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I have the following system of ODEs

$$ \begin{cases} x' = -ax \\ y' = -ay \\ \end{cases} \implies \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -a & 0 \\ 0 & -a \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$

its eigenvalue $\lambda = -a$, and substituting it to $(A - \lambda I)\mathbf{k} = \mathbf{0}$:

$$ (A - \lambda I)k = \begin{bmatrix} -a - (-a) & 0 \\ 0 & -a - (-a) \end{bmatrix} \begin{bmatrix} k_1 \\ k_2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} k_1 \\ k_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \ , $$

which implies that all vectors in $\mathbb{R}^2 \ (\neq \mathbf{0})$ is an eigenvector of $\lambda = -a$.

How do I write its homogeneous solution? Is it just $\mathbf{v} = \mathbf{k}e^{-at}$, where $\mathbf{k}$ is any vector in $\mathbb{R}^2$? But I believe there must be more terms since $\lambda$ is a repeated eigenvalue but I do not know what it could be.

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  • $\begingroup$ I'm not sure what you mean by 'which implies all vectors in $\mathbb{R}^{2}$ is an eigenvector of $\lambda = -a$'. I read it as there are two associated eigenvectors in the eigenspace, $\{ (1, 0)^{T}, (0, 1)^{T} \}$, which you would expect from looking at the ODEs themselves. $\endgroup$ Apr 13, 2021 at 3:10
  • $\begingroup$ Since any vectors in $\mathbb{R}^2$ make the equation true. $\endgroup$
    – sinclair
    Apr 13, 2021 at 3:13
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    $\begingroup$ I understand now, though I'm not sure what you mean by 'there must be more terms because $\lambda$ is a repeated eigenvalue'. The general solution is as $\color{red}{\text{you have written it}}$, and is the same as $\color{blue}{\text{my solution}}$ $$\boldsymbol{v}=\color{red}{(k_{1},k_{2})^{T}e^{-at}}=\color{blue}{k_{1}(1,0)^{T}e^{-at}+k_{2}(0,1)^{T}e^{-at}}$$ The specific values of $k_{1},k_{2}$ are picked out by the initial conditions for the ODEs at time $t = t^{*}$ $$\boldsymbol{v}(t^{*}) = (k_{1},k_{2})^{T} e^{-at^{*}} = (x_{0}, y_{0}) \implies (k_{1},k_{2}) = (x_{0}, y_{0})e^{at^{*}}$$ $\endgroup$ Apr 13, 2021 at 3:53
  • $\begingroup$ @mattos I see, thanks. $\endgroup$
    – sinclair
    Apr 13, 2021 at 4:51

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