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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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HINT A matrix is diagonalizable, if it's normal. –  draks ... Apr 16 '12 at 16:03
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@draks: It's only one implication. If a matrix is normal, then it is diagonalizable. There are non-normal, diagonalizable matrices. Showing (here) that $A$ is not normal doesn't help in proving that it is not diagonalizable. –  Najib Idrissi Apr 16 '12 at 16:21
    
For the eigenvalue $4$, the eigenspace is the span of $(1,3,9)$. Other than that, you are perfectly correct. But see Arturo's answer, you did more work than you had to. –  David Mitra Apr 16 '12 at 16:28
    
@NajibIdrissi: You have it the wrong way around. There are normal matrices that are not diagonalizable: but certainly all diagonalizable matrices are normal, by looking at the diagonal decomposition you should see this. For a normal non-diagonalizable matrix take $M=(a_1 , a_2)$ where the column vectors are $a_1 = (1,1)$ and $a_2 =(-1,3)$. –  Thomas E. Mar 26 at 19:57
    
@ThomasE. en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices draks was saying that normal matrices are diagonalizables (by the spectral theorem). But the question asks to show that a matrix is not diagonalizable. Why are you pinging me on a three year old question? –  Najib Idrissi Mar 26 at 21:38

3 Answers 3

up vote 10 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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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. –  Thomas E. Mar 26 at 19:46

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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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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