Question

Obtain an orthogonal basis with respect to the standard inner product for the subspace of R4 defined by { (x,y,w,z)| x+2y+z+3w=0 , x-y-z=0}.

Answer 1

Answer on Question#40156 - Math - Linear Algebra:

Obtain an orthogonal basis with respect to the standard inner product for the subspace of $\mathbb{R}^4$ defined by $\{(x,y,z,w) \mid x + 2y + z + 3w = 0, x - y - z = 0\}$.

Solution.

Let $S = \{(x, y, z, w) \mid x + 2y + z + 3w = 0, x - y - z = 0\}$. Hence:

$$\left\{ \begin{array}{c} x + 2y + z + 3w = 0 \\ x - y - z = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} 3y + 2z + 3w = 0 \\ x = y + z \end{array} \right. \Rightarrow \left\{ \begin{array}{c} z = -\frac{3}{2}(y + w) \\ x = y + z \end{array} \right. \Rightarrow \left\{ \begin{array}{c} z = -\frac{3}{2}(y + w) \\ x = -\frac{1}{2}(y + 3w) \end{array} \right.;$$

So:

$$S = \left\{ \left( -\frac{y}{2} - \frac{3w}{2}, y, -\frac{3y}{2} - \frac{3w}{2}, w \right) \mid y, w \in \mathbb{R} \right\} = \{ yA + wB \mid y, w \in \mathbb{R} \},$$ $$\text{where } A = \left( -\frac{1}{2}, 1, -\frac{3}{2}, 0 \right), B = \left( -\frac{3}{2}, 0, -\frac{3}{2}, 1 \right);$$

We can conclude that:

$$S = V(A, B) \Rightarrow \dim(S) = 2;$$

$\{A, B\}$ is a basis in $S$. Use Gram-Schmidt process to get an orthogonal basis $\{A', B'\}$:

$$A' = A = \left( -\frac{1}{2}, 1, -\frac{3}{2}, 0 \right);$$ $$B' = B - \frac{(A, B)}{(A, A)} A = \left( -\frac{3}{2}, 0, -\frac{3}{2}, 1 \right) - \frac{\frac{3}{4} + 0 + \frac{9}{4} + 0}{\frac{1}{4} + 1 + \frac{9}{4} + 0} \left( -\frac{1}{2}, 1, -\frac{3}{2}, 0 \right) =$$ $$= \left( -\frac{3}{2}, 0, -\frac{3}{2}, 1 \right) - \frac{6}{7} \left( -\frac{1}{2}, 1, -\frac{3}{2}, 0 \right) = \left( -\frac{15}{14}, -\frac{6}{7}, -\frac{3}{14}, 1 \right);$$

Hence, $\left\{ \left( -\frac{1}{2}, 1, -\frac{3}{2}, 0 \right), \left( -\frac{15}{14}, -\frac{6}{7}, -\frac{3}{14}, 1 \right) \right\}$ is an orthogonal basis in $S$.