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For eigenvalue 1, the eigenspace I got was $span[ 1,0,0]$, and for eigenvalue 4, the eigenspace I got was $span[ 1,-3,9]$. Do they look right?

The reason the matrix is not diagonalizable is because we only have 2 linearly independent eigevectors so we can't span R3 with them, hence we can't create a matrix E with the eigenvectors as its basis. Is that correct, did I word it right?

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up vote 12 down vote accepted

The algebraic multiplicity of $\lambda=1$ is $2$. A matrix is diagonalizable if and only if the algebraic multiplicity equals the geometric multiplicity of each eigenvalues.

By your computations, the eigenspace of $\lambda=1$ has dimension $1$; that is, the geometric multiplicity of $\lambda=1$ is $1$, and so strictly smaller than its algebraic multiplicity. Therefore, $A$ is not diagonalizable.

Note that you don't actually need to compute the eigenspace to determine diagonalizability: you just need to figure out the dimension of the eigenspace. The eigenspace of $\lambda=1$ is the nullspace of $A-I$. Since $$A-I = \left(\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 3 \end{array}\right)$$ has rank $2$, it has nullity $1$, so the dimension of the eigenspace corresponding to $\lambda=1$ is $1$, strictly smaller than the algebraic multiplicity. This suffices to show $A$ is not diagonalizable.

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1  
In your second paragraph you probably meant to say that the geometric multiplicity is strictly smaller than its algebraic multiplicity, instead of repeating the word geometric. – T. Eskin Mar 26 '15 at 19:46
    
This is off topic, but I was wondering how to prove that the equality of algebraic multiplicity and geometric multiplicity implies diagonalizable? – Ignite Jun 9 at 15:40

The eigenvalues are correct, they are $\lambda_1=1$ and $\lambda_2=4$. The algebraic multiplicities are $m(\lambda_1)=2$ and $m(\lambda_2)=1$.

The eigenspace relative to $\lambda_1$ is (as you have correctly found) $V_{\lambda_1} = \langle (1,0,0) \rangle$. The eigenspace relative to $\lambda_2$ is $V_{\lambda_2} = \langle (1,3,9) \rangle$.

As you said, since the $1=\dim V_{\lambda_1} \ne m(\lambda_1) =2$ you can easily conclude that the matrix is not diagonalizable.

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SPOILER

$\left[ A,A^\dagger \right]_- = AA^\dagger -A^\dagger A = \begin{pmatrix}1 & 0& 0\\ 0 & 0& 3\\ 0 & 3& -1 \end{pmatrix}$

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4  
The fact that the matrix is not normal only implies that it is not unitarily diagonalizable. There is no guarentee that the matrix is not normally diagonalizable. – EuYu Apr 16 '12 at 16:30

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