Question

consider the vector space of R³

2.1. Show that <x, y>=x₁y₁+2x₂y₂+x₃y₃, x=$\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix}$ ,y=$\begin{bmatrix} y1 \\ y2 \\ y3 \end{bmatrix}$$\in$ R³

^ 2.2. Are the vectors $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \\ - 1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$ Linearly independent?

2.3. Apply the Grem-Schmidt process to the following subset of R³ $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 1 \\ - 1 \end{bmatrix}$, $\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$ to find an orthogonal basis with the inner product defined in 2.1. for the span of this subset

Answer 1

2.

$\begin{vmatrix} 1 & 1& 1\\ 1 & 1&-1\\ 1&-1&-1 \end{vmatrix}=-2-2=-4$

Since determinant is not equal zero, the vectors are linearly independent.

3.

orthogonal basis:

$v_1=u_1=(1,1,1)$

$v_2=u_2-\frac{\langle u_1,u_2\rangle}{|u_1|^2}u_1=(1,1,-1)-\frac{2}{3}(1,1,1)=(1/3,1/3,-5/3)$

$v_3=u_3-\frac{\langle u_1,u_3\rangle}{|u_1|^2}u_1-\frac{\langle u_3,v_2\rangle}{|v_2|^2}v_2=(1,-1,-1)+\frac{2}{3}(1,1,1)-\frac{4}{9}(1/3,1/3,-5/3)=$

$=(5/3,-1/3,11/27)$

1.

Let $(e_1,e_2,e_3)$ be a basis.

Then:

$x=x_1e_1+x_2e_2+x_3e_3$

$y=y_1e_1+y_2e_2+y_3e_3$

$\langle x,y \rangle =\sum_i\sum_j x_iy_i\langle e_i,e_j\rangle$

if $\langle e_1,e_1\rangle=1,\langle e_1,e_2\rangle=\langle e_1,e_3\rangle=\langle e_2,e_3\rangle=0,\langle e_2,e_2\rangle=2,\langle e_3,e_3\rangle=1$

then $\langle x,y \rangle=$ x₁y₁+2x₂y₂+x₃y₃