Discussion
Let \(P = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{pmatrix}\). Define an inner product on \(\mathbb{R}^3\) with respect to P as
\[ \langle x, y \rangle_P = x^{\top} P y, \]
for all \(x, y \in \mathbb{R}^3\).
Consider the subspace V = span { \(\begin{pmatrix}1\\0\\0\end{pmatrix}\), \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) }. Which one of the following sets is an orthonormal basis of V with respect to \(\langle x, y \rangle_P\)?
Answer: C
The orthonormal basis is \(\{(1, 0, 0)^{\top}, (0, \frac{1}{5}, \frac{1}{5})^{\top}\}\).
Apply Gram-Schmidt with \(\langle x, y \rangle_P = x_1 y_1 + 9x_2 y_2 + 16 x_3 y_3\). First, \(e_1 = (1,0,0)^{\top}\) has \(\langle e_1, e_1 \rangle_P = 1\), so it is already a unit vector. Next remove the component of \(v = (1,1,1)^{\top}\) along \(e_1\):
\[ \begin{aligned} \langle v, e_1 \rangle_P &= 1 \\ w = v - e_1 &= (0, 1, 1)^{\top} \\ \langle w, w \rangle_P &= 9 + 16 = 25 \end{aligned} \]
so \(w\) has length 5 and the unit vector is \((0, \frac{1}{5}, \frac{1}{5})^{\top}\).
\((0,1,1)^{\top}\) is orthogonal to \(e_1\) but has length 5, and \((0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})^{\top}\) is normalised in the ordinary dot product, not in this one. \((0,1,0)^{\top}\) does not even lie in V.
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