Question

Suppose A=PDP^-1, where Dis defined as in (9). Use (3)to show that e^At=Pe^DtP^-1.

Short answer

The result for the given problem is $ePD{P}^{-1}t=I+PDP{P}^{-1}t+P{D}^2{P}^{-1}\frac{t^2}{2!}+......+P{D}^K{P}^{-1}\frac{t^k}{k!}+......=P{e}^{Dt}{P}^{-1}$. The result is verified.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

where I is the unit matrix of order n.

Therefore the infinite matrix power series is

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula,

Verification of (PDP-1)k =PDkP-1.

We are given that,

A=PDP^-1 (1)

and,

$e^{At}=I+At+A^2\frac{t^2}{2!}+....+A^k\frac{t^k}{k!}+.....\left(2\right)$

Substituting (1) into (2).

$e^{PD{P}^{-1}t}=I+PD{P}^{-1}t+{\left(PD{P}^{-1}\right)}^2\frac{t2}{2!}+......+{\left(PD{P}^{-1}\right)}^k\frac{tk}{k!}+......$

Observe that,

$$\begin{gathered} {\left(PD{P}^{-1}\right)}^2=PD{P}^{-1}.PD{P}^{-1}=P{D}^2{P}^{-1} \\ {\left(PD{P}^{-1}\right)}^3={\left(PD{P}^{-1}\right)}^2.PD{P}^{-1}=P{D}^2{P}^{-1}PD{P}^{-1}=P{D}^3{P}^{-1} \\ . \\ . \\ ({PD{P}^{-1})}^k=P{D}^k{P}^{-1} \end{gathered}$$

Hence it is verified that $({PD{P}^{-1})}^k=P{D}^k{P}^{-1}$.

Verification of eAt= PeDtP-1.

From the above step we know,

$e^{PD{P}^{-1}t}=I+PDP{P}^{-1}t+{\left(PD{P}^{-1}\right)}^2\frac{t2}{2!}+......+{\left(PD{P}^{-1}\right)}^k\frac{tk}{k!}+......$

Therefore,

$$\begin{gathered} e^{PD{P}^{-1}t}=I+PDP{P}^{-1}t+P{D}^2{P}^{-1}\frac{t^2}{2!}+......+P{D}^k{P}^{-1}\frac{t^k}{k!}+...... \\ =P\left[1+Dt+D\frac{t^2}{2!}+....+D^k\frac{t^k}{k!}+...\right]P^{-1} \\ =P{e}^{Dt}{P}^{-1} \end{gathered}$$

Hence it is verified that e^At= Pe^DtP^-1.