Codomain and Range of Linear Transformation

[songoku]

Homework Statement

Let $T:\mathbb R^3 \rightarrow \mathbb R^2$ be
$$T \vec a=\begin{bmatrix}
a\\
b-c
\end{bmatrix}$$

(i) Check whether $T$ is an onto mapping of $\mathbb R^3 \rightarrow \mathbb R^2$
(ii) Check whether $T$ a one to one mapping
(iii) State the codomain and range of $T$
(iv) Explain the connection of answer of part (iii) to answer of part (i) and (ii)

Relevant Equations

T is onto mapping if $T \vec x=y$ has solution for all value of $y$

T is one to one mapping if $T \vec x=0$ has only trivial solution

Standard matrix for T is:
$$P=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & -1
\end{bmatrix}$$

(i) Since matrix P is already in reduced row echelon form and each row has a pivot point, $T$ is onto mapping of $\mathbb R^3 \rightarrow \mathbb R^2$

(ii) Since there is free variable in matrix P, T is not one to one mapping

(iii) Codomain is $\mathbb R^2$ and range is also $\mathbb R^2$

(iv) Since $T$ is onto mapping, the codomain will be the same as the range. But I don't understand the connection to answer in part (ii)

Is my answer correct?

Thanks

 

[Office_Shredder]

I think there is no connection to part 2.

Your answers look good to me.

You also can know T is not 1-1 since it maps a 3 dimensional space to a space of at most two dimensions, and it must be onto since you can find two linearly independent vectors in the range (this is not hard to do, e.g. T(1,0,0) vs T(0,1,0)).

 

[songoku]

I think there is no connection to part 2.

Your answers look good to me.

You also can know T is not 1-1 since it maps a 3 dimensional space to a space of at most two dimensions, and it must be onto since you can find two linearly independent vectors in the range (this is not hard to do, e.g. T(1,0,0) vs T(0,1,0)).

If let say the mapping is not onto, what would be the range?

Would it be the same as the question: the range would be $x$ in $\mathbb R^2$ where $x$ is
$$\begin{bmatrix}
a\\
b-c
\end{bmatrix}$$

 

[songoku]

it must be onto since you can find two linearly independent vectors in the range (this is not hard to do, e.g. T(1,0,0) vs T(0,1,0)).

Sorry I don't understand this part. How to relate linear independence to onto mapping?

Thanks

 

[PeroK]

Sorry I don't understand this part. How to relate linear independence to onto mapping?

Thanks

First, I'd say you can show $T$ is onto directly by noting that $\forall (x, y) \in \mathbb R^2$:
$$T(x, y, 0) = (x, y)$$And, you can show $T$ is not 1-1 as$$T(0, 0, 0) = T(0, 1, 1) = (0,0)$$If you know that $T$ is linear, then all you have to do to show that $T$ is onto is to show that $T$ maps two elements of $\mathbb R^3$ to any two linearly indepedent vectors. That follows from knowing that the range is a vector space and other basic results from linear algebra.

 

[Office_Shredder]

I want to emphasize, your answer is totally correct, we're just trying to show another way of doing it that doesn't rely on the echelon form, for educational purposes.

 

[songoku]

I want to emphasize, your answer is totally correct, we're just trying to show another way of doing it that doesn't rely on the echelon form, for educational purposes.

Yes, and to show my gratitude I am trying to listen and learn from all the replies. I have learned about linear independence and also about onto mapping but I don't know the relation between those two.

I am trying to think about this but I am not really sure this is correct:
If $T$ maps two elements of $\mathbb R^3$ to any two linearly independent vectors, it means $\mathbb R^2$ is span of those two linearly independent vectors so all the points in $\mathbb R^2$ can be stated as linear combination of those two linearly independent vectors and this shows that $T \vec x = y$ will have solution for all value of $y$ (so this proves that $T$ is onto mapping)

If you know that $T$ is linear, then all you have to do to show that $T$ is onto is to show that $T$ maps two elements of $\mathbb R^3$ to any two linearly indepedent vectors. That follows from knowing that the range is a vector space and other basic results from linear algebra.

I know $T$ is linear but I don't know how to relate that to onto mapping and linearly independent vectors. I have not learned about vector space, so does it mean I have to learned about vector space first to be able to understand this?

Thanks

 

[PeroK]

I know $T$ is linear but I don't know how to relate that to onto mapping and linearly independent vectors. I have not learned about vector space, so does it mean I have to learned about vector space first to be able to understand this?

You introduced concepts like "row echelon form" and "free variable in the matrix" in your OP. That is explicitly using linear algebra to answer the question. Whereas, in post #5 I used only the basic definitions of 1-1 and onto that apply to all mappings. So, no, you don't need to use linear algebra, and I'm confused now why you didn't simply use the basic definitions of 1-1 and onto?

 

[songoku]

You introduced concepts like "row echelon form" and "free variable in the matrix" in your OP. That is explicitly using linear algebra to answer the question. Whereas, in post #5 I used only the basic definitions of 1-1 and onto that apply to all mappings. So, no, you don't need to use linear algebra, and I'm confused now why you didn't simply use the basic definitions of 1-1 and onto?

Sorry maybe I am not explaining myself clearly

First, I'd say you can show $T$ is onto directly by noting that $\forall (x, y) \in \mathbb R^2$:
$$T(x, y, 0) = (x, y)$$

I understand this. All points in $\mathbb R^2$ can be mapped from $\mathbb R^3$ by taking $z=0$ so the mapping is onto.

And, you can show $T$ is not 1-1 as$$T(0, 0, 0) = T(0, 1, 1) = (0,0)

I also understand this. Since two inputs produce same output, it is not one - one.

If you know that $T$ is linear, then all you have to do to show that $T$ is onto is to show that $T$ maps two elements of $\mathbb R^3$ to any two linearly indepedent vectors.

This is the part that I don't really understand. There are several ways to answer my question in post#1 and this is one of it and if possible, I want to understand this.

$T$ is onto can be proven by showing $T$ maps two elements of $\mathbb R^3$ to any two linearly independent vectors. Why?

Thanks

 

[PeroK]

This is the part that I don't really understand. There are several ways to answer my question in post#1 and this is one of it. $T$ is onto can be proven by showing $T$ maps two elements of $\mathbb R^3$ to any two linearly independent vectors.

Why?

A linear transformation always maps a vector space onto a vector subpsace of the codomain. If we find two linearly independent vectors in the range of $T$, then the range of $T$ must be a 2D vector subspace of $\mathbb R^2$. And, that means the range must be $\mathbb R^2$ itself.

In general, the property of being a linear transformation significantly constrains a function and means that other properties are equivalent to being 1-1 and/or onto. For example, a linear transformation $T$ from the vector space $V$ to any other vector space is 1-1 iff: $$\forall x \in V: \ x \ne 0 \ \Rightarrow \ T(x) \ne 0$$In other words, to show $T$ is 1-1 we only have to check that no non-zero vectors get mapped to the zero vector.

 

[songoku]

Thank you very much for the help and explanation Office_Shredder and PeroK