3 ms·
>A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult. Isn't this a 1 dimensional
by gizmo686 8y ago
>A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult.
Isn't this a 1 dimensional space. Eg. Consider the vector space R2 over R.
If you have the vector (1,0), there is no way to arrive at the vector (1,1) through just scalar multiplication.
- dxbydt 8y agoif you don’t give me basis how’ll i span the space ?
- zazen 8y agoYou talked about having "a member" and getting to every other member with scalar multiplication only. But even given a 2-member basis for R2, how are you planning on using 2 members, with scalar multiplication only? I'm afraid you have badly misremembered this stuff.
- dxbydt 8y agoyou are right.