# Show that U intersection W does not equal ((0,0,0))

1. Oct 13, 2008

### franky2727

show that U intersection W does not equal ((0,0,0)) and hence that U+W is not a direct sum

U being (a,0,a) and W being (c,d,c+2d) and we know that c d and a are elements of R, why are there not a direct sum? is there a rule that prevents any of a c or d from being the zero vector? if this is the case obviously its easy to solve???

2. Oct 13, 2008

### CompuChip

It is only a direct sum if they have no elements in common, that is: you can only write every element from U+W in the form u+w (with u in U and w in W) in one and only one way.
This follows from the first part, namely: that the intersection is not empty. Because, if x is an element in the intersection it can be written both as x + 0 and 0 + x, so the decomposition is not unique.

To show that it is non-empty, you only need to give a single non-zero element that is in both, which is not hard.

3. Oct 13, 2008

### franky2727

sorry that went right over my head, can someone dumb it down a bit please?

4. Oct 13, 2008

### jjou

To show that $$U\cap W\neq\{(0,0,0)\}$$, you need to find some other vector (x,y,z) which belongs in both U and W where at least one of x, y, or z is not zero.

In other words, if $$(a, 0, a) = (c,d,c+2d)$$, what equations can you set up to solve for a, c, and d? Do all three variables have to equal 0?

5. Oct 13, 2008

### jjou

Just a note: unless the above is a typo, it seems that you have a pretty severe misconception about vectors. a, c, and d are elements of $$\mathbb{R}$$, i.e. scalars. Thus none of them can "be the zero vector." The zero vector in this case (since we are working in $$\mathbb{R}^3$$) is the vector (0, 0, 0).