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[Linear Algebra] Linear transformation proof

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Homework Statement



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.
 
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Answers and Replies

  • #2
andrewkirk
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##A## is a subspace of ##V## and thus meets all criteria for being a subspace.
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.
##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]
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.
What exactly does the notation ##T|_C## mean?
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##.
 
  • #3
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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.
 
  • #4
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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?
That would be my guess, too.
 
  • #5
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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.
 
  • #6
vela
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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##.
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.
 
  • #7
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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.
 
  • #8
andrewkirk
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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.
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).
 

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