# Repeated Root Eigenvalues

The question is:

Solve the initial value problem:

\begin{align*} \frac{dx_1}{dt}&=40x_1-6x_2+18x_3,\\ \frac{dx_2}{dt}&=-6x_1+45x_2+12x_3,\\ \frac{dx_3}{dt}&=18x_1+12x_2+13x_3,\\ \end{align*} with \begin{align*} x_1(0) &= 3,\\ x_2(0) &= -4,\\ x_3(0) &= 12\\ \end{align*}

So my A matrix is:
$$A=\pmatrix{40&-6&18\cr-6&45&12\cr18&12&13\cr}$$ and to find the eigenvalues I find the determinant of this matrix:
$$A=\pmatrix{40-\lambda&-6&18\cr-6&45-\lambda&12\cr18&12&13-\lambda\cr}$$

which gives me these eigenvalues and vectors:
$\lambda _1=0 --- v_1=(-3,-2,6)$
$\lambda_{2,3}=49 --- v_2=(2,0,1)$ and $v_3=(-2,3,0)$

The place I'm having trouble is, from this question I know that the general solution for real eigenvalues is:
$$x=c_1e^{a_1t}v_1+c_2e^{a_2t}v_2+c_3e^{a_3t}v_3$$

but what would it be for repeated roots? Is it:
$$x=c_1e^{a_1t}v_1+c_2e^{a_2t}v_2+c_3te^{a_3t}v_3$$

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A simple answer: it depends. You seem to have two linearly independent eigenvectors for your eigenvalue of multiplicity 2 (I assume that your calculations are correct), which means that the general solution formula works. The last formula you wrote is generally wrong. – Artem Dec 2 '12 at 22:06
Note that in this case $A$ is a real symmetric matrix, so it always has a basis of eigenvectors. – Robert Israel Dec 2 '12 at 22:08

Whenever $v$ is an eigenvector of $A$ for eigenvalue $\alpha$, $x = e^{\alpha t} v$ is a solution of $x' = A x$. Here you have three linearly independent eigenvectors, so three linearly independent solutions of that form, and so you can get the general solution as a linear combination of them. The fact that two of the eigenvalues are equal is irrelevant. The only case where you need solutions such as $t e^{\alpha t} v_1 + e^{\alpha t} v_2$ is where there is a "missing eigenvector": the eigenvalue $\alpha$ has geometric multiplicity less than its algebraic multiplicity.
The algebraic multiplicity of eigenvalue $\alpha$ is the exponent of $z-\alpha$ in the factorization of the characteristic polynomial $\det(A-zI)$. The geometric multiplicity is the number of linearly independent eigenvectors of $\alpha$. If the geometric multiplicity is less than the algebraic multiplicity, you don't have enough eigenvectors to produce a general solution using the linear combinations of $e^{\alpha t} v$. – Robert Israel Dec 3 '12 at 0:32