Eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}1 & 1 - i\\1 + i & 0\end{array}\right]$$$

Find the eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}1 & 1 - i\\1 + i & 0\end{array}\right]$$$.

Start from forming a new matrix by subtracting $$$\lambda$$$ from the diagonal entries of the given matrix: $$$\left[\begin{array}{cc}1 - \lambda & 1 - i\\1 + i & - \lambda\end{array}\right]$$$.

The determinant of the obtained matrix is $$$\left(\lambda - 2\right) \left(\lambda + 1\right)$$$ (for steps, see determinant calculator).

Solve the equation $$$\left(\lambda - 2\right) \left(\lambda + 1\right) = 0$$$.

The roots are $$$\lambda_{1} = 2$$$, $$$\lambda_{2} = -1$$$ (for steps, see equation solver).

These are the eigenvalues.

Next, find the eigenvectors.

• $$$\lambda = 2$$$

  $$$\left[\begin{array}{cc}1 - \lambda & 1 - i\\1 + i & - \lambda\end{array}\right] = \left[\begin{array}{cc}-1 & 1 - i\\1 + i & -2\end{array}\right]$$$

  The null space of this matrix is $$$\left\{\left[\begin{array}{c}1 - i\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).

  This is the eigenvector.

• $$$\lambda = -1$$$

  $$$\left[\begin{array}{cc}1 - \lambda & 1 - i\\1 + i & - \lambda\end{array}\right] = \left[\begin{array}{cc}2 & 1 - i\\1 + i & 1\end{array}\right]$$$

  The null space of this matrix is $$$\left\{\left[\begin{array}{c}- \frac{1}{2} + \frac{i}{2}\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).

  This is the eigenvector.

Eigenvalue: $$$2$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}1 - i\\1\end{array}\right]$$$A.

Eigenvalue: $$$-1$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}- \frac{1}{2} + \frac{i}{2}\\1\end{array}\right] = \left[\begin{array}{c}-0.5 + 0.5 i\\1\end{array}\right]$$$A.