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?
Best Answer
It's good so far. To complete, consider $$T : U_1 \times U_2 \to V : ((x_n), (y_n)) \mapsto (z_n),$$ where $$z_n = \begin{cases}x_{n - 2} & \text{if }n > 2 \\ y_n & \text{if } n = 1, 2.\end{cases}$$