I need guidance in solving this problem: Find an orthogonal basis for $\mathbb{R}^3$ containing the set $S = \{v_1, v_2\}$ where $v_1 = \begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}$ and $v_2 =
\begin{bmatrix}
0 \\
-1 \\
2
\end{bmatrix}, \langle v_1, v_2\rangle := a_1b_1 + 4a_2b_2 + a_3b_3$
My approach:
Applying the Gram-Schmidt Method,
\begin{align*}
x_1 &= v_1\\
x_2 &= v_2 – \dfrac{{v_2}^Tx_1}{{x_1}^Tx_1}x_1 = \begin{bmatrix}
0 \\
-1 \\
2
\end{bmatrix} – \dfrac{\begin{bmatrix}
0 & -1 & 2
\end{bmatrix}\begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}}{\begin{bmatrix}
-4 & 1 & 2
\end{bmatrix}\begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}}\begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}\\
&= \begin{bmatrix}
0 \\
-1 \\
2
\end{bmatrix} – \dfrac{3}{21}\begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}\\
&= \begin{bmatrix}
0 \\
-1 \\
2
\end{bmatrix} – \begin{bmatrix}
-4/7 \\
1/7 \\
2/7
\end{bmatrix} = \begin{bmatrix}
4/7 \\
-8/7 \\
12/7
\end{bmatrix}\\\\
\text{Orthogonal basis } &= \text{ span}\left\{x_1, x_2\right\}\\
&= \text{ span}\left\{\begin{bmatrix}
-4 \\
1 \\
2
\end{bmatrix}, \begin{bmatrix}
4/7 \\
-8/7 \\
12/7
\end{bmatrix}\right\}
\end{align*}
Is my solution and conclusion correct?
Best Answer
With respect to the given inner product, you have $\langle v_1,v_2\rangle=0$; in other words, they're orthogonal.
So, find a vector$$u=\begin{bmatrix}a\\b\\c\end{bmatrix}$$which is orthogonal to both and which os not the null vector. That is, solve the system$$\left\{\begin{array}{l}\langle v_1,u\rangle=0\\\langle v_2,u\rangle=0.\end{array}\right.$$Every solution is of the form$$u=\begin{bmatrix}2c\\c\\2c\end{bmatrix}$$and you can take, say,$$u=\begin{bmatrix}2\\1\\2\end{bmatrix}.$$