# Find a basis of U

1. Dec 16, 2009

### Striker2

I have my linear algebra exam coming up but I missed the class on bases. Can anyone show me how this is solved?

2. Consider the subspace U of R4 defined by
U = span{(−1, 1, 0, 2), (1, 0, 0, 1), (2,−1, 1,−1), (0, 1, 0, 3)}
• Find a basis of U.
• Find a basis of the orthogonal complement U.

2. Dec 17, 2009

### HallsofIvy

Staff Emeritus

In any case I am sure your textbook says that a "basis" for a vector space has three properties:

1: Its vectors span the space.
2: Its vectors are independent.
3: The number of vectors in the space is equal to the dimension of the space.
(Actually, the "theorem" is that any two bases for the same space contain the same number of vectors. And then we define that number to be the dimesnsion.)\

And, any two of these imply the third. That is, if you know a set of vectors spans the space and is independent, then it is a basis and the number of vectors in it is the dimension of the space.

If you have a set of independent vectors and the number of vectors in it is equal to the dimension of the space, then it is a basis and spans the space.

If you have a set of vectors that span the space and the number of vectors in it is equal to the dimension of the space, then it is a basis and the vectors are independent.

Here, you are given that this set of vectors spans U. IF they are independent, then they are a basis. If they are not, then some of the vectors can be written terms of the others. Drop those and you haven't lost anything. What remains is a basis.

The vectors are "independent" if the only way to have a(−1, 1, 0, 2)+ b(1, 0, 0, 1)+ c(2,−1, 1,−1)+ d(0, 1, 0, 3)= (0, 0, 0, 0) is to have a= b= c= d. That is the same as (-a+ b+ 2c, a- c+ d, c, 2a+b-c+ 3d)= (0, 0, 0, 0) which means, of course, -a+ b+ 2c= 0, a- c+ d= 0, c= 0,, 2a+b-c+ 3d= 0. Try solving those equations for a, b, c, and d. An obvious solution is a= b= c= d= 0. If that is the only solution, then the vectors given are independent and form a basis. If not, then at least one can be written in terms of the others. Say, perhaps, that c and d can be written interms of a and b. Then you don't need the vectors multiplied by c and d and just those multiplied by a and b are necessary. That would be your basis.

3. Dec 17, 2009

### Striker2

Awesome! Thanks for the help, I understand completely.