In \(\RNrSpc{3}\) let \(V\) be the subvector space generated by \(\Vect{a}\DefEq (1,1,1)\). Describe \(V\) and \(V^{\bot}\), and compute their dimensions.
The vector \(\Vect{a}\) is not the zero vector. Therefore its span \(V\) is a line in \(\RNrSpc{3}\); in fact it is the line through \(\Vect{0}\) in the direction of \(\Vect{a}\). An basis for \(V\) is \(\EuScript{A}\DefEq \Set{ \Vect{a} }\). So
| \(\Dim{V}\) | \(=\) | \(1\) |
The orthogonal complement of \(V\) is a hyperspace: it is the plane through the origin with normal vector \(\Vect{a}\). The dimension formula confirms
| \(\Dim{V^{\bot}}\) | \(=\) | \(\Dim{\RNrSpc{3}} - \Dim{V}\) |
| \(\) | \(=\) | \(3 - 1\) |
| \(\) | \(=\) | \(2\) |
In \(\RNrSpc{5}\) let \(V\) be the subvector space generated by \(\Vect{a}\), \(\Vect{b}\), and \(\Vect{c}\) with
| \(\Vect{a}\) | \(=\) | \((3,5,-2,1,4)\) |
| \(\Vect{b}\) | \(=\) | \((2,-2,2,3,-3)\) |
| \(\Vect{c}\) | \(=\) | \((-1,9,-6,-5,10)\) |
Find the dimension of \(V\) and its orthogonal complement.
We notice that \(\Vect{a}\) and \(\Vect{b}\) are not parallel. So they are linearly independent. Are these vectors also linearly independent of \(\Vect{c}\)? – Solving the vector equation
| \(s\cdot \Vect{a} + t\cdot \Vect{b}\) | \(=\) | \(\Vect{c}\) |
yields the solution \(s=1\) and \(t=-2\); i.e. \(\Vect{c}\) is a linear combination of \(\Vect{a}\) and \(\Vect{b}\). Therefore
| \(V\) | \(=\) | \(\span\Set{ \Vect{a},\Vect{b} }\) |
and \(\EuScript{B} = (\Vect{a},\Vect{b})\) is an ordered basis for \(V\). So
| \(\Dim{V}\) | \(=\) | \(2\) |
The dimension of the orthogonal complement \(V^{\bot}\) may be computed using the dimension formula:
| \(\Dim{V^{\bot}}\) | \(=\) | \(\Dim{\RNrSpc{5}} - \Dim{ V }\) |
| \(\) | \(=\) | \(5 - 2\) |
| \(\) | \(=\) | \(3\) |