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True or false and explain why?: a matrix with characteristic polynomial $\lambda^3 -3\lambda^2+2\lambda$ must be diagonalizable.

First I found the lambda's that make this zero (eigenvalues) and got $0, 1, 2$ but I don't know if having $0$ as an eigenvalue means that the matrix is not diagonalizable? I know that a matrix has $0$ as an eigenvalue if it is not invertible, but I don't know if a matrix needs to be invertible to be diagonalizable? Also if a matrix has complex eigenvalues does that also mean it cannot be diagonalizable?

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    $\begingroup$ Is the matrix with all entries equal to $0$ diagonalisable? $\endgroup$ – Nigel Overmars May 7 '16 at 16:35
  • $\begingroup$ @NigelOvermars The definition for a diagonal matrix is "all non diagnoal entries are zero" so I guess the zero matrix is considered diagonal. $\endgroup$ – idknuttin May 7 '16 at 16:40
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    $\begingroup$ The question in your title ("If zero is an eigenvalue...") and in the question body (the characteristic polynomial is $\lambda(\lambda - 1)(\lambda - 2)$") do not match (in the strong sense that the answers differ). Could you please fix this? $\endgroup$ – Andrew D. Hwang May 7 '16 at 16:45
  • $\begingroup$ I understand now, a matrix is diagonalizable iff the algebraic multiplicity = the geometric multiplicity for each eigenvalue. $\endgroup$ – idknuttin May 7 '16 at 16:59
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    $\begingroup$ You might try recalculating those eigenvalues since one of them is incorrect. $\endgroup$ – amd May 7 '16 at 19:13
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The characteristic polynomial splits as follows $P_M(\lambda)=\lambda(\lambda-1)(\lambda-2)$. The matrix is $3\times 3$ (the degree of the characteristic polynomial) and has three distinct eigenvalues therefore it is diagonalisable.

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  • $\begingroup$ That is based on the assumption that eigenvectors that belong to different eigenvalues are linearly independent? $\endgroup$ – Antoine May 7 '16 at 16:45
  • $\begingroup$ @Antoine That's not an assumption, it is an easy proof. $\endgroup$ – N. S. May 7 '16 at 16:50
  • $\begingroup$ You have a double implication if you use the minimal polynomial instead of characteristic polynomial: a square matrix is diagonalizable if and only if its minimal polynomial factors into DISTINCT LINEAR factors. $\endgroup$ – akech May 7 '16 at 16:57
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    $\begingroup$ @marwalix I am not sure that the author of the question is aware of that, since (it seems to me that) he knows that there are three different eigenvalues (although he claims these are $0$, $2$, $3$), but he is not able to draw your conclusion. Also, I would say that there are things that are much easier to prove. $\endgroup$ – Antoine May 7 '16 at 17:03
  • $\begingroup$ @Antoine I understand marwalix's answer $\endgroup$ – idknuttin May 7 '16 at 17:09
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Edit (better wording)

$0$ as an eigenvalue doesn't hinder the diagonalization, actually there is no eigenvalue that would hinder it.

That doesn't mean that every matrix is diagonalizable, but that the eigenvalues have no influence, at least in $\mathbb{C}$.

Let A be a complex $n x n$ matrix with $A=diag(\lambda_1,...,\lambda_n)$, it is always diagonalizable, no matter what eigenvalues.


If $\lambda \in \mathbb{C}$ belongs to $A \in \mathbb{R}$ you have a problem, since you can't diagonalize it, without using $\mathbb{C}$.

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  • $\begingroup$ So does that mean every matrix is diagonaliable? $\endgroup$ – idknuttin May 7 '16 at 16:41
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    $\begingroup$ I don't think the question is whether the 0 eigenvalue hinders diagonalization, but rather whether it will guarantee it. $\endgroup$ – Morgan Rodgers May 7 '16 at 16:42
  • $\begingroup$ Although I removed my downvote because the question is not very clear. $\endgroup$ – Morgan Rodgers May 7 '16 at 16:44
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    $\begingroup$ @idknuttin No, the algebraic and geometric multiplicity has to be equal for all eigenvalues. $\endgroup$ – Patrick Abraham May 7 '16 at 16:59

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