Null Space Question?

Question

How might I go about proving that, for a linear operator T, null((T-a)(T-b)) = null(T-a) + null(T-b) for distinct scalars a,b?

I have a relatively rudimentary understanding of null spaces and don't really know where to go. Any help?

Answer 1

This appears to be a non-trivial theorem.



It is a direct result of the Jordan Canonical Form (for those familiar with that result).  Write the Jordan Canonical Form for T and examine matrices for the differences (T-aI) and (T-bI) along with the product (T-aI)(T-bI).  Nullity(T-aI) and Nullity(T-bI) are the dimensions of the eigenspaces for eigenvalues "a" and "b" of T, that is the number of zero columns in those difference matrices.  Nullity((T-aI)(T-bI)) is the number of zero columns in the product matrix, which is easily seen to correspond to the zero columns in the Jordan Canonical matrices for (T-aI) and (T-bI).  {Note that this result relies on the fact that "a" and "b" are distinct, so that (T-aI) and (T-bI) have their zero columns in different blocks.}



A more elementary proof can be constructed by carefully examining the kernels(as suggested by D g).



Suppose 0 = (T-a)(T-b)(x).  {That is, x is in the kernel of (T-a)(T-b).}



Hint1:  Consider y = (T-b)(x)/(a-b)

(Why is it important that a and b are distinct?)

Can you show that y is in the kernel some operator, NOT (T-b)?



Hint 2:  Also Consider z = x - y.

Can you show that z is in some kernel?

(Can you see why that factor of (a-b) was put in the above?)



If you can decompose every element of the kernel of (T-a)(T-b)

uniquely as a sum of elements from the kernels of (T-a) and (T-b),

AND that every such sum IS in the kernel of (T-a)(T-b) - the easy part -

then the kernel of (T-a)(T-b) is a direct sum of those other two kernels

and its dimension is the sum of their dimensions (nullities).



Can you show that uniqueness?  What would happen if a vector

was in both the kernel of (T-a) and the kernel of (T-b)?



This might be a challenge assignment for some linear algebra course, so I hope I haven't spoilt it!

Answer 2

These are always interesting  proofs some require alot of forthought   or  alot of  writing this one  may be simple algebra



since T is a linear operator 



Definition: If T∈L(V,W) then the Null Space or Kernel of the linear transformation T is the subset of V defined as null(T)={v∈V:T(v)=0}, that is, the null space of T is the set of vectors from V that are mapped to the zero vector in W under T



your  question 



null((T-a) (T-b)) = null(T-a) + null(T-b)



i think this should be 



null((T -aI)(T -bi)) = null(T-aI) + null(T - bI)



might be able to say











null ((T(T-bI) - aI(T- bI) ) = null(T-aI) + null(T - bI)



null(T(T-bI)) - null(aI(T - bI))



i cant get any farther  i just  am sorry   













// this part is  the  euqation you gave 

I am trying to see how to add a scalar to a linear transform.



my understanding is that  a linear operator is some kind of vector  



my difficulty is the use of the scalar but i think i have it ..



a scalar is NOT  a vector space 



it cant have two values   its not a matrix or  linear transformation .. so  when you use a linear transformation on it such as  NULL



then you get an  EMPTY SET...



NULL(T - a) = NULL(T) because  the NULL of scalar a is  NULL set or nothing



NULL(T - b) = NULL(T)



so NULL(T - a) + NULL(T - b)= 2NULL(T)



NULL ((T - a ) (T - b))

Answer 3

Could you perhaps submit a photo of the question. The notation you’re using seems unclear to me.