Would this be a correct reason for why row reduction method gives linearly independent column vectors of a matrix that are basis of column space

[rishabhn99]

the columns containing the pivot entries of the row reduced matrix correspond to the column vectors of the matrix that are linearly independent,because every column of the row reduced matrix can be expressed as the linear combinations of columns containing pivot entries that means they are of the form Ax=b,where x are the coefficients of linear combinations.and A is the column vectors of row reduced form and b is every other column of row reduced form.since this Ax=b has same solutions under row operations.that is why if we reverse the row operations we get the linearly independent vectors that are the basis for column space.

correction-A is the matrix containing column vectors of row reduced matrix containing the pivot entries.also to aid imagination we can see the the matrix obtained by reversing the row operations from which we had gotten the row reduced matrix as a combination of columns of A and b getting reversed due to same row operations.

 

[blamocur]

I don't understand this post. Even worse, I cannot even tell if there is a question in there since I did not see any question marks.

 

[rishabhn99]

sir,i am trying to justify why the method of row reduction gives the linearly independent basis vectors of the column space of a matrix.and the question is to verify whether my justification is correct.in the method,we row reduce the matrix,and then the columns containing the pivot entries correspond to the columns of the matrix which are linearly independent and form the basis for the column space of a matrix.i am trying to show why this works.in the method,i am saying let us take the columns corresponding to to the pivot entries of the row reduced matrix.let us form a matrix A with these columns.now every other column of the row reduced matrix can be expressed as a linear combination of the columns containing pivot entries.that is evident because in row reduced form,the column containing the pivot entries entries has all 0's except only some 1's.so their linear combination gives every other column of the row reduced matrix.now we can say this situation can be reprsented by Ax=b,where x is the column containing all the coefficients of the linear combinations and b is any random column vector other than the column vectors containing pivot elements,and A is the matrix composed of the pivot entry columns.Now if we perform row operations on this Ax=b,we can get back the original matrix(the one we had obtained from row reduction).because the columns of A are the pivot entry columns and b is any other column of the row reduced matrix.this shows that the columns corresponding to the pivot entries columns can be used in linear combination to form every other column of the matrix.which shows the columns corresponding to the pivot entry columns form the basis for the column space of the matrix.I hope this explanation of what I'm trying to say is clearer.

 

[blamocur]

sir,i am trying to justify why the method of row reduction gives the linearly independent basis vectors of the column space of a matrix

What if the columns of the original matrix aren't linearly independent?