linear algebra- vector-basis-subspace

arbolis · #1 $Old$ 10-10-2008, 10:53 AM

Hi,
Tell whether or not the following vectors are linear independent, if they are l.i., say if they generate $\mathbb{R}^3$ or $\mathbb{R}^4$ and in the contrary case characterize implicitly the subspace generated and give a basis of this subspace.

The vectors are (1,1,2,4),(2,-1,-5,2),(1,-1,-4,0) and (2,1,1,6).
My attempt : I formed and reduced this matrix : $\begin{bmatrix} 1&2&1&2 \\ 1&-1&-1&1 \\ 2&-5&-4&1 \\ 4&2&0&6 \end{bmatrix}$ to this one $\begin{bmatrix} 1&0&-\frac{1}{3}&\frac{4}{3} \\ 0&1&\frac{2}{3}&\frac{1}{3} \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}$ . I concluded by saying that the vectors are linear dependent (because of the 2 rows that are 0, so 2 of the 4 vectors are linear dependent between them) and that the vectors generate $\mathbb{R}^2$ (because 2 rows are reduced) and that a basis of $\mathbb{R}^2$ is $\{ (1,0,0,0),(0,1,0,0) \}$ .
I have some questions : first, is what I've done correct (at least logically)?
Second question : how can I describe implicitly $\mathbb{R}^2$ ?
Third question : is $\{ (1,0,0,0),(0,1,0,0) \}$ a possible basis of $\mathbb{R}^{2}$ ? What about $\{ (1,0,0,0),(0,0,1,0) \}$ ? (I guess yes).
Thank you very much.

watchmath · #2 $Old$ 10-10-2008, 03:48 PM

The first step is correct, they are linearly dependent and the dimension is 2. But this doesn't mean that the subspace is R^2. There are lots of subspace of dimension 2.

Your work basically tells you that you don't need all four vectors to generate the subspace. Two is enough. Can you choose these two vectors?

arbolis · #3 $Old$ 10-10-2008, 06:04 PM

Thank you watchmath!

Quote:

The first step is correct, they are linearly dependent and the dimension is 2. But this doesn't mean that the subspace is R^2. There are lots of subspace of dimension 2.

Yes, you're right.
And about

Quote:

Your work basically tells you that you don't need all four vectors to generate the subspace. Two is enough. Can you choose these two vectors?

I did chose the vectors (1,0,0,0) and (0,1,0,0) because they are the 2 columns with only a 1 and all the other elements are 0. So that was wrong?
Maye I had to consider an amplied matrix... but I thought it wasn't worth it since I thought I had found an easy answer. (that is, I could reduce the rows of the matrix that are not all 0s.)

watchmath · #4 $Old$ 10-10-2008, 07:05 PM

The two unit vectors (that you chose) might be not generators of the subspace even they generate a subspace of dimension 2.

You need to take 2 vectors among the 4 original vectors. Since you know that the subspace has dimension two you can take any two vectors (among 4) for which the two are linearly independent.

arbolis · #5 $Old$ 10-10-2008, 07:10 PM

If I understood well, I can take (1,1,2,4) and (2,-1,-5,2) (the 2 first vectors).

watchmath · #6 $Old$ 10-10-2008, 07:22 PM

Yes in this case you can choose that one.

Now suppose we have a matrix A where the vector columns of A are $c_1,c_2,c_3,c_4$ and suppose by row operation we can reduce A into
$\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$ .

What are the generators of the subspace generated by $c_1,c_2,c_3,c_4$ ?

arbolis · #7 $Old$ 10-10-2008, 09:23 PM

Quote:

Originally Posted by watchmath $View Post$

Yes in this case you can choose that one.

Now suppose we have a matrix A where the vector columns of A are $c_1,c_2,c_3,c_4$ and suppose by row operation we can reduce A into
$\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$ .

What are the generators of the subspace generated by $c_1,c_2,c_3,c_4$ ?

I'd say $c_1$ , $c_2$ and $c_3$ . I can add that I'd say it to be true even if we had the matrix $\begin{pmatrix}1&0&0&89\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$ . I hope I'm right.