1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: [Linear Algebra] Linear transformation proof

  1. Jun 14, 2018 #1
    1. The problem statement, all variables and given/known data

    Let ##V## and ##W## be vector spaces, ##T : V \rightarrow W## a linear transformation and ##B \subset Im(T)## a subspace.

    (a) Prove that ##A = T^{-1}(B)## is the only subspace of ##V## such that ##Ker(T) \subseteq A## and ##T(A) = B##
    (b) Let ##C \subseteq V## be a subspace. Prove that ##A = Ker(T) \oplus C## iff ##T(C) = B## and ##T|_C## is injective.

    The attempt at a solution

    Per usual, I'm stuck on the notation here, but I think I have an idea about where the proof comes from, at least in the first part.

    To organize my information, I know the following:

    ##A## is a subspace of ##V## and thus meets all criteria for being a subspace.
    ##A = T^{-1}(B) | T^{-1} : W \rightarrow V##
    As T is invertible, we can deduce that T is bijective as a function and thus both onto and one-to-one, and also that ##V \cong W##. [I have this proof from my notes and previous work]

    I also have the definition of kernel and the proof relating it to the transformation's injectivity, also from a previous exercise.

    Now, defining the kernel of ##T## :

    ##Ker(T) = \{v \in V : T(v) = 0\} = T^{-1} (\{0\})##
    ##T^{-1} (\{0\}) \in T^{-1} \rightarrow Ker(T) \subseteq A##

    I can prove that more formally, but does the spirit of the exercise even go in that direction? Similarly, is it using the injectivity from ##Ker(T)## that I prove ##T(A) = B## or can I use the definition of ##T^{-1}## to show that if I apply ##T## to ##T^{-1}(w)## I obtain ##\{w\}## and then use injectivity?

    I''ll try to work out the second half of the exercise after the first. What exactly does the notation ##T|_C## mean?

    Thanks as always for any and all assistance.
     
    Last edited: Jun 14, 2018
  2. jcsd
  3. Jun 14, 2018 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you might be expected to prove that it meets those criteria, or show by other means that it's a subspace. To say it meets the criteria because it's a subspace is begging the question of how we know it's a subspace.
    We do not know that T is invertible, and we need to cover the cases where it is not. If the kernel of T is nontrivial then T will not be invertible, and ##T^{-1}## will not be a function. Rather ##T^{-1}(B)## denotes all points in the domain that map to points in ##B##. Also, T is not necessarily injective or bijective.
    The usual meaning would be the function that has domain C and maps points in C to the same points as T does. It is also known as the 'restriction of T to C'.

    Finally, it would help if you clarified the meaning of ##N(T)## and ##Ciff##. My guess is that ##N(T)## refers to the null space of the matrix of ##T##, which is the same as ##Ker\ T##, but it would be odd to use two different names for it in the same problem. I have no ideas about ##Ciff##.
     
  4. Jun 14, 2018 #3
    Sorry, the notation was bugging out. That doesn't read "Ciff" but instead "C iff" and N(T) was an oversight on my part, as I was translating from the Spanish Núcleo de T, which is the Kernel and might just be a Spanish adaptation of the notation denoting the null space? I edited to make the post's formatting clearer for anyone else reading the question.

    I will reply to this post with my attempt at proving tomorrow as it is late here, but I wanted to make sure the problem was described correctly.
     
  5. Jun 15, 2018 #4

    Mark44

    Staff: Mentor

    That would be my guess, too.
     
  6. Jun 15, 2018 #5
    OK, I've approached the problem differently, though I feel a little stuck.

    ##B \subset Im(T)## is a subspace.
    ##Im(T) = \{w \in W : w = T(v)\}##
    ##B \subset Im(T) = \{b \in W : b = T(a), a \in V\}##
    ##Ker(T) = \{v \in V : T(v) = 0\}##

    ##A = T^{-1}(B) \rightarrow A = \{a \in V : T(a) = b\}##

    Now, is ##A## a subspace?
    ##T^{-1}(b_1+b_2) = (a_1+a_2) = a_1 + a_2 = T^{-1}(b_1) + T^{-1}(b_2)##
    ##T^{-1}(c \cdot b_1) = (c \cdot a_1) = c \cdot (a_1) = c \cdot (T^{-1}(b_1))##
    ##T^{-1}(0) = 0, 0_v \in A##

    I conclude that ##A## is indeed a subspace. I can also see that ##Ker(T)## is in ##A##.

    How do I prove that ##A## is the only subspace which meets these conditions?

    Thanks for all help.
     
  7. Jun 15, 2018 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You haven't shown A is a subspace. You want to show, for example, that if ##a_1## and ##a_2## are in A, then ##a_1+a_2## is in A. You seem to be arguing if ##b_1## and ##b_2## are in B, then ##b_1+b_2## is in B.

    You can't treat ##T^{-1}## as a function. For example, saying ##T^{-1}(0) = 0## isn't correct. If Ker(T) is not trivial, there are other elements in the domain which map to 0.
     
  8. Jun 15, 2018 #7
    I'm sorry, among other things I tried was to show that if ##T^{-1}## is a transformation then it would be linear. But again, I treated it as a function there.

    I understand how to show ##T^{-1}## is a subspace (although I'll need to ponder for a moment how to write that the 0 vector belongs there), but I don't know how to show that A is the only subspace which meets the conditions posed in the exercise. I'm a bit stuck, any hint would be appreciated.
     
  9. Jun 15, 2018 #8

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You need to show that for any subspace different from A, the conditions are met. If a subspace B is not equal to A then either it contains vectors not in A or A contains vectors not in B. So first assume there is a vector ##\vec b\in B-A## and show one of the conditions is not met (or assume the conditions are met and deduce a contradiction). Then assume there is a vector ##\vec a\in A-B## and show one of the conditions is not met (or assume the conditions are met and deduce a contradiction).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted