Systems of differential eqns (including eigenvalues and eigenvectors)

[hannahszx]

I am struggling with understanding this question and I am unsure where to start. Any help would be great.

 

[anon_hedgepig]

Here's a hint: What you want to do is find the two variables x and y, so form a column vector with x and y, as well as with x' and y', such that your vector X comprises x and y. Then, use what you know about matrix multiplication (in reverse) to rewrite the system and determine what matrix A is.

 

[HallsofIvy]

You have the two linear differential equations,
\(\displaystyle \frac{dx}{dt}= -7x+ 7y\)
\(\displaystyle \frac{dy}{dt}= 9x- 5y\).

Now, do you know what matrices are and how to multiply matrices? If so then you know that \(\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}ax+ by \\ cx+ dy\end{bmatrix}\).

Comparing those, you should see that you can write the two equations as
\(\displaystyle \frac{d\begin{bmatrix} x \\ y\end{bmatrix}}{dt}= \begin{bmatrix}-7x+ 7y \\ 9x- 5y \end{bmatrix}= \begin{bmatrix}-7 & 7 \\9 & -5\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}\).

So \(\displaystyle A= \begin{bmatrix}-7 & 7 \\9 & -5\end{bmatrix}\).

"Eigenvalues" for a matrix, A, are numbers, \(\displaystyle \lambda\), such that \(\displaystyle Av= \lambda v\) for some non-zero v. That means that \(\displaystyle Av- \lambda v= (A- \lambda I)v= 0\) has a non-zero solution. Obviously if \(\displaystyle A- \lambda\) has an inverse, we could multiply by that inverse to get v= 0. Since there are non-zero solutions, there must not be an inverse so the determinant must be 0.

We must have \(\displaystyle \left|\begin{array}{cc}-7-\lambda & 7 \\ 9 & -5-\lambda \end{array}\right|= (-7-\lambda)(-5-\lambda)- 63= \lambda^2+ 12\lambda- 28= 0\).
That can be factored as \(\displaystyle (\lambda- 2)(\lambda+ 14)= 0\) so the eigenvalues are \(\displaystyle \lambda= 2\) and \(\displaystyle \lambda= -14\).

Because 2 is an eigenvalue, there must exist non-zero vectors \(\displaystyle v= \begin{bmatrix}x \\ y\end{bmatrix}\) such that
\(\displaystyle \begin{bmatrix}-7 & 7 \\9 & -5\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix}-7x+ 7y \\ 9x- 5y\end{bmatrix}= 2\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}2x \\ 2y \end{bmatrix}\).

So we must have -7x+ 7y= 2x and 9x- 5y= 2y. Both of those reduce to 9x= 7y. Taking x=7, y= 9, one eigenvector, corresponding to eigenvalue 2, is \(\displaystyle \begin{bmatrix} 7 \\ 9 \end{bmatrix}\) and all multiples are eigenvectors.

Because -14 is an eigenvalue, there must exist non-zero vectors \(\displaystyle v= \begin{bmatrix}x \\ y\end{bmatrix}\) such that
\(\displaystyle \begin{bmatrix}-7 & 7 \\9 & -5\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix}-7x+ 7y \\ 9x- 5y\end{bmatrix}= -14\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}14x \\ -14y \end{bmatrix}\).

So we must have -7x+ 7y= -14x and 9x- 5y= -14y. Both of those reduce to x= -y. Taking x=1, y= -1, one eigenvector, corresponding to eigenvalue -14, is \(\displaystyle \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) and all multiples are eigenvectors.