If we allow for complex eigenvalues, our previous statement that a square nxn matrix always has $\leq$ becomes false. If we allow for complex Eigenvalues, then we find that every nxn matrix has n eigenvalues (don't have to be unique, but they have n of them).

For example, if we allow $A = \begin{bmatrix} 0&-1\\1&0 \end{bmatrix}$, we can see that its eigenvalues cannot have real values.

\begin{align} \lambda^2 + 1 = 0 \\ \lambda = \imath \text{ and } -\imath \end{align}

If we are in $C^2$ (the complex plane if dimension 2), we can construct the plane with the vectors $\begin{bmatrix} 1\\ \imath \end{bmatrix}$ and $\begin{bmatrix} 1\\ -\imath \end{bmatrix}$. If we allow A to operate on $C^2$, we get

\begin{align} \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} \begin{bmatrix} 1\\ \imath \end{bmatrix} = \begin{bmatrix} -\imath \\ 1 \end{bmatrix} = -\imath \begin{bmatrix} 1\\ \imath \end{bmatrix}.\\ \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} \begin{bmatrix} 1\\-\imath \end{bmatrix} = \begin{bmatrix} \imath\\1 \end{bmatrix} = \imath \begin{bmatrix} 1\\-\imath \end{bmatrix}. \end{align}

So $\imath$ and $-\imath$ have the eigenvectors $\begin{bmatrix} 1\\-\imath \end{bmatrix}$ and $\begin{bmatrix} 1\\ \imath \end{bmatrix}$ RESPECTIVELY.

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