Eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}5 & 11\\11 & 25\end{array}\right]$$$
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Find the eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}5 & 11\\11 & 25\end{array}\right]$$$.
Solution
Start from forming a new matrix by subtracting $$$\lambda$$$ from the diagonal entries of the given matrix: $$$\left[\begin{array}{cc}5 - \lambda & 11\\11 & 25 - \lambda\end{array}\right]$$$.
The determinant of the obtained matrix is $$$\lambda^{2} - 30 \lambda + 4$$$ (for steps, see determinant calculator).
Solve the equation $$$\lambda^{2} - 30 \lambda + 4 = 0$$$.
The roots are $$$\lambda_{1} = 15 - \sqrt{221}$$$, $$$\lambda_{2} = \sqrt{221} + 15$$$ (for steps, see equation solver).
These are the eigenvalues.
Next, find the eigenvectors.
$$$\lambda = 15 - \sqrt{221}$$$
$$$\left[\begin{array}{cc}5 - \lambda & 11\\11 & 25 - \lambda\end{array}\right] = \left[\begin{array}{cc}-10 + \sqrt{221} & 11\\11 & 10 + \sqrt{221}\end{array}\right]$$$
The null space of this matrix is $$$\left\{\left[\begin{array}{c}- \frac{10 + \sqrt{221}}{11}\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).
This is the eigenvector.
$$$\lambda = \sqrt{221} + 15$$$
$$$\left[\begin{array}{cc}5 - \lambda & 11\\11 & 25 - \lambda\end{array}\right] = \left[\begin{array}{cc}- \sqrt{221} - 10 & 11\\11 & 10 - \sqrt{221}\end{array}\right]$$$
The null space of this matrix is $$$\left\{\left[\begin{array}{c}\frac{-10 + \sqrt{221}}{11}\\1\end{array}\right]\right\}$$$ (for steps, see null space calculator).
This is the eigenvector.
Answer
Eigenvalue: $$$15 - \sqrt{221}\approx 0.133931252681494$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}- \frac{10 + \sqrt{221}}{11}\\1\end{array}\right]\approx \left[\begin{array}{c}-2.260551704301682\\1\end{array}\right]$$$A.
Eigenvalue: $$$\sqrt{221} + 15\approx 29.866068747318506$$$A, multiplicity: $$$1$$$A, eigenvector: $$$\left[\begin{array}{c}\frac{-10 + \sqrt{221}}{11}\\1\end{array}\right]\approx \left[\begin{array}{c}0.442369886119864\\1\end{array}\right]$$$A.