$$A=\begin{bmatrix} i & 0 & 0\\ 0 & 0 & i\\ 0&i&0 \end{bmatrix}$$ 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.
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- $\begingroup$ Indeed, you can. $\endgroup$Abezhiko– Abezhiko2023-10-14 11:07:38 +00:00Commented Oct 14, 2023 at 11:07
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1 Answer
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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 !
- $\begingroup$ So can I use {1,1,1} and {0,1,1} as the 2 eigenvectors of the matrix for eigenvalue i? $\endgroup$TshrD23– TshrD232023-10-15 15:40:01 +00:00Commented Oct 15, 2023 at 15:40