Eigenvalues and eigenvectors of [31014]\left[\begin{array}{cc}3 & -10\\1 & -4\end{array}\right]

The calculator will find the eigenvalues and eigenvectors of the square 22x22 matrix [31014]\left[\begin{array}{cc}3 & -10\\1 & -4\end{array}\right], with steps shown.

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Find the eigenvalues and eigenvectors of [31014]\left[\begin{array}{cc}3 & -10\\1 & -4\end{array}\right].

Solution

Start from forming a new matrix by subtracting λ\lambda from the diagonal entries of the given matrix: [3λ101λ4]\left[\begin{array}{cc}3 - \lambda & -10\\1 & - \lambda - 4\end{array}\right].

The determinant of the obtained matrix is (λ1)(λ+2)\left(\lambda - 1\right) \left(\lambda + 2\right) (for steps, see determinant calculator).

Solve the equation (λ1)(λ+2)=0\left(\lambda - 1\right) \left(\lambda + 2\right) = 0.

The roots are λ1=1\lambda_{1} = 1, λ2=2\lambda_{2} = -2 (for steps, see equation solver).

These are the eigenvalues.

Next, find the eigenvectors.

  • λ=1\lambda = 1

    [3λ101λ4]=[21015]\left[\begin{array}{cc}3 - \lambda & -10\\1 & - \lambda - 4\end{array}\right] = \left[\begin{array}{cc}2 & -10\\1 & -5\end{array}\right]

    The null space of this matrix is {[51]}\left\{\left[\begin{array}{c}5\\1\end{array}\right]\right\} (for steps, see null space calculator).

    This is the eigenvector.

  • λ=2\lambda = -2

    [3λ101λ4]=[51012]\left[\begin{array}{cc}3 - \lambda & -10\\1 & - \lambda - 4\end{array}\right] = \left[\begin{array}{cc}5 & -10\\1 & -2\end{array}\right]

    The null space of this matrix is {[21]}\left\{\left[\begin{array}{c}2\\1\end{array}\right]\right\} (for steps, see null space calculator).

    This is the eigenvector.

Answer

Eigenvalue: 11A, multiplicity: 11A, eigenvector: [51]\left[\begin{array}{c}5\\1\end{array}\right]A.

Eigenvalue: 2-2A, multiplicity: 11A, eigenvector: [21]\left[\begin{array}{c}2\\1\end{array}\right]A.