Sylvester Method

 

 

Sylvester's Formula

Suppose that A is a diagonalizable \( n \times n \) matrix; this means that all its eigenvalues are not defective and there exists a basis of n linearly independent eigenvectors. Now assume that its minimal polynomial (the polynomial of least possible degree that anihilates the matrix) is a product of simple terms:

\[ \psi (\lambda ) = \left( \lambda - \lambda_1 \right) \left( \lambda - \lambda_2 \right) \cdots \left( \lambda - \lambda_k \right) , \]
where \( \lambda_1 , \lambda_2 , \ldots , \lambda_k \) are distinct eigenvalues of the matrix A ( \( k \le n \) ).

Let \( f(\lambda ) \) be a function defined on the spectrum of the matrix A. The last condition means that every eigenvalue λi is in the domain of f, and that every eigenvalue λi with multiplicity mi > 1 is in the interior of the domain, with f being (mi — 1) times differentiable at λi. We build a function \( f\left( {\bf A} \right) \) of diagonalizable square matrix A according to James Sylvester, who was an English lawyer and music tutor before his appointment as a professor of mathematics in 1885. To define a function of a square matrix, we need to construct k Sylvester auxiliary matrices for each distinct eigenvalue \( \lambda_i , \quad i= 1,2,\ldots k: \)

\[ {\bf Z}_{\lambda_i} = {\bf Z}_{i} = \dfrac{\left( {\bf A} - \lambda_1 {\bf I} \right) \cdots \left( {\bf A} - \lambda_{i-1} {\bf I} \right) \left( {\bf A} - \lambda_{i+1} {\bf I} \right) \cdots \left( {\bf A} - \lambda_k {\bf I} \right)}{(\lambda_i - \lambda_1 ) \cdots (\lambda_i - \lambda_{i-1} ) (\lambda_i - \lambda_{i+1} ) \cdots (\lambda_i - \lambda_k )} . \]
Now we define the function \( f\left( {\bf A} \right) \) according to the formula:
\[ f\left( {\bf A} \right) = \sum_{i=1}^k f(\lambda_i )\,{\bf Z}_i . \]

Each Sylvester's matrix is a projection matrix on eigenspace of the corresponding eigenvalue.

Example. Consider a 2-by-2 matrix

\[ {\bf A} = \begin{bmatrix} 4&5 \\ -1& -2 \end{bmatrix} \]
that has two distinct eigenvalues \( \lambda_1 =3 \) and \( \lambda_2 =-1 . \) The eigenvectors corresponding these eigenvalues are
\[ {\bf v}_1 = \left\langle -5 , \ 1 \right\rangle^T \qquad {\bf v}_2 = \left\langle 1 , \ -1 \right\rangle^T , \]

