Where have I made my mistake in calculating $P^{-1}AP$?

Question

I have

$$Q = \begin{pmatrix} -\mu & \mu \\ \lambda & -\lambda \end{pmatrix}$$

and I want to work out the value of $\mathbb{P}(t) = \exp(Qt)$

So I diagonalised $Q$ and then worked out the exponential of the diagonal matrix. I got this to be:

$${Q}t = \pmatrix{-\mu t &\mu t \\ \lambda t & -\lambda t} = \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1}^{-1} \cdot \pmatrix{0 & 0 \\ 0 & -t(\lambda + \mu)} \cdot \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1}.$$

So using the middle matrix, I got

$$\exp (Qt) = \pmatrix{1 & 0 \\ 0 & \exp(-t(\lambda + \mu))} = \pmatrix{1 & 0 \\ 0 & \exp(T)},$$

where $T = -t(\lambda + \mu)$.

Then, using $\exp (P^{-1}AP) = P^{-1}e^AP$, I was supposed to get

$$\mathbb{P}(t) = \exp({Q} t) = \frac{1}{\lambda + \mu}\pmatrix{\lambda + \mu \exp(T) & \mu - \mu \exp(T) \\ \lambda - \lambda\exp(T) & \mu + \lambda\exp(T)}.$$

This is what I did. To first work out $P^{-1}$ I got

$$P^{-1} = \frac{1}{1 + \frac{\mu}{\lambda}} \pmatrix{1 & \frac{\mu}{\lambda} \\ -1 & 1} = \frac{\lambda}{\lambda + \mu} \pmatrix{1 & \frac{\mu}{\lambda} \\ -1 & 1}$$

Then doing $P^{-1}e^A$ gave me

$$\frac{\lambda}{\lambda + \mu} \pmatrix{1 & \frac{\mu}{\lambda} \exp (T) \\ -1 & \exp (T)}$$

Then doing this times $P$ gave me

$$\frac{\lambda}{\lambda + \mu} \pmatrix{1 + \frac{\mu}{\lambda} \exp (T) & -\frac{\mu}{\lambda} + \frac{\mu}{\lambda} \exp (T) \\ -1 + \exp (T) & \frac{\mu}{\lambda} + \exp (T)}$$

Multiplying through by $\lambda$ gives me

$$\frac{\lambda}{\lambda + \mu} \pmatrix{\lambda + \mu \exp (T) & - \mu - \mu \exp (T) \\ - \lambda + \exp(T) & \mu + \lambda \exp (T)}$$

Clearly it's started going wrong in the matrix before this but I can't see where I've made my mistakes. Can someone help please?

Thank you

Answer 1

When you calculate $\exp(tQ)$, you don't need to diagonalize $tQ$. Instead, you can diagonalize $Q=V\Lambda V^{-1}$, where the column vectors of $V$ are the eigenvectors of $Q$. (When you don't remember whether $Q=V\Lambda V^{-1}$ or $Q=V^{-1}\Lambda V$ is correct, think about $QV=V\Lambda $.)

Once you have $Q=V\Lambda V^{-1}$, $tQ=V(t\Lambda) V^{-1}$. Then you would get $$ \exp(tQ)=V\exp(t\Lambda)V^{-1} $$ where $\exp(t\Lambda)$ is easy to get.

For a degenerate $2\times 2$ matrix $Q$, you don't need to calculate the eigenvalues by the characteristic polynomial. One of them must be $\lambda_1=0$, and another one is $\lambda_2=\operatorname{trace}(Q)$.

Answer 2

Kaish in in first post you write correct, now in this post not. $Qt=PAP^{-1}$. So my hint must be: you can use $e^{PAP^{-1}}=Pe^AP^{-1}$

Calculating the determinant we have

$(-\mu-A)(-\lambda-A)-\mu\lambda=0$ $\implies$ $A^2+(\mu+\lambda)A=0$ $\implies$ A=0 or $A=-(\mu+\lambda)$

can you find the eigenvector from here?