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So I have the Eigenspace of:

$$ \begin{align*} -5x -y + 3z &= 0 \\ -18x -3y + 9z &= 0 \\ -16x -3y + 9z &= 0 \end{align*} $$ I get the solution $y = 3z$, and hence the vector for a basis $(0,3,1)$ but I know I need a second vector to complete the basis. And the only other solution I can find is $x = 0$ and I don't think that's very useful? The only thing I can think of is if we use the fact that you can't find the other vector for the basis, it means it's not diagonalisable.

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You'll need a total of 3 vectors for a basis. – icurays1 Dec 1 '12 at 16:48
yeah I know, but that's for the Eigenspace of 1, I need 2 from the Eigenspace of 0. – Adam Dec 1 '12 at 16:53
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This system of equations turns out to be non diagonalizable, which means that there does not exist a basis of eigenvectors that span the entire space. In other words, the eigenspaces are each one dimensional, and the eigenvector you found is the only linearly independent one for the eigenvalue $0$. This illustrates the difference between algebraic and geometric multiplicites - the algebraic multiplicity of an eigenvalue $\lambda$ is the number of times it appears, while the geometric multiplicity is the dimension of the corresponding eigenspace. In your case, you find an expression for all eigenvectors with eigenvalue zero: $x=0$ and $y=3z$. Since this describes a line, the dimension of this eigenspace is 1.

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You won't get 2 from the Eigenspace of 0. Algebraic Multiplicity of 0 is 2. The dimension eigenspace of $0$ is the dimension of the null space of $A-\lambda I=A$. The rank of the matrix is 2, the dimension of the null space is 1. Thus, you can have only 1 linearly independent vector in the eigenspace of 0.

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