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I saw below theorem in my test book(by friedberg)

Theorem : a linear operator $T$ on $n$-dimensional vector space $V$ over field $K$is diagonalizable "if and only if" both of the condition holds.

1) the characteristic polynomial of $T$ splits over $K$

2) algebraic multiplicity of each eigenvalue is equal to geometric multiplicity.

While in other textbook I saw theorem that,

Theorem: a linear operator $T$ on $n$-dimensional vector space $V$ over field $K$is diagonalizable "if and only if" condition (2) of above theorem holds.

So which of these two theorems are true? First one or second one?

Further,

If 'first one' is true then can anyone given me example of linear operator $T$ on $n$-dimensional space which is not diagonalizable but satisfies condition (2). Thank you.

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    $\begingroup$ The second theorem is certainly false as stated, and as implied by the first theorem, any linear operator whose characteristic polynomial does not split over $K$ is a counterexample. Although perhaps $K$ is implicitly assumed to be algebraically closed, in which case it's fine. $\endgroup$ Commented Sep 28, 2017 at 3:10
  • $\begingroup$ Thank u so much for your help. No one helped me for this question(I asked it before 2days). Further, can you give me an example which shows that second theorem is false. $\endgroup$ Commented Sep 28, 2017 at 3:13
  • $\begingroup$ Do you know how to write down a linear operator with any characteristic polynomial? $\endgroup$ Commented Sep 28, 2017 at 3:14
  • $\begingroup$ Yes I know that... $\endgroup$ Commented Sep 28, 2017 at 3:15
  • $\begingroup$ And do you know how to write down a (separable) polynomial which doesn't split over some field? $\endgroup$ Commented Sep 28, 2017 at 3:16

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The second theorem is certainly false, any real matrix that does not have real eigenvalues is a counterexample.

If you add the condition

(3) ... and the sum of the geometric multiplicities of the eigenvalues is equal to the dimension of the space

then diagonalizability is equivalent to (2)(3).

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