SOLUTION: Find a basis for the span of the given vectors [1 -1 0], [-1 0 1], [0 1 -1] I reduced it and got stuck after that. I am supposted to use the properties (zero martrix and such) or

Algebra ->  Matrices-and-determiminant -> SOLUTION: Find a basis for the span of the given vectors [1 -1 0], [-1 0 1], [0 1 -1] I reduced it and got stuck after that. I am supposted to use the properties (zero martrix and such) or      Log On


   

Question 106542: Find a basis for the span of the given vectors
[1 -1 0], [-1 0 1], [0 1 -1]
I reduced it and got stuck after that. I am supposted to use the properties (zero martrix and such) or something else? I am just stuck and have no clue as to what I am looking for. Our text is custum and does not have an example of this but it does have examples of finding the basis of row space, column space, and null cpace of a matrix. Is this the same thing?
Please Help!
Thank you in advance!!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Remember to find a basis, we need to find which vectors are linear independent. So take the set

and form the matrix

%28matrix%283%2C3%2C1%2C-1%2C0%2C-1%2C0%2C1%2C0%2C1%2C-1%29%29

Now use Gaussian Elimination to row reduce the matrix

%28matrix%283%2C3%2C1%2C-1%2C0%2C0%2C1%2C-1%2C-1%2C0%2C1%29%29 Swap rows 2 and 3


%28matrix%283%2C3%2C1%2C-1%2C0%2C0%2C1%2C-1%2C0%2C-1%2C1%29%29 Replace row 3 with the sum of rows 1 and 3 (ie add rows 1 and 3)


%28matrix%283%2C3%2C1%2C-1%2C0%2C0%2C1%2C-1%2C0%2C0%2C0%29%29 Replace row 3 with the sum of rows 2 and 3 (ie add rows 2 and 3)


%28matrix%283%2C3%2C1%2C0%2C-1%2C0%2C1%2C-1%2C0%2C0%2C0%29%29 Replace row 1 with the sum of rows 1 and 2 (ie add rows 1 and 2)


Now the matrix in reduced row echelon form. Notice the matrix only has 2 pivot columns (which are the first two columns). This means the first two columns of the original matrix are linearly independent. Since the third column does not have a pivot, it is dependent on the first two columns


So to form a basis, simply pull out the linearly independent columns of the original set of vectors to get the set

this set will span the original set (since taking out a dependent vector does not change the span). Also since the set is linearly independent, this set forms a basis (since both properties are satisfied)

So the basis is:



If this isn't what you're looking for, just let me know.