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.