Question

Hello can you help me with this problem please =D

which is the smallest subspace of R2 that contains A={(x,y) in R2 :x>=0 & y>=0} ???

Answer 1

maybe a line like ax + by = 0 where a,b are scalars in R...

Answer 2

because given a vector (a,b) in R^2 every multiple t*(a,b), where t is a real number, fills the line that's the subspace of R^2 with dimension 1 (the free variable t)...

Answer 3

thanks!!

Answer 4

The question is looking for a subspace, and by definition a subspace has to be a vector space itself, meaning it has to be closed under vector addition and scalar multiplication. The set of points on the x-y plane for which x >= 0 and y >= 0 (the upper-right quadrant of the x-y plane) is not itself a subspace, since it is not closed under scalar multiplication: (1, 1) is in the set, but -2 * (1, 1) = (-2, -2) is not.

Since you can multiply any point (x1, y1) in the set by a negative number, any subspace containing the set has to also include all points (x2, y2) for which x2 <= 0 and y2 <= 0--in other words it has to include the lower-left quadrant of the x-y plane.

But that's not all, since the subspace also has to be closed under vector addition. So if, say, (1, 2) is in the subspace and (-2, -1) has to be in it as well, then so does (1, 2) + (-2, -1) = (-1, 1), which is a point in the upper-left quadrant of the x-y plane. Similarly, if (2, 1) is in the subspace and (-1, -2) is also, then so does (2, 1) + (-1, -2) = (1, -1), which is a point in the lower-right quadrant.

We now see that the subspace has to contain all four quadrants of the x-y plane. So the smallest subspace of R^2 that contains A={(x,y) in R2 :x>=0 & y>=0} is R^2 itself.

Answer 5

Here's a drawing to show this better

Answer 6

that's amazing!!! thank you so much!!!

Answer 7

Maybe I am missing something, but isn't the zero vector

0
0

the smallest subspace that meets this criteria?