# Thread: Vector valued DE system

1. ## Vector valued DE system

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

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

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

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

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

2. ## Re: Vector valued DE system

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

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

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

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

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

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

4. ## Re: Vector valued DE system

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