respectively. Since the minimal polynomail is \( \psi (\lambda ) = (\lambda -3)(\lambda +1) \) is a product of simple factors, we build Sylvester's auxiliary matrices:
\[ {\bf Z}_3 = \frac{{\bf A} +1}{3+1} = \frac{1}{4} \begin{bmatrix} 5&5 \\ -1& -1 \end{bmatrix} \quad\mbox{and} \quad {\bf Z}_{-1} = \frac{{\bf A} -3}{-1-3} = \frac{1}{4} \begin{bmatrix} -1&-5 \\ 1&5 \end{bmatrix} . \]
These matrices are projectors on eigenspaces:
\[ {\bf Z}_3^2 = {\bf Z}_3 \qquad \mbox{and} \qquad {\bf Z}_{-1}^2 = {\bf Z}_{-1} . \]
Let us take an arbitrary vector, say \( {\bf x} = \langle 2, \ 2 \rangle^T , \) and expand it with respect to eigenvectors:
\[ {\bf x} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} = a\, {\bf v}_1 + b\,{\bf v}_2 = a \begin{bmatrix} -5 \\ 1 \end{bmatrix} + b \begin{bmatrix} 1 \\ -1 \end{bmatrix} , \]
where coefficients a and b need to be determined. Of course, we can find them by solving the system of linear equations:
\[ \begin{split} 2 &= a(-5) + b , \\ 2 &= a - b . \end{split} \]
However, the easiet way is to apply Sylvester's auxiliary matrices:
\begin{align*} {\bf Z}_3 \, {\bf x} &= \frac{1}{4} \begin{bmatrix} 5&5 \\ -1& -1 \end{bmatrix} \, \begin{bmatrix} 2\\ 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 5&5 \\ -1& -1 \end{bmatrix} \, \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 10 \\ 2 \end{bmatrix} = \begin{bmatrix} 5\\ -1 \end{bmatrix} = -{\bf v}_1 , \\ {\bf Z}_{-1}\, {\bf x} &= \frac{1}{4} \begin{bmatrix} -1&-5 \\ 1&5 \end{bmatrix} \, \begin{bmatrix} 2\\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ 3 \end{bmatrix} = -3\, {\bf v}_2 . \end{align*}
Therefore, coefficients are \( a =-1 \) and \( b =-3 . \) So we get the expansion:
\[ {\bf x} = \begin{bmatrix} 2 \\ 2 \end{bmatrix} = - {\bf v}_1 - 3\,{\bf v}_2 = - \begin{bmatrix} -5 \\ 1 \end{bmatrix} -3 \begin{bmatrix} 1 \\ -1 \end{bmatrix} . \]
We can define some functions, for instance, \( f(\lambda ) = e^{\lambda \,t} \) and \( g(\lambda ) = \cos \left( \lambda \,t\right) . \) Indeed,
\begin{align*} f({\bf A}) &= e^{3t}\, {\bf Z}_3 + e^{-t}\, {\bf Z}_{-1} = e^{3t}\, \frac{1}{4} \begin{bmatrix} 5&5 \\ -1& -1 \end{bmatrix} + e^{-t} \, \frac{1}{4} \begin{bmatrix} -1&-5 \\ 1&5 \end{bmatrix} , \\ g({\bf A}) &= \cos (3t) \, {\bf Z}_3 + \cos\, t\, {\bf Z}_{-1} = \frac{\cos \,3t}{4} \begin{bmatrix} 5&5 \\ -1& -1 \end{bmatrix} + \frac{\cos\, t}{4} \begin{bmatrix} -1&-5 \\ 1&5 \end{bmatrix} .\end{align*}
sage: M.rank()
sage: M.right_nullity()

Example 1.7.1: Consider the \( 3 \times 3 \) matrix \( {\bf A} = \begin{bmatrix} 1&4&16 \\ 18&20&4 \\ -12&-14&-7 \end{bmatrix} \) that has three distinct eigenvalues
A = {{1,4,16},{18,20,4},{-12,-14,-7}}
Eigenvalues[A]
Out[2]= 9, 4, 1
To determine a function of matrix, \( f ({\bf A} ) , \) we first find Sylvester's auxiliary matrices:
sage: M.rank()
sage: M.right_nullity()

 

Example 1.7.2: Consider the \( 3 \times 3 \) matrix \( {\bf A} = \begin{bmatrix} -20&-42&-21 \\ 6&13&6 \\ 12&24&13 \end{bmatrix} \) that has two distinct eigenvalues
Eigenvalues[A]
Out[2]= 4, 1, 1
Then we define the resolvent:
resolvent = FullSimplify[Inverse[lambda*IdentityMatrix[3] - A]]
Out[2]= {{(-25 + lambda)/( 4 - 5 lambda + lambda^2), -(42/(4 - 5 lambda + lambda^2)), -(21/( 4 - 5 lambda + lambda^2))}, {6/(4 - 5 lambda + lambda^2), ( 8 + lambda)/(4 - 5 lambda + lambda^2), 6/( 4 - 5 lambda + lambda^2)}, {12/(4 - 5 lambda + lambda^2), 24/( 4 - 5 lambda + lambda^2), (8 + lambda)/(4 - 5 lambda + lambda^2)}}
Using the resolvent, we define Sylvester's auxiliary matrices (that are actiually projector matrices on eigenspace)
Z1 = FullSimplify[(lambda - 1)*resolvent] /. lambda -> 1
Out[3]= {{8, 14, 7}, {-2, -3, -2}, {-4, -8, -3}}
Z4 = FullSimplify[(lambda - 4)*resolvent] /. lambda -> 4
Out[4]= {{-7, -14, -7}, {2, 4, 2}, {4, 8, 4}}
sage: M.rank()
sage: M.right_nullity()

 

Example 1.7.3:
sage: M.rank()
sage: M.right_nullity()