Proving a matrix is diagonaziable.

[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?

thanks in advance.

 

[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.

 

[MathematicalPhysicist]

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

 

[morphism]

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

 

[mathwonk]

do you know what a jordan form is?

 

[MathematicalPhysicist]

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

you mean that it is decomposed of linear polynomials, and thus the matrix is diagonizable.
but how to translate it to here.
i mean I am given that d=dimKer(H-xI)^2<=1, now if H-xI=0 then we have that minimal poly is linear product, but i don't see how to arrive at this, if i prove that dimKer(H-xI)=n then I am 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?

 

[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.

 

[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.

 

[matt grime]

but the maximum size of jordan block is determined by the multicipity of the eigen value in the minimal polynomial.

yes.

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,

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

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,

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

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.

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.