If \(V\) and \(W\) are subvector spaces of \(\RNrSpc{n}\), then their intersection \(\Intrsctn{V}{W}\) is also a subvector space of \(\RNrSpc{n}\).
If \(V\) and \(W\) are subvector spaces of \(\RNrSpc{n}\), then their intersection \(\Intrsctn{V}{W}\) is also a subvector space of \(\RNrSpc{n}\).
To shorten notation, set \(U\DefEq \Intrsctn{V}{W}\). Then \(U\) contains the \(\Vect{0}\)-vector, as it is contained in both \(V\)and \(W\). So it remains to show that \(U\) is closed under vector addition and scalar multiplication
To see that \(U\) is closed under vector addition, let \(\Vect{x}\) and \(\Vect{y}\) be in \(U\). This means that both vectors belong to \(V\) and to \(W\). Now, both \(V\) and \(W\) are closed under vector addition. So \(\Vect{x} + \Vect{y}\) belongs to \(V\) and to \(W\) and, hence, to \(U\).
To see that \(U\) is closed under scalar multiplication, let \(\Vect{x}\) in \(U\) and \(t\) in \(\RNr\) be arbitrary. So \(\Vect{x}\) belongs to \(V\) and to \(W\). As both \(V\) and \(W\) are closed under scalar multiplication, \(t \Vect{x}\) also belongs to \(V\) and to \(W\) and, hence, to \(U\).
This proves that the intersection \(\Intrsctn{V}{W}\) satisfies all requirements of a subvector space.