You are given the vectors \(\Vect{a}\), \(\Vect{b}\), and \(\Vect{c}\) in \(\RNrSpc{3}\) with
| \(\Vect{a}\) | \(=\) | \((1,1,1)\) |
| \(\Vect{b}\) | \(=\) | \((0,-1,1)\) |
| \(\Vect{c}\) | \(=\) | \((1,1,0)\) |
Verify that these vectors form a basis of \(\RNrSpc{3}\), then orthonormalize them.
We begin by checking that the given vectors are linearly independent. we use the determinant.
\[ \det \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & -1& 1 \\ 1 & 1 & 0 \end{array} \right] = \det \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 1 &-1 & 0 \\ 1 & 1 &-1 \end{array} \right] = 1 \neq 0 \]So the vectors are linearly independent. Then we know that three linearly independent vectors in the 3-dimensional space \(\RNrSpc{3}\) always form a basis. This answers the first part of the problem.
We use the Gram-Schmidt orthonormalization method to turn the given basis of \(\RNrSpc{3}\) into an orthonormal basis of \(\RNrSpc{3}\).
| \(\Vect{u}\) | \(\DefEq \) | \(\frac{\Vect{a}}{ \Norm{ \Vect{a} } }\) |
| \(\) | \(=\) | \(\frac{1}{\sqrt{3}}\cdot (1,1,1)\) |
To orthonormalize \(\Vect{b}\) with respect to \(\Vect{u}\), we encounter the expression
| \(\left[ \DotPr{ (0,-1,1) }{ \left( \tfrac{1}{\sqrt{3}}\cdot (1,1,1) \right) }\right]\cdot \tfrac{1}{\sqrt{3}}\cdot (1,1,1)\) | \(=\) | \(0\) |
So we find
| \(\Vect{v}\) | \(=\) | \(\frac{ \Vect{b} }{ \Norm{ \Vect{b} } }\) |
| \(\) | \(=\) | \(\frac{1}{ \sqrt{2} }\cdot (0,-1,1)\) |
To orthonormalize \(\Vect{c}\) with respect to \(\Vect{u}\) and \(\Vect{v}\) we encounter the expressions
| \(\left[ \DotPr{ (1,1,0) }{ \left( \tfrac{1}{\sqrt{3}}\cdot (1,1,1) \right) }\right]\cdot \tfrac{1}{\sqrt{3}}\cdot (1,1,1)\) | \(=\) | \(\frac{2}{3}\cdot (1,1,1)\) |
| \(\left[ \DotPr{ (1,1,0) }{ \left( \tfrac{1}{\sqrt{2}}\cdot (0,-1,1) \right) }\right]\cdot \tfrac{1}{\sqrt{2}}\cdot (0,-1,1)\) | \(=\) | \(-\frac{1}{2}\cdot (0,-1,1)\) |
Therefore the orthonormalization of \(\Vect{c}\) with respect to \(\Vect{u}\) and \(\Vect{v}\) is
| \(\Vect{w}\) | \(\DefEq \) | \(\frac{ (1,1,0) - \tfrac{2}{3}\cdot (1,1,1) + \tfrac{1}{2}\cdot (0,-1,1) }{ \Norm{ (1,1,0) - \tfrac{2}{3}\cdot (1,1,1) + \tfrac{1}{2}\cdot (0,-1,1) } }\) |
| \(\) | \(=\) | \(\frac{ \tfrac{1}{6}\cdot (2,-1,-1) }{ \Norm{ \tfrac{1}{6}\cdot (2,-1,-1) } }\) |
| \(\) | \(=\) | \(\frac{1}{\sqrt{6}}\cdot (2,-1,-1)\) |
Thus \(\Vect{u}\), \(\Vect{v}\), and \(\Vect{w}\) as found above form the Gram-Schmidt orthonormalization of \(\Vect{a}\), \(\Vect{b}\), and \(\Vect{c}\).