Finding a Subspace W of R^4 for Direct Sum V(+)W

[student82]

V is a subspace of R^4
V={(x, -y, 2x+y, x-2y): x,y E R}

1) extend {(2,-1,5,0)} to a basis of V.

2) find subspace W of R^4 for which R^4= direct sum V(+)W.


solution:

1)the dimension of V is 2.therefore i need to add one more vector to (2,-1,5,0).
the 2nd vector is (1,0,2,1).
therefore the basis is {(2,-1,5,0),(1,0,2,1)}.

i want to know whether my answer is correct.


2)dim of W is 2.
so I've to extend the basis for V by just adding any two vectors in R4, making sure that they don't become linearly dependent.

in this case I'm not able to find the basis.should i take the standard basis i.e. (1,0,0,0) or (0,1,0,0) or(0,0,1,0) as the first vector.

 

[Defennder]

I believe your answer to 1) is correct. The resulting basis appears to span V.

I'm not entirely certain what you mean for 2), but I interpret the question to mean: "Extend the basis found in part 1) to span R^4". I hope I got it correct.

For 2), the strategy you should follow is: Create a 2x4 matrix A with the 2 vectors from 1) as row vectors. Reduce the matrix to reduced row echelon form by Gauss-Jordan elimination. Once you get to this stage, consider the 4 standard bases (1,0,0,0), (0,1,0,0) (0,0,1,0) and (0,0,0,1). To span R^4, you need to add 2 of these vectors to your reduced-row echelon form of A, so that you may reduce A to the identity matrix. Once you write it all out, it shouldn't too difficult to see which of the standard basis vectors you need to add as row vectors to A in order to be able to reduce to I.

 

[student82]

is (0,1,0,0) the correct answer?

 

[student82]

sorry
is {(0,0,1,0),(0,1,0,0)} the correct answer.

 

[Defennder]

Yeah I think your answer works. By the way, I realized my "hint" earlier was wrong, apparently (1,0,0,0) and (0,0,1,0) works as well. There should be more than one correct answer here.

 

[student82]

thanks