# Show that $U_1 \times U_2$ is isomorphic to $U_1 + U_2$ But $U_1 + U_2$ is not a direct sum

Give an example of a vector space $$V$$ and subspaces $$U_1, U_2 \subset V$$ Such that $$U_1 \times U_2$$ is isomorphic to $$U_1 + U_2$$ but $$U_1 + U_2$$ is not a direct sum. Hint: $$V$$ must be infinite-dimensional by 3.78 in the textbook.

Here is what I have so far.

Suppose $$V \in \mathbb{R}^{\infty}$$ and $$U_1 = \mathbb{R}^{\infty}$$ and $$U_2 = \{x_1, x_2, 0, 0, \dots \}$$. Well clearly $$U_1 + U_2$$ is not a direct sum because $$U_1 \cap U_2 \neq 0$$. To show that $$U_1 \times U_2$$ is isomorphic to $$U_1 + U_2$$, we need to show that there is an isomorphic linear map $$T: U_1 \times U_2 \rightarrow U_1 + U_2$$.

Here I am not sure how to show that it is invertible and subjective. Can I get a hint in the direction to go? Thanks

Edit: Update on proof

Injectivity - To show injectivity, I want to show that null$$T = \{0\}$$. If we shift the elements in $$U_1 + U_2$$ over to the left twice, then there won't be any duplicates of $$x_1, x_2$$ and null$$T = \{0\}$$ because

$$T(x_1, x_2, \dots) = (x_1, x_2, x_3, \dots)$$

Surjectivity - I don't really have an idea of what to do here. I wrote out what vectors look like on both sides, but don't really see an obvious step forwards.

$$[(x_1, x_2, x_3 \dots),(x_1, x_2, 0, 0, \dots)] =(x_1, x_2, x_3, \dots) + (x_1, x_2, 0, 0, \dots)$$

It seems that $$x_1, x_2, x_3, \dots$$ are already all in $$U_1 \times U_2$$ and $$U_1 + U_2$$ and nothing needs to be done here?

It's good so far. To complete, consider $$T : U_1 \times U_2 \to V : ((x_n), (y_n)) \mapsto (z_n),$$ where $$z_n = \begin{cases}x_{n - 2} & \text{if }n > 2 \\ y_n & \text{if } n = 1, 2.\end{cases}$$
• Not quite. Remember, $T$ takes elements of $U_1 \times U_2$, all of which are pairs of sequences (one of which contains at most $2$ non-zero terms). To show injectivity, you need to show that if $(z_n)$ (as defined above) is the zero sequence, them $((x_n), (y_n))$ is the pair containing the $0$ sequence twice. The surjectivity proof too could use some work. If I give you an arbitrary sequence $(z_n) \in V$ (which is equal to $U_1 + U_2$), what would be the unique pair of sequences $((x_n), (y_n)) \in U_1 \times U_2$ such that $T((x_n), (y_n)) = (z_n)$? Commented Oct 7, 2019 at 3:51
• @TheoBendit I believe the $T$ you constructed is not injective. For instance $T((1, 1, 0, 0,\dots), (0, 0, 0, 0,\dots)) = 0$. Commented Mar 2, 2021 at 7:26