Linear algebra: span Which of the following vectors span R2(1x2 matrix)? (a) [1 2], [-1 1] (b) [0 0], [1 1], [-2 -2] Do I make the matrices into row reduced form in order to solve these? Like for (a), would it be like [1 2; -1 1], so [1 0; 0 1]?

I think both span it.

Sure the second one is not lin ind but it still spans

The second set does not span R2

Why not?

Notice that [0,0] and [-2, -2] are both scalar multiples of [1, 1]. This means that the second set of vectors only spans the set of vectors [a, a] where a is some number in R.

How do you know if the 1xn matrices span it? I am confused

If it was an augmented matrix, I would determine whether it spans or not by its consistency but I don't know how to deal with this question..

Oh yeah I didnt catch that. But if [0 0] was [1,2] then it would span r^2 right?

If [0, 0] was [1, 2] it would span. @he66666 I can determine that the first set spans R2 because I can see that R2 is a two-dimensional vector space, and [1, 2] and [-1, 1] are two linearly independent vectors. This means that they form a basis, and so they span R2.

What if you have a set of three vectors that are lin independent and are two dimensional. Does that set also span R^2?

If the vectors are contained in R^3, I believe that the lin. independent vectors span a vector space that's isomorphic to R^2.

For example, take the vectors [1 1 1] and [1 1 2]. These are obviously linearly independent, but they are not contained in R^2, so they can't actually span R^2. They would instead span some subspace of R^3 that is isomorphic to R^2.

I didn't learn independent vectors yet. But we learned that we should investigate the consistency of the linear system to determine whether a specific vector v belongs to span S, and using the reduced row form to see whether it spans. Is there another way to solve it?

For the first one, I think you would want to rref [1 -1;2 1]. If one of the two rows is 0 after you rref it, they should be linearly independent (aka not in the span of the other).

OS just crashed twice for me x.x By showing that the other vectors are scalar multiples of [1 1], you are showing that they are in the same span.

perhaps another way to see it is that the vector [0,0] has no span, so we need only consider the other two vectors

Thanks guys :)

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