# [Linear Algebra] Linear transformation proof

• iJake
In summary, the problem asks for a subspace of a vector space which is both injective and bijective, has the same domain as the transformation, and is the only such subspace. The solution seems to imply that the only subspace which satisfies these conditions is ##A##.

## 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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iJake said:
##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##.

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.

iJake said:
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.

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.

iJake said:
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.

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.

iJake said:
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).

## 1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another in a way that preserves the algebraic structure of both spaces. In simpler terms, it is a function that takes in a vector and outputs another vector while maintaining certain properties, such as scaling, rotation, and translation.

## 2. What are the properties of a linear transformation?

The properties of a linear transformation include preserving addition, scalar multiplication, and the zero vector. This means that the transformation of the sum of two vectors is equal to the sum of the individual transformations, the transformation of a scalar multiple of a vector is equal to the scalar multiple of the transformation of the vector, and the transformation of the zero vector is equal to the zero vector.

## 3. How do you prove that a function is a linear transformation?

To prove that a function is a linear transformation, you must show that it satisfies the properties of a linear transformation. This can be done by using the definitions of addition, scalar multiplication, and the zero vector, and showing that they hold true for the given function. Additionally, you can also use theorems and properties of linear transformations to assist in the proof.

## 4. What is the importance of linear transformations?

Linear transformations have many applications in various fields, such as physics, engineering, computer graphics, and data analysis. They allow us to represent and manipulate data in a more efficient and meaningful way, and can help us solve complex problems by breaking them down into simpler, more manageable parts.

## 5. Can you provide an example of a linear transformation?

One example of a linear transformation is a rotation in two-dimensional space. This can be represented by a function that takes in a 2D vector and outputs a rotated version of that vector. The properties of addition, scalar multiplication, and the zero vector are preserved in this transformation, making it a linear transformation.