# 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.