Vector Space, dimensions and kernal rank

[zcomputer5]

Please could someone help me with this question, thank you.

Find dim[Ker(D^2 -D: P_3(F_3) ==>P_3(F_3))]

Where dim is dimension, Ker is kernal

D is the matrix
0100
0020
0003
0000

D^2 is the derivative of D is it equals

0020
0006
0000
0000

And F_3 is the field subscript3

so D^2 -D

Should equal

0220
0010
0000
0000

But where do I go from here? I have tried reducing this matrix to row echelon form however this doesn't seem logical, do you have any ideas on finding the dimension of the kernal?

THANK YOU

 

[adriank]

I will write T = D² - D.

Well, the dimension of the image space of T is the number of independent columns in a matrix of T (indeed, the image space is spanned by the columns of the matrix); from what you have found, clearly dim(im T) = 2 (you can be sure of this by reducing the matrix to column echelon form). But since the domain of T, P₃(F₃), has dimension 4, we have 4 = dim(im T) + dim(ker T).

Equivalently, dim(ker T) is the number of zero columns in a column echelon form of a matrix of T.

 

[Defennder]

so D^2 -D

Should equal

0220
0010
0000
0000

Don't think this is right. Recheck your working again.

But where do I go from here? I have tried reducing this matrix to row echelon form however this doesn't seem logical, do you have any ideas on finding the dimension of the kernal?

THANK YOU

One way to do it, as adrian suggested, would be to find dim(R(T)) and then use the dimension theorem to find dim(ker(T)) or nullity. The other way would be to let a vector v = (w x y z)^T. Then solve the homogenous system of linear equations Tv = 0, to find a basis for nullspace(T). Then simply count the number of vectors in that basis for the answer.

 

[adriank]

Don't think this is right. Recheck your working again.

It is correct. Remember that we're working in the finite field F₃.

 

[Defennder]

Yeah, you're right. Silly me.