# Still stuck on orthogonalisation

#### Hamlings

Hi all

Still stuck on a problem involving Gram-Schidmt process, to find an orthogonal basis for T that includes the vector (1, 0, 0, 2).

set T = {(x, y, z, 2x + y − 3z) : x, y, z ∈ R}.

Thanks.

#### Greens

First, we need a basis for $T$. This can be found by finding the matrix representation of $T$ and using the linearly independent columns.

We know $T: \mathbb{R}^3 \mapsto \mathbb{R}^4$. Let $\alpha$ and $\beta$ be the standard bases for $\mathbb{R}^3$ and $\mathbb{R}^4$ respectively.

$\displaystyle [T]_{\alpha}^{\beta} = \left( \; \; [T(1,0,0)]_{\beta} \; \; [T(0,1,0)]_{\beta}\; \; [T(0,0,1)]_{\beta} \; \; \right)$

$\displaystyle = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 2 & 1 & -3 \end{bmatrix}$

This matrix is already in Echelon Form. Since the columns are linearly independent, we know that those column vectors form a basis for $T$

For the G.S process, we are seeking the orthogonal basis $\{w_1, w_2 , w_3\}$. We start with the required vector $(1,0,0,2)$, which is conveniently included in the basis.

$\displaystyle w_1 = v_1 = (1,0,0,2)$

$\displaystyle w_2 = (0,1,0,1) - \frac{2}{5}(1,0,0,2) = \left(\frac{-2}{5} , 1 , 0 , \frac{1}{5} \right)$

$\displaystyle w_3 = (0,0,1,-3) - \frac{-6}{5}(1,0,0,2) - \frac{-3}{30}(-2,5,0,1) = \left(1,\frac{1}{2},1,\frac{-1}{2} \right)$

#### Hamlings

Thank you very much.