Question

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 \[ \langle x,y\rangle_P = x^{\top}Py, \quad \text{for all } x,y \in \mathbb{R}^3. \] Consider the subspace \(V = \text{span}\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\right\}\). Which one of the following sets is an orthonormal basis of \(V\) with respect to \(\langle x,y\rangle_P\)?

• A. \(\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix}\right\}\)
• B. \(\left\{\begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix}\right\}\)
• C. \(\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\frac{1}{5}\\\frac{1}{5}\end{pmatrix}\right\}\)
• D. \(\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}\end{pmatrix}\right\}\)

Hint:

Work out $\langle v,v\rangle_P = v_1^2+9v_2^2+16v_3^2$ for each candidate vector and check it equals 1, then check the cross term is 0.

Answer 1

The Correct Option is C

Step 1: Understand the weighted inner product.
Here \(P=\text{diag}(1,9,16)\), so for vectors \(x=(x_1,x_2,x_3)\) and \(y=(y_1,y_2,y_3)\), the inner product is \(\langle x,y\rangle_P = x_1y_1 + 9x_2y_2 + 16x_3y_3\). A set is orthonormal for this inner product if each vector has \(\langle v,v\rangle_P=1\) and any two distinct vectors satisfy \(\langle v,w\rangle_P=0\).

Step 2: Normalize the first spanning vector.
Take \(v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\). Then \(\langle v_1,v_1\rangle_P = 1(1)(1)=1\), so \(v_1\) is already a unit vector under this inner product. Set \(e_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\).

Step 3: Apply Gram-Schmidt to the second spanning vector.
Take \(v_2=\begin{pmatrix}1\\1\\1\end{pmatrix}\). Its component along \(e_1\) is \(\langle v_2,e_1\rangle_P = 1(1)(1)+9(1)(0)+16(1)(0)=1\).
Subtract this component: \(w_2 = v_2 - 1\cdot e_1 = \begin{pmatrix}0\\1\\1\end{pmatrix}\).
Find its \(P\)-norm: \(\langle w_2,w_2\rangle_P = 1(0)^2+9(1)^2+16(1)^2 = 9+16=25\), so \(\|w_2\|_P=5\).
Normalizing, \(e_2 = \dfrac{w_2}{5} = \begin{pmatrix}0\\\frac{1}{5}\\\frac{1}{5}\end{pmatrix}\).

Step 4: Check the wrong options.
Option (A) uses \(w_2\) without dividing by its norm 5, and \(\langle w_2,w_2\rangle_P=25\ne 1\), so it is not normalized and is wrong.
Option (B) replaces \(e_1\) with \(\begin{pmatrix}0\\1\\0\end{pmatrix}\), which has \(\langle \cdot,\cdot\rangle_P = 9 \ne 1\), so it fails to be a unit vector and is wrong.
Option (D) divides the second vector by \(\sqrt2\) instead of 5, using ordinary Euclidean length instead of the \(P\)-weighted length, giving \(\langle \cdot,\cdot\rangle_P=9(\frac12)+16(\frac12)=12.5\ne 1\), so it is also wrong.

Final Answer:
The orthonormal basis with respect to \(\langle x,y\rangle_P\) is \(\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\frac{1}{5}\\\frac{1}{5}\end{pmatrix}\right\}\). \[ \boxed{\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\frac{1}{5}\\\frac{1}{5}\end{pmatrix}\right\}} \]