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Let $V$ be a direct sum of two subspaces $U,W$. Prove that $U\times W$ is isomorphic to $V$?

My problem lies in the fact the book has not provided any formal definition or theorem about $U\times W$, I mean about the product of Vector Spaces, on this case subspaces. As far as I can see the $ker(T)=0$, because $0\times 0$ is preserved but I cannot talk about dimensions.

Questions:

Could anyone provide me a definition of $U\times W$?

How can I prove $U\times W$ is isomorphic to V?

Thanks in advance!

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The space $U\times W$ consists of all ordered pairs $(u,w)$, where $u\in U$ and $w\in W$. The operations are given by coordinate wise addition and scalar multiplication.

Now define $T:U\times W\to V$ by $T(u,w)=u+w$. We have $$T(\lambda u_1+u_2,\lambda w_1+w_2)=\lambda(u_1+w_1)+u_2+w_2=\lambda T(u_1,w_1)+T(u_2,w_2).$$ So $T$ is linear.

Since $V=U+W$, given $v\in V$ there exist $U\in U$ and $w\in W$ with $v=u+w=T(u,w)$. So $T$ is onto.

If $T(u,w)=0$, then $u+w=0$. Thus $u=-w$. Because $U\cap W=\{0\}$, $u=w=0$. So $T$ is one-to-one.

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  • $\begingroup$ Please Could you explain me how is $U\cap W=\{0\}$? $\endgroup$ – Pedro Gomes Jun 23 '17 at 11:09
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    $\begingroup$ Definition of "direct sum". $\endgroup$ – Martin Argerami Jun 23 '17 at 12:30
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The product $U \times W$ for any sets $U$ and $V$ is defined as $U \times W = \{(u,w) \ | \ u\in U, w \in W\}$. If $U$ and $V$ are vector spaces over the same field, their Cartesian product is always a vector space by defining $$ (I).\ \ (u_1,w_1) + (u_2,w_2) = (u_1+u_2,w_1+w_2), \ \ (II). \ \ a(u,w) = (au,aw), $$ where $a$ is a scalar. If $V = U \oplus W$ then (as you point out) $U \cap W = 0$. You can easily show $\dim (U\times W) = \dim U + \dim W$. It is also true that by identifying $U$ with $\{(u,0) \ | \ u \in U\}$ and $W$ with $\{(0,w) \ | \ w \in W\}$ we may regard $U$ and $W$ as subspaces of $U \times W.$

Does this help?

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  • $\begingroup$ $\dim (U\times W) = \dim U + \dim W$ regardless of whether or not $U\cap W=\{0\}$. We need $U\cap W=\{0\}$ to show that $\dim(U+W)=\dim(U)+\dim(W)$. $\endgroup$ – Aweygan Jun 21 '17 at 18:32
  • $\begingroup$ @Aweygan Correct. Fixed. $\endgroup$ – Mortified Through Math Jun 21 '17 at 18:37
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    $\begingroup$ There is no need to deal with dimensions (which could be infinite). All one needs is to show that the obvious map $(u,w)\longmapsto u+v$ is trivially an isomorphism. $\endgroup$ – Martin Argerami Jun 21 '17 at 18:42
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Hint: $U\times W$ is by definition the set of pairs $(u,w)$, $u\in U,\;w\in W$, endowed with operations componentwise $$(u,w)+(u',w')=(u+u', w+w')\quad \lambda (u,w)= (\lambda u,\lambda w),\qquad\text{ (external direct sum)}$$ You can define a linear map \begin{align}U\times W&\longrightarrow V\\ (u,w)&\longmapsto u-w\end{align} and show it is an isomorphism.

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  • $\begingroup$ I imagine (but am not sure) that you would need to introduce the factor space $U \oplus W / W$ and show $U \simeq U \oplus W / W$ to rigorously prove your isomorphism...I'm not sure OP is at that level. Though there could certainly be something I'm missing. $\endgroup$ – Mortified Through Math Jun 21 '17 at 18:40
  • $\begingroup$ Not really: if $V$ is finite-dimensional: both spaces have the same dimension, hence it is enough to show the linear map is injective (or surjective). For a general vector space, you have to define the inverse linear map, which s not very hard. $\endgroup$ – Bernard Jun 21 '17 at 18:46

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