Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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?

share|improve this question
    
HINT A matrix is diagonalizable, if it's normal. –  draks ... Apr 16 '12 at 16:03
2  
@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

3 Answers 3

up vote 5 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 geometric 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.

share|improve this answer

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.

share|improve this answer

SPOILER

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

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.