Eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}17 & 3\\3 & 9\end{array}\right]$$$

Find the eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}17 & 3\\3 & 9\end{array}\right]$$$.

Start from forming a new matrix by subtracting $$$\lambda$$$ from the diagonal entries of the given matrix: $$$\left[\begin{array}{cc}17 - \lambda & 3\\3 & 9 - \lambda\end{array}\right]$$$.

The determinant of the obtained matrix is $$$\left(\lambda - 18\right) \left(\lambda - 8\right)$$$ (for steps, see determinant calculator).

Solve the equation $$$\left(\lambda - 18\right) \left(\lambda - 8\right) = 0$$$.

The roots are $$$\lambda_{1} = 18$$$, $$$\lambda_{2} = 8$$$ (for steps, see equation solver).

These are the eigenvalues.

Next, find the eigenvectors.

• $$$\lambda = 18$$$

  $$$\left[\begin{array}{cc}17 - \lambda & 3\\3 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}-1 & 3\\3 & -9\end{array}\right]$$$

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

  This is the eigenvector.

• $$$\lambda = 8$$$

  $$$\left[\begin{array}{cc}17 - \lambda & 3\\3 & 9 - \lambda\end{array}\right] = \left[\begin{array}{cc}9 & 3\\3 & 1\end{array}\right]$$$

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

  This is the eigenvector.

Eigenvalue: $$$18$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}3\\1\end{array}\right]$$$A.

Eigenvalue: $$$8$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}- \frac{1}{3}\\1\end{array}\right]\approx \left[\begin{array}{c}-0.333333333333333\\1\end{array}\right]$$$A.