Characteristic polynomial of a square matrix

The characteristic polynomial of a square matrix is defined as the determinant of the matrix $ M $ of order $ n $ minus $ \lambda $ times the identity matrix of the same order, denoted as $ Id_n $. It is expressed as: $$ p_M (\lambda) = \det(M - \lambda \cdot Id_n) $$

The characteristic polynomial of a matrix is an essential tool for finding its eigenvalues.

Example

Consider, for example, a 2x2 matrix $ M $:

$$ M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} $$

To determine the characteristic polynomial of this matrix, we substitute $ M $ and the 2x2 identity matrix $ Id_2 $ into the formula:

$$ p_M (\lambda) = \det(M - \lambda \cdot Id_2) $$

$$ p_M (\lambda) = \det \left( \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \lambda \cdot Id_2 \right) $$

$$ p_M (\lambda) = \det \left( \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \lambda \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) $$

After multiplying $ \lambda $ by the identity matrix, we get:

$$ p_M (\lambda) = \det \left( \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} \right) $$

$$ p_M (\lambda) = \det \begin{pmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{pmatrix} $$

Now, expanding the determinant yields:

$$ p_M (\lambda) = (1 - \lambda)(4 - \lambda) - (2 \cdot 3) $$

$$ p_M (\lambda) = 4 - 4 \lambda - \lambda + \lambda^2 - 6 $$

which simplifies to:

$$ p_M (\lambda) = \lambda^2 - 5\lambda + 2 $$

The characteristic polynomial for matrix $ M $ can now be used to find its eigenvalues.