Jorge Valle

Finding the eigenvalues of a matrix

Finding the eigenvalues of a matrix can be done by solving the characteristic equation.

Matrix used

Here's the matrix I will be referring to.

$$ A = \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix} $$
Figure 1: The matrix we're using.

Solving the characteristic equation

To find the eigenvalues of the above A, we solve the characteristic equation, which is defined as follows.

$$ \det(A - \lambda I) = 0 $$
Figure 2: Definition of characteristic equation.

In this case, $I$ is simply the identity matrix. Swapping in $A$ and $I$ yields the following.

$$ \det( \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} ) = 0 \\ $$
Figure 3: Let's swap in the identity matrix.

$1$ times $λ$ is just $λ$, so we can clarify further.

$$ \det( \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} ) = 0 $$
Figure 4: Further clarifying.

Then, we do the simple subtraction.

$$ \det( \begin{bmatrix} 2 - \lambda & 3 \\ 3 & 2 - \lambda \end{bmatrix} ) = 0 $$
Figure 5: We subtract.

Remembering how to find the determinant, we can then express this as follows.

$$ \begin{align} ((2 - \lambda) \times (2 - \lambda)) - (3 \times 3) & = 0 \\ 4 - 2\lambda - 2\lambda + \lambda^2 - 9 & = 0 \\ -5 - 4\lambda + \lambda^2 & = 0 \end{align} $$
Figure 6: Finding the determinant.

This can be rearranged to a tidier expression.

$$ \lambda^2 - 4\lambda - 5 = 0 $$
Figure 7: Cleaning up.

This is now the characteristic polynomial, with solutions at 5 and -1: these are the eigenvalues of $A$.

Numpy implementation

This is the method for returning eigenvalues in Numpy. Note that the similar numpy.linalg.eig() method returns both the eigenvalues and the eigenvectors.