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Find the general solution of the homogeneous ordinary differential equation $\dot{\mathbf{x}}=A\mathbf{x}$.

$$A=\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$

I know how to answer these kind of questions, but the specific example above I am struggling with because I'm not sure I'm finding the correct geometric multiplicity for the repeated eigenvalue. I have the eigenvalues $\lambda=4,1,1$. Hence the eigenvalue $\lambda=1$ has algebraic multiplicity of 2, so I need to find the geometric multiplicity.

I have

$$A-\lambda I=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$$

Now, normally I would spot the number of independent rows/columns and conclude the geometric multiplicity from this. But it seems this matrix has 0 independent rows/columns, which would make the geometric multiplicity 3, but this violates the fact that the geometric multiplicity can not be greater than the algebraic multiplicity. Where is my thinking going wrong?

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$$A-\lambda I=\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}$$

Now, normally I would spot the number of independent rows/columns and conclude the geometric multiplicity from this. But it seems this matrix has 0 independent rows/columns, which would make the geometric multiplicity 3, but this violates the fact that the geometric multiplicity can not be greater than the algebraic multiplicity. Where is my thinking going wrong?

Why do you think this matrix has $0$ independent rows? Clearly the first row "on its own" is independent... Subtract the first row from the second and third to get:

$$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 \\ 0&0 & 0 \\ 0 & 0& 0 \end{pmatrix}$$ So you're looking for solutions to the homogeneous system with (only) the equation $x+y+z=0$. The geometric multiplicity is $2$ as well and two linearly independent eigenvectors are, for example, $(1,-1,0)$ and $(1,0,-1)$.

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  • $\begingroup$ I was under the impression a row is independent if it isn't a linear combination of the other rows. If the first row is independent, doesn't that make all the rows independent? And then the GM would be 0? $\endgroup$ – user390960 Jan 11 '17 at 10:01
  • $\begingroup$ One non-zero vector alone, is always linearly independent. Take for example $(1,2,3)$ and $(2,4,6)$. These vectors are clearly linearly dependent because they are multiples of one another. But that doesn't change the fact that either of those vectors alone is linearly independent. $\endgroup$ – StackTD Jan 11 '17 at 10:04
  • $\begingroup$ But in this case are we considering them all alone? $\endgroup$ – user390960 Jan 11 '17 at 10:05
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    $\begingroup$ No, but the question you have to ask is: "out of these 3 vectors (you consider every row as a vector), how many are linearly independent"?. The answer is 1. $\endgroup$ – StackTD Jan 11 '17 at 10:07

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