Find all vectors in $\mathbb{R}^5$ such that $S$ is an orthogonal set. Express answer as a linear span.

Question

Find all vectors (x1, x2, x3, x4, x5) in $\mathbb{R}^5$ such that S = {(1, 1, 1, 2, 0) , (0, 0, −2, 1, 0) , (0, 0, 0, 0, 1) , (x1, x2, x3, x4, x5)} is an orthogonal set. Express your answer as a linear span.

Clue was given to form a homogeneous system and solve it. I have tried solving

$$ \left[\begin{array}{rrrrr|r} 1 & 1 & 1 & 2 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0\\ \end{array}\right] $$

Attempt at a solution:

$$ x_2 = s ,\,\,\, x_4 = t , \,\, where\,\, s,t \in \mathbb{R}, \\ x_1 = -s - \frac{5t}{2} , \, \, \, x_3 = \frac{t}{2}, \,\,\,x_5 = 0 \\\therefore\\(x_1,x_2,x_3,x_4,x_5) = (-s - \frac{5t}{2},\,\,\,s,\,\,\,\frac{t}{2},\,\,\,t,\,\,\,0) $$

But how do I answer in terms of linear span form here on?

Any help would be greatly appreciated !

EDIT:

With help from @Gerry's comments I got to

$$ (x_1,x_2,x_3,x_4,x_5) = span{\{(-1,1,0,0,0),(\frac{-5}{2},0, \frac{1}{2},1,0)\} } $$

Answer 1

My path to getting a solution.

We know that for orthogonal sets, any distinct pair has their dot product equals to zero ie $ u_i.u_j = 0 \,\, for \,\,\,i \neq j $

The matrix way of doing that is simply $$ u_iu_j^T $$ Set up the system $$ \left[\begin{array}{rrrrr} 1 & 1 & 1 & 2 & 0 \\ 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{rrrrr} x_1 & x_2 & x_3 & x_4 & x_5 \\ \end{array}\right]^T = 0 $$

Solve augmented matrix $$ \left[\begin{array}{rrrrr|r} 1 & 1 & 1 & 2 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0\\ \end{array}\right] $$ Results $$ x_2 = s ,\,\,\, x_4 = t , \,\, where\,\, s,t \in \mathbb{R}, \\ x_1 = -s - \frac{5t}{2} , \, \, \, x_3 = \frac{t}{2}, \,\,\,x_5 = 0 \\\therefore\\(x_1,x_2,x_3,x_4,x_5) = (-s - \frac{5t}{2},\,\,\,s,\,\,\,\frac{t}{2},\,\,\,t,\,\,\,0) $$ Convert to linear span => Solution : $$ (x_1,x_2,x_3,x_4,x_5) = span{\{(-1,1,0,0,0),(\frac{-5}{2},0, \frac{1}{2},1,0)\} } $$

Answer 2

The vector $\begin{bmatrix}1&-1&0&0&0\end{bmatrix}$ is obviously orthogonal to all vectors given. Then using the adjugate matrix, as in this answer, we get a last orthogonal vector, $\begin{bmatrix}5&5&-2&-4&0\end{bmatrix}$.

Thus, an orthogonal basis for the subspace orthogonal to the three vectors given in the question is $$ \begin{bmatrix}1&-1&0&0&0\\5&5&-2&-4&0\end{bmatrix} $$