Let $W_i (i =1,2)$ be subspaces of $V$. $W_1 \cap W_2 =0$. If $U$ is a subspace of $V$, is it true that $U \cap (W_1 \oplus W_2) = (U \cap W_1) \oplus (U \cap W_2)$?

If it is true, is there any way to give a formal proof? If it is not, is there a hint for constrcting a contradiction or giving a counter example?

  • $\begingroup$ No, it is not true. $\endgroup$ – Charlie Frohman Jul 31 at 23:13

It is false. Take $V=\mathbb R^2$, $W_1=\mathbb R\times\{0\}$, $W_2=\{0\}\times\mathbb R$, and $U=\{(x,x)\mid x\in\mathbb R\}$. Then $U\cap W_1=U\cap W_2=\{0\}$, but $U\cap(W_1\bigoplus W_2)=U\neq\{0\}$.

  • $\begingroup$ So this counterexample can be considered as a $y=x$ linear on a $x-y$ plane where $W_1$ is x-axis and $W_2$ is y-axis? $\endgroup$ – WaterBro Jul 31 at 23:16
  • $\begingroup$ That's a way of looking at it, yes. $\endgroup$ – José Carlos Santos Jul 31 at 23:18

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