Exercise 3.E.9

Answers

Say that $A_{1} ⋂ A_{2}$ is not Let

A_{1} = v_{1} + U_{1}

and

A_{2} = v_{2} + U_{2}

Where $U_{1}, U_{2}$ are subspaces of $V$ Now, define $a, b$ such that

a = v_{1} + u_{1} = v_{2} + w_{1}

and

b = v_{1} + u_{2} = v_{2} + w_{2}

Where $u_{1}, u_{2} \in U_{1}$ and $w_{1}, w_{2} \in U_{2}$ Therefore, by Exercise 8 of 3.E,

\begin{matrix} \begin{aligned} λa + (1 - λ) b & \in A \\ λ (v_{1} + u_{1}) + (1 - λ) (v_{1} + u_{2}) & \in A \\ λ v_{1} + λ u_{1} + v_{1} + u_{2} - λ v_{1} - λ u_{2} & \in A \\ v_{1} + λ u_{1} - λ u_{2} + u_{2} & \in A \\ v_{1} + U & = A \end{aligned} \end{matrix}

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Arden_Shi

2025-06-03 16:07