
   • So let's just use $(a,b)inmathbb R^2$ 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 $mathbb R^{1times2}$ 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 $(1,2.5,0)inmathbb R^3$? It should still be in $mathbb R^2$.


• The difference is that $(1,2.5)inmathbb R^2$ and $(1,2.5,0)inmathbb R^3$.