Proving a matrix is diagonaziable.

1. Jul 7, 2007

MathematicalPhysicist

im given a matrix H from M_n(C) (the space of nxn matrices above the comples field).
and we know that for every a in C, dim(ker(H-aI)^2)<=1.
prove that H is diagonizable.

obviously if i prove that its characteristic poly exists then bacuase every poly above C can be dissected to linear factors, then also its minimal poly can be dissected to linear factors and thus H is diagonaizable, but how to do it?

2. Jul 8, 2007

matt grime

Every matrix has a characteristic poly, so your argument is invalid. The characteristic poly doesn't tell you anything about diagonalizability. The minimal poly may tell you something useful.

3. Jul 8, 2007

MathematicalPhysicist

care to elaborate a bit more?
i mean what exactly the minimial polynomial can help here?

4. Jul 8, 2007

morphism

If the minimal polynomial has no repeated roots, then...

5. Jul 8, 2007

mathwonk

do you know what a jordan form is?

6. Jul 10, 2007

MathematicalPhysicist

you mean that it is decomposed of linear polynomials, and thus the matrix is diagonizable.
but how to translate it to here.
i mean im given that d=dimKer(H-xI)^2<=1, now if H-xI=0 then we have that minimal poly is linear product, but i dont see how to arrive at this, if i prove that dimKer(H-xI)=n then im done, but how to do it?

p.s
mathwonk, i do know what is jordan noraml form is, but i don't see here relevancy?

7. Jul 10, 2007

matt grime

Why don't you see the relevance of JNF? It gives the answer instantly. A matrix is not diagonalizable if and only if there is a jordan block of size greater than 1. But the condition you have precludes that happening.

8. Jul 10, 2007

MathematicalPhysicist

but the maximum size of jordan block is determined by the multicipity of the eigen value in the minimal polynomial.
oh, wait a minute, bacuase it's above C, then it means that the minimal poly is product of linear polys, so obviously we don't have higher degrees, and
Ker(H-aI) is a subspace of Ker(H-aI)^2, so dim(Ker(H-aI))<=1, so we have just one block at most, but every Ker(H-aI)^2 its dim is <= 1, so obviously we don't have here dome subspace which nullifies the entire C^n, but still we ned to have information on the minimal poly to gain information about the maximal size of the block.

9. Jul 10, 2007

matt grime

yes.

No. There is nothing that states that the linear factors of the minimal poly of a matrix are distinct, irrespective of the field.

Absolutely not. If there is one block, then it can't be diagonalizable.

I have no idea what you mean.

If we have a Jordan block of size 2, or more, and e-value t, then there are vectors x and y such that Mx=tx, and My=ty+x. That is the definition of a Jordan block. Now show that x and y are killed by (M-t)^2.