Matrix Functions

Let JJ be the Jordan matrix J=[λ1λ1λ1λ]=λIn+BnJ = \begin{bmatrix} \lambda & 1 & & \\ & \lambda & 1 & & \\ & & \ddots & \ddots & \\ & & & \lambda & 1 \\ & & & & \lambda \\ \end{bmatrix} = \lambda {I}_{n} + {B}_{n}

where

Bn=[0101010]{B}_{n} = \begin{bmatrix} 0 & 1 & & & \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ & & & & 0 \\ \end{bmatrix}

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

Solution

Observations

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

1) Bn2=[00100110000]{{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

Bnn1=[1]{{B}_{n}}^{n-1} = \begin{bmatrix} & & & & 1 \\ & & & & \\ & & & & & \\ & & & & & \\ & & & & & \\ & & & & & \\ \end{bmatrix}

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

Hence Bn{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 JJ is computed as follows: Jk=(λIn+Bn)k=λkIn+(k1)λk1Bn+(k2)λk2Bn2+...{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: Jk=[λkλkλkλk]+[0kλk10kλk10kλk10]+...{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} + ...

=[λkkλk1(kn1)λkn+1λk(kn2)λkn+2kλk1λk] = \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(η)In+f(η)J+12!f(η)J2+...f(J) = f(\eta){I}_{n} + f'(\eta)J +\frac{1}{2!} f''(\eta){J}^{2} + ...

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

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

f(J)=[f(η)f(η)f(η)f(η)]+[f(η)λf(η)f(η)λf(η)f(η)f(η)λ]+...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)=[f(λ)f(λ)f(n1)(λ)(n1)!f(λt)f(n2)(λ)(n2)!f(λ)f(λ)]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
4 years, 12 months ago

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