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I have the following exercise which I have problems with

Exercise

Use the fact that the eigenvalues of a square matrix $A$, are solutions of the characteristic equation to show that $e^{\lambda_i t}$ for all eigenvalues $\lambda = \lambda_1, \lambda_2 \cdots \lambda_n$ can be expressed as $$e^{\lambda t} = \alpha_0(t)+\alpha_1(t)\lambda+\cdots+\alpha_{n-1}(t)\lambda^{n-1}$$

Solution

If the matrix $A$ has eigenvalues $λ = λ_1, λ_2,\cdots,λ_n$, with the characteristic polynomial $$\lambda^{n}+a_{n-1}\lambda^{n-1}+\cdots+a_0=0$$

the eigenvalues of $At$ are $λt$ and are solutions of $$ (\lambda t)^n +(a_{n-1}t)(\lambda t)^{n-1}+\cdots+a_0 t^n = 0 $$ The expansion of $e^{λt}$ is $$ e^{\lambda t} = 1+\lambda t+\frac{1}{2}(\lambda t)^2 +\cdots+\frac{1}{k!}(\lambda t)^k, ~~~~ k \rightarrow \infty $$ Using the characteristic polynomial, the term $(λt)^n$ can be replaced by an expression where the highest power of $(λt)$ is $n − 1$ $$(\lambda t)^n = -((a_{n-t}t)(\lambda t)^{n-1}+\cdots +a_0t^n = E_{n-1}((\lambda t)^{n-1}) $$ Similarly $(\lambda t)^{n+1}$ is $$ (\lambda t)^{n+1}=E_n((\lambda t)^n) = E_n(E_{n-1}((\lambda t)^{n-1})) $$ All further terms can be replaced by expressions where the highest power of $λt$ is $n − 1$, so $$e^{\lambda t} = \alpha_0 (t)+\alpha_1(t)\lambda+\cdots+\alpha_{n-1}(t)\lambda^{n-1}$$ The $\alpha$'s are functions of $t$ because each coefficient of $(λt)^k$ in the expressions $E$ involves $t$.

Question

What do they mean with expansion of $e$?

What does the capital letter $E(())$ stand for?

What does they mean with "can be replaced by expressions where the highest power of $λt$ is $n−1$"?

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  • $\begingroup$ The expansion of $e^{\lambda t}$ is just the Taylor series. The $E$ is just trying to be shorthand to say that $(\lambda t)^n$ can be expressed as a polynomial in terms less than $n$. Any term of index $\geq n$ can be reduced (iteratively), and hence the result of a polynomial of degree $n-1$ $\endgroup$ – Scott Burns Dec 9 '16 at 0:32
  • $\begingroup$ Could you please explain to me how they get from $E_n((\lambda t)^n)$ to $E_n(E_{n-1}((\lambda t)^{n-1}))$ ? $\endgroup$ – autoship Dec 9 '16 at 9:42
  • $\begingroup$ Yes, it's just a substitution. $(\lambda t)^n = E_{n-1}((\lambda t)^{n-1})$. This is not the greatest notation - it would be better read as $E_{n-1}(\lambda t)$ $\endgroup$ – Scott Burns Dec 9 '16 at 17:40
  • $\begingroup$ Thanks for the help. Could you help me with what they mean with "All further terms can be replaced by expressions where the highest power of λt is n−1"? I do not understand the step in between. $\endgroup$ – autoship Dec 10 '16 at 19:41

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