Tool to calculate the characteristic polynomial of a matrix. The characteristic polynomial of a matrix M is computed as the determinant of (X.I-M).
Characteristic Polynomial of a Matrix - dCode
Tag(s) : Matrix
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The characteristic polynomial (or sometimes secular function) $ P $ of a square matrix $ M $ of size $ n \times n $ is the polynomial defined by $$ P_M(x) = \det(M - x.I_n) \tag{1} $$ or $$ P_M(x) = \det(x.I_n - M) \tag{2} $$ with $ I_n $ the identity matrix of size $ n $ (and det the matrix determinant).
The 2 possible values $ (1) $ and $ (2) $ give opposite results, but since the polynomial is used to find roots, the sign does not matter.
The equation $ P = 0 $ is called the characteristic equation of the matrix.
The characteristic polynomial $ P $ of a matrix, as its name indicates, characterizes a matrix, it allows in particular to calculate the eigenvalues and the eigenvectors.
If $ M $ is a diagonal matrix with $ \lambda_1, \lambda_2, \ldots, \lambda_n $ as diagonal elements, then the computation is simplified and $$ P_M(x) = (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_n) $$
If $ M $ is a triangular matrix with $ \lambda_1, \lambda_2, \ldots, \lambda_n $ as diagonal elements, then as for diagonal matrix, the computation is simplified and $$ P_M(x) = (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_n) $$
The calculation of the characteristic polynomial of a square matrix of order 2 can be calculated with the determinant of the matrix $ [ x.I_2 - M ] $ as $$ P_M(x) = \det [ x.I_2 - M ] $$
The polynomial can also be written with another formula using the trace of the matrix $ M $ (noted Tr): $$ P_{M_2}(x) = \det( x.I_2 - M ) = x^2 - \operatorname{Tr}(M)x+ \det(M) $$
Example: $$ M=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \\ \Rightarrow x.I_n - M = \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} x-1 & -2 \\ -3 & x-4 \end{bmatrix} \\ \Rightarrow \det(x.I_n - M) = (x-1)(x-4)-((-2)\times(-3)) \\ \Rightarrow P_M(x) = x^2-5x-2 $$
Calculation of the characteristic polynomial of a square 3x3 matrix can be calculated with the determinant of the matrix $ [ x.I_3 - M ] $ as $$ P_M(x) = \det [ x.I_3 - M ] $$
Example: $$ M = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} $$ $$ [ x.I_3 - M ] = x \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - M = \begin{bmatrix} x-a & -b & -c \\ -d & x-e & -f \\ -g & -h & x-i \end{bmatrix} $$ $$ P_M(x) = \det [ x.I_3 - M ] = -a e i+a e x+a f h+a i x-a x^2+b d i-b d x-b f g-c d h+c e g-c g x+e i x-e x^2-f h x-i x^2+x^3 $$
It is also possible to use another formula with the Trace of the matrix $ M $ (noted Tr): $$ P_{M_3}(x) = -x^3 + \operatorname{Tr}(M)x^2 + \frac{1}{2} \left( \operatorname{Tr}^2(M) - \operatorname{Tr}(M^2) \right) x + \frac{1}{6} \left( \operatorname{Tr}^3(M) + 2\operatorname{Tr}(M^3) - 3\operatorname{Tr}(M)\operatorname{Tr}(M^2) \right) $$
Calculation of the characteristic polynomial of an order 4 square matrix can be calculated with the determinant of the matrix $ [ x.I_4 - M ] $ as $$ P_M(x) = \det [ x.I_4 - M ] $$
Example: $$ M = \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix} $$ $$ [ x.I_4 - M ] = x \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} - M = \begin{bmatrix} x-a & b & c & d \\ e & x-f & g & h \\ i & j & x-k & l \\ m & n & o & x-p \end{bmatrix} $$ $$ P_M(x) = \det [ x.I_4 - M ] = a f k p-a f k x-a f l o-a f p x+a f x^2-a g j p+a g j x+a g l n+a h j o-a h k n+a h n x-a k p x+a k x^2+a l o x+a p x^2-a x^3-b e k p+b e k x+b e l o+b e p x-b e x^2+b g i p-b g i x-b g l m-b h i o+b h k m-b h m x+c e j p-c e j x-c e l n-c f i p+c f i x+c f l m+c h i n-c h j m+c i p x-c i x^2-c l m x-d e j o+d e k n-d e n x+d f i o-d f k m+d f m x-d g i n+d g j m-d i o x+d k m x-d m x^2-f k p x+f k x^2+f l o x+f p x^2-f x^3+g j p x-g j x^2-g l n x-h j o x+h k n x-h n x^2+k p x^2-k x^3-l o x^2-p x^3+x^4 $$
The characteristic polynomial is unique for a given matrix. There is only one way to calculate it and it has only one result.
On the other hand, two different matrices can give the same characteristic polynomial.
A matrix $ M $ and its matrix transpose $ M^T $ have the same characteristic polynomial.
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