There is no linear operator $T$ on $\mathbb{C}^2$ st $(\alpha|T(\alpha)=0$ for all the vectors

Question

I have been able to prove this but my question is what does it signify geometrically?

Here inner product is the standard inner product. Does it signify there is no linear operator $T$ such that it can rotate the vector $\alpha$ in 90 degrees?Why does it happen? How do I generalize this?

Answer 1

For $\mathbb{R}^2,$ the rotation $T = \begin{pmatrix} 0 & 1\\-1& 0 \end{pmatrix}$ is the only solution, but it is easy to show that this does not work on $\mathbb{C}^2.$ I assume this is your solution, and what it shows is that, indeed, a ninety degree rotation does not map every complex vector to an orthogonal one. This is because the matrix is actually diagonalizable over the complex numbers (eigenvalues $\pm i.$)