Let u and v be vectors in R2 It can be shown that the set P

Let u and v be vectors in R^2. It can be shown that the set P of all points in the parallelogram determined by u and iJ has the form au + bv, for 0 lessthanorequalto a lessthanorequalto 1, 0 lessthanorequalto b lessthanorequalto 1. Let T: R^2 rightarrow R^2 be a linear transformation. Explain why the image of a point in P under the transformation T lines in the parallelogram determined by T(u) and T(v).

Solution

Use the definition of what a subspace means. It is first closed under vector addition, has an identity, and is closed under scalars. First show closure under vector addition which is basically proving this: If x,yUVx+yU+Vx,yUVx+yU+V. Now the identity is trivial to show. Since UUand VV are finite subspaces it follows by definition that... And for scalar multiplication. Let rRrRwhere rr will be a scalar. Let xUVxUxUVxU and xVxV. It follows that since UU and VV are subspaces then rxUrxU and rxVrxUVrxVrxUV. Hence we conclude...

For part b) Same thing. Let xU+VxU+V. First show closure. That is if x,yU+Vx,yU+V where x=a+bx=a+b and y=c+dy=c+d where a,cUa,cU and b,dVb,dV then we see x+y=(a+b)+(c+d)x+y=(a+b)+(c+d)which is still an element of the set U+VU+V. For the identity its easy. I will denote 00 to be the identity both in UU and VV since it follows by definition of a subspace. Thus 0=0+0U+V0=0+0U+V And lastly scalar multiplication which I will leave up to you.

For part c) Recall what dimension means and remember U+VU+V is a set.

I forgot to put arrows on top of the elements but they are all vectors. So you can place arrows for clarity.