Is it true that: $(U + W)\cap S = U \cap S + W\cap S$?, where $U,W,S \subseteq V$ where $V$ is a vectorial space.

by C. Cristi   Last Updated October 16, 2018 20:20 PM - source

Is it true that $(U + W)\cap S = U \cap S + W\cap S$?, where $U,W,S$ $\subseteq V$ where $V$ is a vectorial space.

My attempt:

$$\begin{array}{l}U=\{\lambda(\begin{array}{c}u_1\\\vdots\\u_n\end{array})\;\vert\:\lambda\in\mathbb{R}\}\\W=\{\mu(\begin{array}{c}w_1\\\vdots\\w_n\end{array})\;\vert\:\mu\in\mathbb{R}\}\\S=\{\alpha(\begin{array}{c}s_1\\\vdots\\s_n\end{array})\vert\;\alpha\in\;\mathbb{R}\}\\U\cap S=\{\lambda(\begin{array}{c}u_1\\\vdots\\u_n\end{array})\;-\alpha(\begin{array}{c}s_1\\\vdots\\s_n\end{array})=0\vert\lambda,\;\alpha\in\;\mathbb{R}\}\\W\cap S=\{\mu(\begin{array}{c}w_1\\\vdots\\w_n\end{array})\;\;-\alpha(\begin{array}{c}s_1\\\vdots\\s_n\end{array})=0\vert\mu,\;\alpha\in\;\mathbb{R}\}\\U\cap S\:+\;W\cap S\;=\;\{\lambda(\begin{array}{c}u_1\\\vdots\\u_n\end{array})+\mu(\begin{array}{c}w_1\\\vdots\\w_n\end{array})-\alpha(\begin{array}{c}s_1\\\vdots\\s_n\end{array})=0\vert\lambda,\mu,\;\alpha\in\;\mathbb{R}\}\;=\;(U+W)\cap S\end{array}$$

Answers 1

The statement is not true. The comment gives a counterexample.

But I suspect you mean $U, V, W$ are subspaces. In this case, here is another counterexample in $\mathbb R^2$: Take $U = \text{span}((1,0))$ and $W = \text{span}((0,1))$, $S=\text{span}((1,1))$. Then $U+V = \mathbb R^2$, $(U+V)\cap S=S$ but $U\cap S = \{0\}$ and $V \cap S =\{0\}$.

October 16, 2018 20:13 PM

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