Eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}2 & 5\\-2 & -4\end{array}\right]$$$

Find the eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}2 & 5\\-2 & -4\end{array}\right]$$$.

Start from forming a new matrix by subtracting $$$\lambda$$$ from the diagonal entries of the given matrix: $$$\left[\begin{array}{cc}2 - \lambda & 5\\-2 & - \lambda - 4\end{array}\right]$$$.

The determinant of the obtained matrix is $$$\lambda^{2} + 2 \lambda + 2$$$ (for steps, see determinant calculator).

Solve the equation $$$\lambda^{2} + 2 \lambda + 2 = 0$$$.

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

These are the eigenvalues.

Next, find the eigenvectors.

• $$$\lambda = -1 - i$$$

  $$$\left[\begin{array}{cc}2 - \lambda & 5\\-2 & - \lambda - 4\end{array}\right] = \left[\begin{array}{cc}3 + i & 5\\-2 & -3 + i\end{array}\right]$$$

  The null space of this matrix is $$$\left\{\left[\begin{array}{c}- \frac{3}{2} + \frac{i}{2}\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).

  This is the eigenvector.

• $$$\lambda = -1 + i$$$

  $$$\left[\begin{array}{cc}2 - \lambda & 5\\-2 & - \lambda - 4\end{array}\right] = \left[\begin{array}{cc}3 - i & 5\\-2 & -3 - i\end{array}\right]$$$

  The null space of this matrix is $$$\left\{\left[\begin{array}{c}- \frac{3}{2} - \frac{i}{2}\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).

  This is the eigenvector.

Eigenvalue: $$$-1 - i$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}- \frac{3}{2} + \frac{i}{2}\\1\end{array}\right] = \left[\begin{array}{c}-1.5 + 0.5 i\\1\end{array}\right]$$$A.

Eigenvalue: $$$-1 + i$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}- \frac{3}{2} - \frac{i}{2}\\1\end{array}\right] = \left[\begin{array}{c}-1.5 - 0.5 i\\1\end{array}\right]$$$A.