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Eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}\frac{4}{5} & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10}\end{array}\right]$$$

The calculator will find the eigenvalues and eigenvectors of the square $$$2$$$x$$$2$$$ matrix $$$\left[\begin{array}{cc}\frac{4}{5} & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10}\end{array}\right]$$$, with steps shown.

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Find the eigenvalues and eigenvectors of $$$\left[\begin{array}{cc}\frac{4}{5} & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10}\end{array}\right]$$$.

Solution

Start from forming a new matrix by subtracting $$$\lambda$$$ from the diagonal entries of the given matrix: $$$\left[\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right]$$$.

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

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

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

These are the eigenvalues.

Next, find the eigenvectors.

  • $$$\lambda = 1$$$

    $$$\left[\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right] = \left[\begin{array}{cc}- \frac{1}{5} & \frac{3}{10}\\\frac{1}{5} & - \frac{3}{10}\end{array}\right]$$$

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

    This is the eigenvector.

  • $$$\lambda = \frac{1}{2}$$$

    $$$\left[\begin{array}{cc}\frac{4}{5} - \lambda & \frac{3}{10}\\\frac{1}{5} & \frac{7}{10} - \lambda\end{array}\right] = \left[\begin{array}{cc}\frac{3}{10} & \frac{3}{10}\\\frac{1}{5} & \frac{1}{5}\end{array}\right]$$$

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

    This is the eigenvector.

Answer

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

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