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I read a theorem that says squared matrix $A_{n\times n}$ is diagonalizable iff there is a set of $n$ linearly independent vectors ,each of which is an eigen-vector of $A_{n\times n}$ .

I understand from the theorem that if $A_{n\times n}$ has at least two dependent eigen-vectors then $A_{n\times n}$ is not diagonalizable. Is it true?and according to this I want to understand when a matrix has linearly dependent eigenvectors ? (eigenvectors is a set of vectors) In other words if there is'nt any set of linearly independent set of eigenvectors for a matrix, then does that mean that the matrix is not diagonalizable?

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  • $\begingroup$ Can't even begin to understand your question. $\endgroup$ – Git Gud Oct 14 '14 at 12:18
  • $\begingroup$ can you explain where is not obvious? $\endgroup$ – eHH Oct 14 '14 at 12:20
  • $\begingroup$ What is the definition of a dependent eigenvector? $\endgroup$ – 5xum Oct 14 '14 at 12:22
  • $\begingroup$ I mean linearly dependent $\endgroup$ – eHH Oct 14 '14 at 12:23
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    $\begingroup$ You are probably asking something else: You want to know, when an eigenspace of a matrix has dimension bigger than one… Is that what you're asking for? $\endgroup$ – frog Oct 14 '14 at 12:32
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The theorem states, correctly, that IF the matrix $A$ has $n$ linearly independent eigenvectors, then $A$ is diagonalizable.

That does NOT mean that if $A$ has two dependent eigenvectors, that it is not diagonalizable. In fact, if $A$ has at least one eigenvector $x$ such that $Ax=\lambda x$, then $2x$ is also an eigenvector of $A$ since $A(2x)=2Ax=2\lambda x=\lambda (2x)$. And since $\{x,2x\}$ is not a linearly independent set, this means that $A$ has two linearly dependent vectors. Does that mean $A$ is not diagonalizable? Of course not!

You actually made two logical mistakes.

You know that if statement $p$ ($A$ has $n$ linearly dependent eigenvectors) is true, then statement $q$ ($A$ is diagonalizable) is true, so in other words $p\implies q$. Then you found another statement, $c$ ($A$ has two linearly independent eigenvectors).

Then, you made two mistakes:

  1. You claim that if $c$ is true, then $p$ is not true (in fact, any matrix with at least one eigenvector has two dependent eigenvectors, but it can still have $n$ linearly dependent eigenvectors. In fact, whenever $p$ is true, $c$ is also true!
  2. You then wanted to conclude that since $p$ is not true, then $q$ is not true. This is a very common mistake, but if $p\implies q$, then $\neg p$ does not imply $\neg q$! For example, the statement "Rex is a dog" being true means that the statement "Rex has fur" is true, however, the statement "Rex is not a dog" does not mean that "Rex has no fur" is true!
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  • $\begingroup$ I think what he is asking is that if you are not able to find a set of linearly independent set of eigenvectors for a matrix, then does that mean that your matrix is not diagonalizable? $\endgroup$ – Parth Thakkar Oct 14 '14 at 13:16
  • $\begingroup$ @ParthThakkar OP says "I understand from the theorem that if $A_{n\times n}$ has at least two dependent eigen-vectors then $A_{n\times n}$ is not diagonalizable" and this statement is not true and is adressed in my answer. If OP meant something else, I think he can explain that himself $\endgroup$ – 5xum Oct 14 '14 at 13:19
  • $\begingroup$ Yeah sure, but I was just trying to help. One more reason for giving in my own interpretation was that I myself do not have an answer for that - that is, what I believe he is asking. $\endgroup$ – Parth Thakkar Oct 14 '14 at 13:22
  • $\begingroup$ @ParthThakkar For that, I think the Jordan form should be helpful... $\endgroup$ – 5xum Oct 14 '14 at 13:23
  • $\begingroup$ @ParthThakkar Sure you know the answer. Just argue by contraposition."If a matrix is diagonalizable, then you can find 'sufficient' linearly independent eigenvectors". $\endgroup$ – Git Gud Oct 14 '14 at 13:24

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