# Thread: Vector valued DE system

1. ## Vector valued DE system

In the last equal sign of the last line below, why is $\displaystyle - \lambda z = \ddot{z}$? Thanks.

Given the DE system
$\displaystyle \ddot{\mathbf{x}}=-kA\mathbf{x}$

where $\displaystyle \mathbf{x}$ is n-vector,
$\displaystyle A$ is an nxn matrix, and
$\displaystyle \mathbf{v}$ is a an eigenvector of A,

let $\displaystyle \mathbf{x}=z\mathbf{v}$, then

$\displaystyle \ddot{\mathbf{x}}=-kA\mathbf{x}=-kA(z\mathbf{v})=-k(A\mathbf{v})z=-k(\lambda\mathbf{v})z=k(-\lambda z)\mathbf{v}=k\ddot{z}\mathbf{v}$.

2. ## Re: Vector valued DE system

Originally Posted by MSUMathStdnt
In the last equal sign of the last line below, why is $\displaystyle - \lambda z = \ddot{z}$? Thanks.

Given the DE system
$\displaystyle \ddot{\mathbf{x}}=-kA\mathbf{x}$

where $\displaystyle \mathbf{x}$ is n-vector,
$\displaystyle A$ is an nxn matrix, and
$\displaystyle \mathbf{v}$ is a an eigenvector of A,

let $\displaystyle \mathbf{x}=z\mathbf{v}$, then

$\displaystyle \ddot{\mathbf{x}}=-kA\mathbf{x}=-kA(z\mathbf{v})=-k(A\mathbf{v})z=-k(\lambda\mathbf{v})z=k(-\lambda z)\mathbf{v}=k\ddot{z}\mathbf{v}$.
Look back to the first of the line. What the first to the next to last term shows is that $\displaystyle k(-\lamba z)\mathbf{v}= \ddot{\mathbf{x}}$$\displaystyle = \ddot{\mathbf{zv}}= k\ddot{z}v because v is a constant. 3. ## Re: Vector valued DE system Originally Posted by HallsofIvy Look back to the first of the line. What the first to the next to last term shows is that \displaystyle k(-\lamba z)\mathbf{v}= \ddot{\mathbf{x}}$$\displaystyle = \ddot{\mathbf{zv}}= k\ddot{z}v$ because v is a constant.
I'm not sure what you're trying to tell me. I understand that $\displaystyle \mathbf{v}$ is constant. But how does what you wrote mean that $\displaystyle - \lambda z = \ddot{z}$?

Is it because $\displaystyle - \lambda z = -\lambda I_n z$, ($\displaystyle I$ being the identity matrix), which would make the right side of this into the left side of the given system with $\displaystyle \lambda$ as $\displaystyle k$ and $\displaystyle I$ as $\displaystyle A$?

4. ## Re: Vector valued DE system

Yes,. if for any (non-zero) vector, v, av= bv, then a= b.