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Could anybody, please, explain to me, how to solve system of 3 differential equations, when it has triple eigenvalue?

I mean... we solved these equations by creating a matrix $A$ of the system and finding eigenvalues $\lambda_1,\lambda_2,\lambda_3$. Then we found eigenvectors $\overrightarrow{v_1},\overrightarrow{v_2},\overrightarrow{v_3}$ by computing result for the matrix $A-\lambda_iE$. Then there was a formula for result $u_i = c_i\cdot e^{\lambda_i\cdot t}\cdot \overrightarrow{v_i}$. And $x =u_1+u_2+u_3$.

And if any eigenvalue were doubled, the second time, when eigenvector was searched we put to the matrix on the right-handed side the eigenvector from the first solution ($A-\lambda_2E=\overrightarrow{v_1}$) and computed results.

But what if the eigenvalue is triple? What then?

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up vote 5 down vote accepted

Suppose $\lambda$ is an eigenvalue of your $3 \times 3$ matrix $A$ with algebraic multiplicity $3$. The geometric multiplicity (i.e. the number of linearly independent eigenvectors) could be either $1$, $2$ or $3$.

If it is $3$, you have solutions of the form $c_1 e^{\lambda t} v_1 + c_2 e^{\lambda t} v_2 + c_3 e^{\lambda t} v_3$.

If it is $2$, take a vector $v_1$ such that $(A - \lambda E) v_1 \ne 0$ (almost any random vector will do), and let $v_2 = (A - \lambda E) v_1$. Then $v_2$ will be an eigenvector. Take another eigenvector, linearly independent of $v_2$, and call it $v_3$. Then the solutions are $c_1 e^{\lambda t} (v_1 + t v_2) + c_2 e^{\lambda t} v_2 + c_3 e^{\lambda t} v_3$.

If the geometric multiplicity is $1$, take a vector $v_1$ such that $(A - \lambda E)^2 v_1 \ne 0$ (again, almost any random vector will do), and let $v_2 = (A - \lambda E) v_1$ and $v_3 = (A - \lambda E) v_2$. Then the solutions are $c_1 e^{\lambda t} (v_1 + t v_2 + \frac{t^2}{2} v_3) + c_2 e^{\lambda t} (v_2 + t v_3) + c_3 e^{\lambda t} v_3$.

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