Question

find the eigenvalues for the system of equations.
x'(t) = [-7 0 0 0
8 -3 4 0
1 0 -5 0
2 1 4 -1 ] X(t)
Thanks in advance!

Answer 1

Would this be like
\(A=\begin{bmatrix}-7&0&0&0\\8&-3&4&0\\1&0&-5&0\\2&1&4&-1\end{bmatrix}\)
and
\(A\vec x=\lambda\vec x\)
where all \(\lambda\) that make this true are called eigenvalues?

I guess that \(x'(t)=\lambda\vec x(t)\), but I haven't seen the notation before.

What you want to do is find the determinant
\(\det(\lambda I-A)=0\)

That's because
\(A\vec x=\lambda\vec x\)
\(\implies \lambda\vec x-A\vec x=0\)
\(\implies (\lambda I-A)\vec x=0\)
which has a non-zero solution for \(\vec x\) if the determinant of the multiplying matrix is not zero.

If you need more help, I think there is a linear algebra group on OpenStudy! This is the ODE section. Take care!

Answer 2

When you find the determinant, it will have \(\lambda\) in it. But you set the determinant to zero and solve for \(\lambda\).