Upon trying to solve this particular system , I've encountered a few problems. $$ y'=5y+4z $$ $$z'=-4y-3z$$

After solving for eigenvalues the quadratic yielded a double root at $\lambda=1$ . But I got stuck after getting at $a=-b$ . This prevented me from solving any further because I eventually arrived at $0=1$ .

My method looks exactly like presented on this site adapted to the case in which $\lambda$ is a double solution.

I'm very interested in a solution.


This is a case where the algebraic multiplicity of $\lambda$ (the order of the root of the characteristic polynomial, in this case $2$) is less than the geometric multiplicity (the number of linearly independent eigenvectors).

You can find one eigenvector easily:

$$ A - I = \begin{pmatrix} 4 & 4 \\ -4 & -4 \end{pmatrix} $$

so $v_1 = (1,-1)$ is an eigenvector. But the above matrix only has a one dimensional kernel (equivalently: it has rank $1$), so you won't find another linearly independent eigenvector.

Do you know how to find generalized eigenvectors? You want to find $v_2$ so that $(A - I)v_2 = v_1$. The choice isn't unique, but $v_2 = (\frac14,0)^T$ is an obvious choice.

Then, the general solution is $x(t) = c_1 v_1 e^{\lambda t} + c_2 e^{\lambda t}(v_1 t + v_2)$. If this isn't familiar, you should review the derivation of this solution.


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