Undergrad Prove that V is the internal direct sum of two subspaces

[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 = {0_V}, 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?

 

[andrewkirk]

What does it mean to be unique?

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.

 

[Austin Chang]

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.

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?

 

[Stephen Tashi]

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 = {0_V}, 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?

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.

 

[andrewkirk]

So for your example if it was not unique U2 = U1

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.

 

[Austin Chang]

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.

Thanks! That was a good example.

 

[Austin Chang]

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.

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?

 

[andrewkirk]

What do you mean by a vector space of dimension at least 1?

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