# I Prove that V is the internal direct sum of two subspaces

1. Feb 6, 2017

### Austin Chang

Let V be a vector space. If U 1 and U2 are subspaces of V s.t. U1+U2 = V and U1 and U1∩U2 = {0V}, then we say that V is the internal direct sum of U1 and U2. In this case we write V = U1⊕U2. Show that V is internal direct sum of U1 and U2if and only if every vector in V may be written uniquely in the form v1+v2 with v1∈U1 and v2∈ U2.

What does it mean to be unique? Does it matter if it is unique?

2. Feb 6, 2017

### andrewkirk

It means that for any vector $u\in V$, if $w_1+w_2=u=v_1+v_2$, for $w_1,v_1\in U_1$ and $w_2,v_2\in U_2$, then $v_1=w_1$ and $v_2=w_2$.

Yes, the uniqueness matters. If the two subspaces overlap, there is more than one way of writing a vector in $V$ as a sum of two vectors one from each subspace. And vice versa, if the representation as a sum is not unique, the two subspaces must overlap.

3. Feb 6, 2017

### Austin Chang

So for your example if it was not unique U2 = U1 and you can write w1 in U2 and w2 in U1? therefore w2+w1 is not unique anymore because the same things came from different subspaces?

4. Feb 6, 2017

### Stephen Tashi

Perhaps an example of non-uniqueness will make this clear. You can represent 12 as the product of two factors $12 = x_1 x_2$ but the two factors are not unique since they and be chosen to have various different values - e.g. (12)(1), (4)(3), (2)(6) etc.

As whether uniqueness matters - that's a subjective question. Uniqueness is often a convenient property to have. For example, the rigorous approach to defining "a" zero z_1 in arithmetic would be to define it by the property that for all numbers x, z1 + x = x. It is convenient that there is a unique number, denoted by "0", that has this property. If there were several unequal zeroes, arithmetic would be more complicated.

5. Feb 6, 2017

### andrewkirk

No, that would be going too far. U2 and U1 will overlap nontrivially, but that doesn't mean they are identical. All it means that their intersection is a vector space of dimension at least 1.

6. Feb 6, 2017

### Austin Chang

Thanks!! That was a good example.

7. Feb 6, 2017

### Austin Chang

What do you mean by at least a vector space of dimension 1? I don't think I've gotten that far to understand it completely. Could you elaborate?

8. Feb 6, 2017

### andrewkirk

A vector space that is not just $\{0_V\}$.