I've two questions in this case. One is technical and the other one is a little more philosophical.
I've the following autonomous system of linear differential equations: $$ \dot{x} = A_{\infty}x$$ where the spectrum of $A$ (i.e. the set of its eigenvalues) is given by $\{s_i\}_{i=1}^{\infty}\subset \mathbb{C}^{0}$ where $\mathbb{C}^0$ is the set of pure imaginary complex numbers (zero real part). In addition, $s_l \neq s_p \,\,\forall \,\,l,p \in\{1,2, \dots\}$, i.e. each geometric multiplicity is equal to 1.
I would like to prove that I can always write my system as: $$ \dot{x}=\begin{bmatrix} A_\nu & 0 \\ 0 & \hat{A}_{\infty} \end{bmatrix} \begin{bmatrix} x_{\nu} \\ \hat{x}\end{bmatrix}$$ where $A_{\nu}$ is a finite-dimensional matrix with spectrum $\{s_i\}_{i=1}^{\nu}$ and $\hat{A}_{\infty}$ an infinite-dimensional matrix with spectrum $\{s_i\}_{i=\nu+1}^{\infty}$ ($x_{\nu}$ and $\hat{x}$ accordingly).
So, here comes the question:
- First of all, can I call $A_{\infty}$ an "infinite-dimensional matrix"? Or is this extremely wrong and I should not call it "matrix" at all?
- Secondly, If $A_{\infty}$ is finite-dimensional, then splitting the system as in the second equation can always be done, since the geometric multiplicity and algebraic multiplicity of each eigenvalue are always 1 and hence, it is diagonalizable. Can I say the same for this infinite-dimensional case??
Thanks!!!!!
You’re supposing that your infinite dimensional space is having a countable dimension. But it could even have a dimension of highest cardinal than the countable one.
Apart from that, you can always find a basis with your eigenvectors and complements it to a basis of the full space. With this, you can write your linear application in two parts as you did. Even if you don’t use a matrix notation.