# Diagonalizability test

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.

• 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. – Qiaochu Yuan Sep 28 '17 at 3:10
• 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. – Akash Patalwanshi Sep 28 '17 at 3:13
• Do you know how to write down a linear operator with any characteristic polynomial? – Qiaochu Yuan Sep 28 '17 at 3:14
• Yes I know that... – Akash Patalwanshi Sep 28 '17 at 3:15
• And do you know how to write down a (separable) polynomial which doesn't split over some field? – Qiaochu Yuan Sep 28 '17 at 3:16