Silly question: Why is it in the diagonalization process we can choose our eigenvector to be anything we want?

Question

Suppose we want to diagonalize the matrix $A = \begin{bmatrix}0& 1\\ -1& 0\end{bmatrix}$

Then using $det(\lambda I - A) = 0$

We find our eigenvalues to be $\lambda_{1,2} = \pm j$

Solving for the $\lambda_1 = j$ eigenvector, we find that

$V_{1} = [V_{11}, V_{12}] = [V_{11}, jV_{11}]$, where $V_1$ is the eigenvector corresponding to the eigenvalue $\lambda_1$

Now my professor simply equates 1 to be $V_{11}$ to get $V_1 = [1, j]$

WHAT? But why!? Can someone explain what is the logical mental process that says you can so freely choose $V_{11}$ to be 1?

Answer 1

There are infinitely many eigenvectors corresponding to one eigenvalue, just check:
If $x$ is an eigenvector corresponding to an eigenvalue $\lambda$ of a matrix $A$, then isn't $\mu x$ also an eigenvector corresponding to $\lambda$?