If \(L, M\) and \(N\) are vector subspaces of \(V\) show that $$ \operatorname{LO}(M+(L \cap N))=L \cap M+L \cap N $$ but it is not truc en general that $$ L \cap(M+N)=\operatorname{Ln} M+L \cap N $$

To prove this, we need to show two things: that every element in \(L \oplus (M + (L \cap N))\) is in \(L \cap M + L \cap N\), and vice versa.Let's take a vector \(v\) in \(L \oplus (M + (L \cap N))\). This means \(v = l + m\) for some \(l\) in \(L\) and \(m\) in \((M + (L \cap N))\). Now since \(m\) is in the sum, this means \(m = m1 + l1\), for some \(m1\) in \(M\) and \(l1\) in \(L \cap N\). Combining these, we get that \(v = l + m1 + l1\), and since both \(l + l1\) and \(m1\) are in \(L\) and \(M\) respectively, it's clear that \(v\) is in \(L \cap M + L \cap N\).For the converse, we show that every vector \(v\) in \(L \cap M + L \cap N\) is in \(L \oplus (M + (L \cap N))\). If \(v\) is in the sum, this means \(v = l1 + m1 + l2\), and we can rewrite this as \(v = (l1 + l2) + m1\), which shows that \(v\) is in \(L \oplus (M + (L \cap N))\).

For this, we need to show that there are cases where \(L \cap (M + N)\) doesn't equal \(L \cap M + L \cap N\). Let's consider the case where \(L = \{0\}\), and \(M = N\). Therefore, \(L \cap (M + N) = L = \{0\}\), while \(L \cap M + L \cap N = \{0\} + \{0\} = \{0\}\). This shows that in this particular case, the two sides aren't equal.