# difference between bases for matrix and row/column space

it is taken from following link

http://www2.kenyon.edu/Depts/Math/Paquin/PracticeExam1Solns.pdf

where this matrix is reduced to row echelon form and is found both row space basis and column space basis,but for row space,we see that basis for row space is different basis from column space,also some lecturer use term basis for matrix itself,for example suppose that first and third row gives us non zero determinant,$|3*4-(-1)*4|=16$ and suppose that all others are zero,then lecturer says that basis are $[3,4 ,0, 7]$ and $[-1,4,0,3]$

so does these terms are same?i mean what is different according to our matrix between basis of whole matrix,row space and column space?please help me

• Are you referring to the determinants of the minors of the matrix? It is not clear which determinants you are talking about. – Sergio Parreiras Dec 15 '13 at 16:08
• i mean generally,to find basis for matrix they are trying to use this rows as basis,whose minors determinant are not zero – dato datuashvili Dec 15 '13 at 16:09

There are several basis you can choose for a vector space.

Say $M$ is your matrix. Then $M\,\mathbb R^4$ is a vector space and since $\det(M)\neq 0$ it has dimension 4, that is any of its bases has four vectors.

Notice that any $v$ in this space can be written as $v=Mw$ for some $w$ so $v$ is a linear combination of the columns of the matrix, which means that the four vector columns of $M$ are a basis for the space.

Now say $\hat{M}$ is $M$ is in reduced column echelon form. Then the four vector columns of $\hat{M}$ are also a basis for the space.

We also have that $\mathbb{R}^4\, M$ is a vector space of dimension 4 since again the determinant of $M$ is not zero. Look at the space as the set of vectors $v$ such that $v=w \,M$ for some $w\in\mathbb{R}^4$. Any such $v$ is a linear combination of the row vectors of $M$. The same comments I made above with the respect the column echelon form also applies here with the row echelon form. The four row vectors of the row echelon matrix will be a basis for $\mathbb{R}^4\, M$.