[Linear Algebra] Linear transformation proof

[iJake]

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.

 

[andrewkirk]

$A$ 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 $Ciff$.

 

[iJake]

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.

 

[Mark44]

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.

 

[iJake]

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.

 

[vela]

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.

 

[iJake]

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.

 

[andrewkirk]

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).