Determine the matrice of $e^{At}$ with respect to a base in $\mathbb{R}^{6}$

Question

Let $A$ be a constant matrice in $\mathbb{R}^{6\times 6}$, $v_i$, $1\leq i \leq 6$ linearly independent vectors and $\lambda_{i}$, $1\leq i\leq 6$ real constants such that $Av_1=\lambda_1v_1$, $Av_2=\lambda_2v_1$, $Av_3=\lambda_3v_3+\lambda_4v_4$, $Av_4=-\lambda_4v_3+\lambda_3v_4$, $Av_5=\lambda_5v_5-\lambda_6v_6$, $Av_6=\lambda_6v_5+\lambda_5v_6$. I am asked, as the title says, to find the matrice of $e^{At}$ with respect to a base in $\mathbb{R}^{6}$.

The base that I have considered is obviously the standard basis.

I have to calculate $e^{At}v_i=\sum_{k=0}^{\infty}\frac{t^kA^kv_i}{k!}$, $\forall i \in \{1,2,\dots,6\}$. It is pretty easy to find $e^{At}v_1$ and $e^{At}v_2$, because $$e^{At}v_1=\sum_{k=0}^{\infty}\frac{t^kA^kv_1}{k!}=Id\cdot v_1+tAv_1+\frac{t^2A^2v_1}{2!}+\dots+\frac{t^nA^nv_1}{n!}+\dots$$ but since we know that $Av_1=\lambda_1v_1$, we can deduce that $A^2v_1=A\lambda_1v_1=\lambda_1Av_1=\lambda_1^{2}v_1$; in general, $A^kv_1=\lambda_1^{k}v_1$, and so the above expression gets simplified to $$e^{At}v_1=\sum_{k=0}^{\infty}\frac{t^kA^kv_1}{k!}=v_1+\lambda_1tv_1+\frac{\lambda_1^{2}}{2!}t^2v_1+\dots =e^{t\lambda_1}v_1=\begin{pmatrix}e^{t\lambda_1}\\0\\0\\0\\0\\0\end{pmatrix}$$ doing a similar exercise for $e^{At}v_2$, we get (correct me if I'm wrong) $$e^{At}v_2=\begin{pmatrix}\lambda_2(e^{\lambda_1t}-1)\\1\\0\\0\\0\\0\end{pmatrix}$$ So as a summary, I have that $$e^{At}=\phi(t)=\begin{pmatrix}e^{\lambda_1t}&\lambda_2(e^{\lambda_1t}-1)&\\0&1&\\0&0&\\0&0&\dots\\0&0&\\0&0&\end{pmatrix}$$ The problem comes with the other vectors of the basis. Since $Av_3$ depends on $v_3$ an also $v_4$ the products of the power series $A^{i}v_3$ they all have a part which depends on $v_3$ and $v_4$ $$A^iv_3=f_i(\lambda_3,\lambda_4)\cdot v_3+h_i(\lambda_3,\lambda_4)\cdot v_4$$ this also happens with $Av_4$. This is what I have come up with at the moment. We know $Av_3=\lambda_3v_3+\lambda_4v_4$, I will try to deduce $A^{k}v_3$, $\forall k\in\mathbb{N}$ $$A^2v_3=A(\lambda_3v_3+\lambda_4v_4)=\dots=(\lambda_3^2-\lambda_4^2)v_3+2\lambda_3\lambda_4v_4$$ $$A^3v_3=A((\lambda_3^2-\lambda_4^2)v_3+2\lambda_3\lambda_4v_4)=\dots=\lambda_3(\lambda_3-3\lambda_4^2)v_3+\lambda_4(3\lambda_3^2-\lambda_4^2)v_4$$ $$A^4v_3=\dots=(\lambda_3^2(\lambda_3-3\lambda_4^2)-\lambda_4^2(3\lambda_3^2-\lambda_4^2))v_3+(\lambda_4\lambda_3(\lambda_3-3\lambda_4^2)+\lambda_3\lambda_4(3\lambda_3^2-\lambda_4^2))v_4$$ I cannot find a closed from for $A^kv_3$, $\forall k\in \mathbb{N}$, not to talk about the other products. Can anyone help me find $e^{At}$?

Thank you, any help will be very much appreciated. The exercise also asks me to find the solution to the Cauchy Problem $x'=Ax, \hspace{2mm}x(0)=v_2+v_3$ and $x'=Ax+b(t), \hspace{2mm}x(0)=v_2+v_3$ where $b(t)=tv_1+v_2$ which in the first case I should use the fact that if the system is homogenious the solution is given by $$\varphi(t)=e^{At}x(0)$$ and in the second case (which the system is not homogenious), we use $$\varphi(t)=e^{At}[x(0)+\int_{0}^{t}e^{-As}b(s)ds]$$ am I right?

Answer 1

I think the following works although there might be a different shortcut. Write your third and fourth system as: $$\begin{cases} (A-\lambda_3 I)v_3 = &\lambda_4 v_4 \\ (A-\lambda_3 I)v_4 = -&\lambda_4v_3 \end{cases}$$ which inductively implies: $$(A-\lambda_3I)^{2k+1}v_3 = \lambda_4^{2k+1}(-1)^kv_4 \text{ and } (A-\lambda_3 I)^{2k} = \lambda_4^{2k}(-1)^kv_3.$$ Then after you do the computation, I think you will end up having: $$e^{At}v_3 = e^{\lambda_3t}(\sin(t\lambda_4)v_4 + \cos(t\lambda_4)v_3)$$ and you can do a similar thing with the remaining two columns.