Use the Gram-Schmidt Process to construct an orthogonal set of vectors using an inner product

Question

This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.

Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:

Column vectors: $[1, 2]$, $[1, 1]$

Inner Product: $\langle u, v \rangle = u^TAv,$

where $A= \begin{bmatrix}6&4\\4&6\end{bmatrix}$.

Answer 1

Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$: $$ u_2 = v_2 - \mathsf{proj}_{u_1}(v_2) = v_2 - \frac{\langle v_2,u_1\rangle}{\langle u_1,u_1\rangle}u_1. $$ We compute $$ \langle v_2,u_1\rangle = \pmatrix{1&1}\pmatrix{6&4\\4&6}\pmatrix{1\\2}= 30 $$ and $$ \langle u_1,u_1\rangle = \pmatrix{1&2}\pmatrix{6&4\\4&6}\pmatrix{1\\2}= 46, $$ so $$ u_2 = (1,1) - \frac{15}{23}\cdot(1,2) = \left(\frac8{23}, -\frac7{23}\right). $$

Answer 2

Let $V$ be an inner product space, with the inner product defined as $\langle u, v \rangle = u^T A v$, where $A= \begin{bmatrix}6&4\\4&6\end{bmatrix}$. The set $\{ (1, 2), (1, 1) \}$ is a basis for $V$, since the vectors are linearly independent and span $V$.

Let $v_1 = (1, 2)$.

$v_2 = (1, 1) - \dfrac{\langle (1 , 1), (1, 2) \rangle}{\langle (1, 2), (1, 2) \rangle} v_1 = \dfrac{\langle (1 , 1), (1, 2) \rangle}{\langle (1, 2), (1, 2) \rangle} (1, 2) = (1,1) - \dfrac{30}{46}(1, 2) = \left( \dfrac{8}{23} , -\dfrac{7}{23} \right)$

Therefore, $\left \{ (1, 2), \left( \dfrac{8}{23} , -\dfrac{7}{23} \right) \right \}$ is an orthogonal basis for $V$.

Check:

$\left \langle (1, 2), \left( \dfrac{8}{23}, \dfrac{-7}{23} \right) \right \rangle = \dfrac{112}{23} - \dfrac{112}{23} = 0$, as required.

Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf