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Matrix Functions

Let \(J\) be the Jordan matrix \[J = \begin{bmatrix} \lambda & 1 & & \\ & \lambda & 1 & & \\ & & \ddots & \ddots & \\ & & & \lambda & 1 \\ & & & & \lambda \\ \end{bmatrix} = \lambda {I}_{n} + {B}_{n}\]

where

\[{B}_{n} = \begin{bmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ & & & & 0 \\ \end{bmatrix}\]

Determine the matrix \(f\left(J\right)\) for some differentiable function \(f\).

Solution

Observations

If we experiment by exponentiating the matrix \({B}_{n}\), we will discover two properties:

1) \[{{B}_{n}}^{2} = \begin{bmatrix} 0 & 0 & 1 & & \\ & 0 & 0 & 1 & \\ & & \ddots & \ddots \\ & & & & & 1 \\ & & & & 0 &0 \\ & & & & 0 &0 \\ \end{bmatrix}\]

and the process continues until

\[{{B}_{n}}^{n-1} = \begin{bmatrix} & & & & 1 \\ & & & & \\ & & & & & \\ & & & & & \\ & & & & & \\ & & & & & \\ \end{bmatrix}\]

2) \({{B}_{n}}^{k} = {0}_{n}\) for \(k \ge n\).

Hence \({B}_{n}\) is called the shift matrix (you shift the 1-diagonal, building up a trail of zeroes the more you exponentiate). Property 2 illustrates the intrinsic nilpotent property of shift matrices.

Step 1

The exponent of \(J\) is computed as follows: \[{J}^{k} = ({\lambda {I}_{n} + {B}_{n}})^{k} = {\lambda}^{k}{I}_{n} + \left( \begin{matrix} k \\ 1 \end{matrix} \right) {\lambda}^{k-1}{B}_{n} + \left( \begin{matrix} k \\ 2 \end{matrix} \right) {\lambda}^{k-2} {{B}_{n}}^{2} + ...\]

(essentially the binomial expansion).

To illustrate the mechanics: \[{J}^{k} = \begin{bmatrix} {\lambda}^{k} & & & \\ &{\lambda}^{k} & & \\ & & \ddots & & \\ & & & {\lambda}^{k} & \\ & & & & {\lambda}^{k} \\ \end{bmatrix} + \begin{bmatrix} 0 & k {\lambda}^{k-1} & & & \\ & 0 & k {\lambda}^{k-1} & & \\ & & \ddots & \ddots & \\ & & & 0 & k {\lambda}^{k-1} \\ & & & & 0 \\ \end{bmatrix} + ... \]

\[ = \begin{bmatrix} {\lambda}^{k} & k {\lambda}^{k-1} & \cdots & \left( \begin{matrix} k \\ n-1 \end{matrix} \right) {\lambda}^{k-n+1} \\ &{\lambda}^{k} &\ddots &\left( \begin{matrix} k \\ n-2 \end{matrix} \right) {\lambda}^{k-n+2} \\ & &\ddots &\vdots \\ & & & k {\lambda}^{k-1} \\ & & & {\lambda}^{k} \\ \end{bmatrix}\]

Step 2

We expand the matrix function \[f(J) = f(\eta){I}_{n} + f'(\eta)J +\frac{1}{2!} f''(\eta){J}^{2} + ...\]

for some point \(\eta\) where the derivatives of \(f\) (with respect to \(\lambda\)) exist.

We substitute the exponents of \(J\) (giving it the shifting property) to the expanded matrix function \(f\), which is actually a sum of matrices.

\[f(J) = \begin{bmatrix} f(\eta) & & & \\ &f(\eta) & & \\ & & \ddots & & \\ & & & f(\eta) & \\ & & & & f(\eta) \\ \end{bmatrix} + \begin{bmatrix} f'(\eta)\lambda &f'(\eta) & & \\ & f'(\eta)\lambda & f'(\eta) & \\ & & \ddots & \ddots & \\ & & & f'(\eta) &\\ & & & f'(\eta)\lambda \\ \end{bmatrix} + ... \]

Collapsing this sum will yield the matrix

\[f(J) = \begin{bmatrix} f(\lambda) & f'(\lambda) & \cdots & \frac{{f}^{(n-1)}(\lambda)}{(n-1)!} \\ &f(\lambda t) &\cdots &\frac{{f}^{(n-2)}(\lambda)}{(n-2)!} \\ & &\ddots & \vdots \\ & & & f'(\lambda) \\ & & & f(\lambda) \\ \end{bmatrix}\]

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 9 months ago

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