There is this Corollary that says:

Let $U_1$ and $U_2$ be subspaces of vector space $V$, then $U_1+U_2$ is also a subspace of $V$.

I looked for proof and i found this:

$$ U_1=(x_1, y_1) $$

$$ U_2=(x_2, y_2) $$

$$ x=(x_1+x_2) x_1\in U_1, x_2\in U_2. $$

$$ y=(y_1+y_2) y_1\in U_1, y_2\in U_2 $$

$$ x+y=x_1+x_2+y_1+y_2=(x_1+y_1)+(x_2+y_2) \in U_1+U_2 $$

I know that subspaces are closed under addition but I can't relate it to what i found. Any explanation?

  • $\begingroup$ Another approach would be to show that $U_1 + U_2$ is the image of the linear map $A : V \to V$ defined by $A(v) = P_{U_1}(v) + P_{U_2}(v)$, where $P_{U_1}$ and $P_{U_2}$ are projections onto $U_1$ and $U_2$ respectively. The image of a linear map is always a subspace. $\endgroup$ – Bungo Sep 26 '17 at 19:26

The proof you found is not complete. You have to check all axioms for a linear space.

The lines just show that $U_1 + U_2$ is closed under addition. Set $U=U_1 + U_2$ and let $a,b\in U$. We have to show $a+b\in U$. Since $a\in U$, there exists $a_1\in U_1$ and $a_2\in U_2$ such that $a=a_1 + a_2$. Similar for $b$. Hence we can write $$a+ b = a_1 + a_2 + b_1 + b_2.$$ Since $a_i, b_i \in U_i$ and the operation $+$ is commutative we further have $$a_1 + a_2 + b_1 + b_2 = a_1 + b_1 + a_2 + b_2$$ and since the $U_i$ are linear spaces, $a_1 + b_1 \in U_1$ and $a_2 + b_2 \in U_2$. But this shows $a+b \in U$.

All you have to do now is to carefully check the remaining axioms. You will see that the properties of $U_1$ and $U_2$ transfer to $U$.

Hint: Instead of checking all axioms you should use a so called subspace criterium/definition. See here.

| cite | improve this answer | |
  • $\begingroup$ In the courses I have seen, it would have been considered inappropriate to check, e.g. associativity, since this shows that the standard subspace inheritance properties are not understood. Closure under addition and scaling (or something similar) is definitely the best route. $\endgroup$ – Zach Boyd Sep 26 '17 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.