transformation (1,0) and (-1,0) to the orthogonal vectors using unitary matrices Does a unitary matrix, witch transform $\begin{pmatrix} 1\\0 \end{pmatrix}$ and $\begin{pmatrix} -1\\0 \end{pmatrix}$ to the orthogonal vectors,  exists? If it is needed more dimensions can be added.
 A: Assuming that “unary” means “unitary”, the answer is negative. Let $U$ be a $2\times2$ unitary matrix. Then $U.\binom10$ and $U.\binom{-1}0$ are symmetric non-null vectors and therefore they cannot be orthogonal.
A: As José Carlos Santos has noted, that the vectors are parallel (antiparallel in fact) prevents a unitary transformation of them being orthogonal. In fact, the inner product preservation $(Ua)^\dagger Ub=a^\dagger b$ proves a more general result: the new vectors are orthogonal iff the old ones are.
A: If you are free to change representation and increase dimensionality then yes you can. For example $$\cases{[1,0]^T\to[1,0,0,0]^T\\ [0,1]^T\to[0,0,1,0]^T\\ [-1,0]^T\to[0,1,0,0]^T\\ [0,-1]^T\to[0,0,0,1]^T}$$
This may look strange, but it really isn't so weird. If you just analyze how matrix multiplication works on the matrix $$\left[\begin{array}{cc}a&b\\b&a\end{array}\right]$$
b acts as negative part of real and a as positive in the sense that $$\left[\begin{array}{cc}0&b\\b&0\end{array}\right]^2 = \left[\begin{array}{cc}b^2&0\\0&b^2\end{array}\right]$$
or in other words $b^2= (-b)^2$
$$\left[\begin{array}{cc}0&b\\b&0\end{array}\right]\left[\begin{array}{cc}a&0\\0&a\end{array}\right] = \left[\begin{array}{cc}0&ab\\ab&0\end{array}\right]$$
or in other words $(-b)a = -ab$
When you start adding numbers, that's when the interesting stuff starts. You will need to build further linear combinations of the matrix elements to preserve uniqueness of the representation, but you actually can keep it 100% linear in the new bigger space.
