Find the eigenvectors of the given matrix.

Question

$$A=\begin{bmatrix} i & 0 & 0\\ 0 & 0 & i\\ 0&i&0 \end{bmatrix}$$ The eigenvalues are i and -i where $i=(-1)^{1/2}$. The eigenvectors corresponding to the eigenvalue i calculated by Mathematica are $\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}$. However, I want to ask if I can replace $\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$ with $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ and declare $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}$ as the two eigenvectors since both of these vectors satisfy the equation $Ax=i x$ and form a linearly independent set.

Answer 1

Yes!, you can add two eigenvectors, since the Eigenspace is a subspace.

Indeed, the Eigenspace corresponding to the eigenvalue $i$ is, $E_i=\{x : (A-iI)x=0\}$ where $x=\begin{pmatrix} x_1 & x_2 & x_3 \end{pmatrix}^T $

Solving $(A-iI)x=0$, we get $x_2=x_3$

So, $$E_i=\Bigg\{x_1 \begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}+x_2 \begin{pmatrix} 0\\ 1 \\ 1 \end{pmatrix}: x_1,x_2 \in F \Bigg\} $$

By choosing $x_1=x_2=1$, you get the mentioned vector of $1$'s !