I'm new to linear algebra and I'm having a hard time getting past this problem:
given the basis vectors in $\Bbb{R}^4$
$v^{(0)} = \begin{bmatrix} 0.5 \\ 0.5 \\ 0.5 \\ 0.5 \\ \end{bmatrix}, \\ v^{(1)} = \begin{bmatrix} 0.5 \\ 0.5 \\ -0.5 \\ -0.5 \\ \end{bmatrix}, \\ v^{(2)} = \begin{bmatrix} 0.5 \\ -0.5 \\ 0.5 \\ -0.5 \\ \end{bmatrix}$
which are mutually orthonormal, find a vector $v^{(3)}$ to form the complete basis $V = \{v^{(0)}, v^{(1)}, v^{(2)}, v^{(3)}\} $
In $\Bbb{R}^3$ I could use a cross product to find the third orthogonal vector, but in $\Bbb{R}^4$ I have no idea how to approach this. I think the vector should be something like this:
$v^{(3)} = \begin{bmatrix} -0.5 \\ -0.5 \\ 0.5 \\ 0.5 \\ \end{bmatrix}$
I just don't know how to actually arrive at this mathematically.
First, let's find a vector that is linearly independent form $ v^{(0)},v^{(1)},v^{(2)}$. You can easily check that $ u = (-1,1,1,1)^{T}$ satisfies this condition. In fact resolving the linear system $\alpha v^{0} +\beta v^{1} + \gamma v^{2} + \delta u = 0$ you'll find $\alpha = \beta = \gamma = \delta = 0$.
Then, in order to use Gram-Schimdt process, compute the scalar products with the orthonormal vectors you already have. In this case $u \cdot v^{(0)} = 1$ and $ u \cdot v^{(1)} = u \cdot v^{(2)} = -1$.
The vector defined by $ u-v^{(0)}+v^{(1)}+v^{(2)}$ in therefore orthogonal to $v^{(0)}, \dots ,v^{(2)}$ by construction and normalizing you get $v^{(3)} = \frac{1}{2}(-1,1,1,-1)^{T}$.