# 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
6 years, 9 months ago

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