# Vector Spaces and Subspaces

1. Sep 12, 2006

### JasonJo

i've been having some trouble with my linear algebra homework and im wondering if you guys could give me some insight or tips on these problems:

Let v be any vector from V, and let a be any real number such that av=0. Show that either a=0 or v=0.
- i was thinking about assuming the hypothesis that av=0, and then proving the implication by showing that a=0 and v=0 are valid solutions, and then that no other distinct solutions exist. but i ran into trouble with proving the "uniqueness" of the two solutions. ****any other methods or approaches would be great to hear

Similar to the first one, if av=v, show that a=1 or v=0. kinda stuck on this one as well.

Let P(R) be the vector space of polynomials in z of degree at most 2 with real coefficients. Thus P(R) = {a + bz + cz^2: a,b,c are all real numbers}
- give an example of a subset U of P(R) that is closed under scalar multiplication but is not a subspace.
^ ok this one is giving me a problem. if scalar multiplication is closed, this means additive inverses exist. this means that the zero vector is also in this subset. i
- give an example of a subspace U of P(R) that is proper, ie not empty and not the entire space
- find another subspace W such that U(direct sum)W = P(R)

Suppose U1, U2, U3 are subspaces such that V= U1+U2+U3, formulate a condition in terms of suitable intersections of U1, U2 and U3 such that V = U1(Direct Sum)U2(direct sum)U3. and then generalize for k subspaces.

to me, it's pretty tough stuff....

2. Sep 12, 2006

### Hurkyl

Staff Emeritus
For your first two, think of how you would do it if you were doing ordinary algebra with real numbers.

3. Sep 12, 2006

### fourier jr

warning: i haven't actually done this problem. but the first thing that came to mind (because of the next problem actually) is that av=0 => (a-0)v=0. dunno if that will actually work but it might be somewhere to start.

i dont know if you know anything about fields but since R is a field (& thus an integral domain) av=v => (a-1)v=0 & use the first part ie (a-1)=0 or v=0. i hope that's right :uhh:

you're given that for U one of the subspace conditions is definitely satisfied. make up a U where the other condition isn't satisfied! it will probably take some tinkering to come up with one.
it's not important that they used the space of quadratic polynomials in the problem. they could have chosen the spaces of cubics, quartics, quintics etc etc it makes no difference.....
use the previous part

this is almost immediate from the definition of direct sum. by definition what is the only thing that two direct summands have in common? (what is in the intersection?)

if it were easy everybody would do it :tongue:

4. Sep 12, 2006

### HallsofIvy

Staff Emeritus
If v is not 0, then there exist a basis containing v. What does that tell you about Ax=0?