# All kinds of Row Space of a matrix

For example, determine a basis for the row space of $$A=\begin{pmatrix} 1& -1& 1& 3& 2\\ 2& -1& 1& 5& 1\\ 3& -1& 1& 7& 0\\ 0& 1& -1& -1& -3 \end{pmatrix}$$

Reduce A to the row-echelon form

$$\begin{pmatrix} 1& -1& 1& 3& 2\\ 0& 1& -1& -1& -3\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0 \end{pmatrix}$$

Therefore, the basis of row space of A is $(1,-1,1,3,2)$, and $(0,1,-1,-1,-3)$

Since any row-echelon form of A is a basis for its row space, so if we reduce A to the reduced row-echelon form

$$\begin{pmatrix} 1& 0& 0& 2& -1\\ 0& 1& -1& -1& -3\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0 \end{pmatrix}$$ so the basis of row space of A can also be $(1,0,0,2,-1)$, and $(0,1,-1,-1,-3)$

I have three questions

1.Does the original rows $(1,-1,1,3,2)$, and $(2,-1,1,5,1)$ in A also form a basis for row space? Because I think I read from somewhere that "the pivot columns do not, necessarily, form a basis for column space. However, the cor responding columns in the original matrix do", I 'm sure if the corresponding rows can also form a basis for row spaces

1. Why pivot columns do not, necessarily, form a basis for column space while pivot rows can forma a basis for row space?

2. How can I see the linear combinaton relationship in each rows in A without transposing A by making each row into column then do the usunal methods?

To see that the pivot columns don't work, just look at an example, e.g., $$A=\pmatrix{1&1\cr1&1\cr}$$ has column space generated by $(1,1)$, while the reduced form $$\pmatrix{1&1\cr0&0\cr}$$ has column space generated by $(1,0)$. Each type of elementary row operation preserves the row space, but not necessarily the column space (think about it!).