MHB Violation of Linear Transformation?

[bwpbruce]

This is a solution that I observed from my textbook to a linear transformation problem:

$\textbf{Problem}$
Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation. Suppose $\{\textbf{u}, \textbf{v}\}$ is a linear independent set, but $\{T\textbf{(u)}, T\textbf{(v)}\}$ is a linearly dependent set. Show that $T\textbf{(x) = 0}$ has a nontrivial solution.

$\textbf{Solution}$
Suppose that $\{\textbf{(u)}, \textbf{(v)}\}$ is a linearly independent set in $\mathbb{R}^n$ and yet $T\textbf{(u)}$ and $\textbf{(v)}$ are linearly dependent. Then there exist weights $c_1, c_2$, not both zero, such that $c_1\textbf{(u)} + c_2\textbf{(v)} = \textbf{0}$ . Because $T$ is linear,
$T(c_1\textbf{(u)} + c_2\textbf{(v)}) = \textbf{0}$. That is, the vector $\textbf{x} = c_1\textbf{u} + c_1\textbf{u}$ satisfies $T\textbf{(x) = 0}$. Furthermore, $\textbf{x}$ cannot be the zero vector, since that would mean that a nontrivial linear combination of u and v is zero, which is impossible because $\textbf{u}$ and $\textbf{v}$ are linearly independent. Thus, the equation $T\textbf{(x) = 0}$ has a nontrivial solution.

Isn't $T$ not linear since $\textbf{x} \ne \textbf{0}$?

Property iii of the Definition of Linear Transformation states $T(\textbf{(0)} = \textbf{0}$ so something is contradictory here.

 

[Evgeny.Makarov]

The solution has to be corrected as follows.

Then there exist weights $c_1,c_2$, not both zero, such that $c_1\textcolor{red}{T}(\mathbf{u})+c_2\textcolor{red}{T}(\mathbf{v})=\mathbf{0}$.

Now concerning your question:

Isn't $T$ not linear since $\textbf{x} \ne \textbf{0}$?

Property iii of the Definition of Linear Transformation states $T(\textbf{(0)} = \textbf{0}$ so something is contradictory here.

Could you describe more precisely where you see a contradiction?

 

[bwpbruce]

I suppose you don't agree that there is a contradiction. I don't see how $T$ satisfies property iii if $\textbf{x} \ne \textbf{0}$. If $\textbf{x} \ne \textbf{0}$, then property iii of the definition of linear transformation is violated. Unless someone can demonstrate to me otherwise. If no explanation or demonstration is provided, I'll just feel left in the dark.

 

[Evgeny.Makarov]

I suppose you don't agree that there is a contradiction.

A good and distinguishing feature about mathematics is that you can actually convince another person if you write your proof in sufficient detail.

I don't see how $T$ satisfies property iii if $\textbf{x} \ne \textbf{0}$.

This property only claims that $T(\mathbf{0})=\mathbf{0}$; it says nothing about $T(\mathbf{x})$ where $\mathbf{x}\ne\mathbf{0}$.

 

[bwpbruce]

A good and distinguishing feature about mathematics is that you can actually convince another person if you write your proof in sufficient detail.

This property only claims that $T(\mathbf{0})=\mathbf{0}$; it says nothing about $T(\mathbf{x})$ where $\mathbf{x}\ne\mathbf{0}$.

Well, but according to the solution provided, $T\textbf{x}$ does equal $\textbf{0}$. Which means $\textbf{x}$ should also be $\textbf{0}$. I'm asking for clarification on the matter. I'm hoping you can help. If you want, you can provide a proof of how it is not contradictory.

 

[bwpbruce]

Here's my claim as simply as I can put it. If $\textbf{x} \ne \textbf{0}$, then it is not possible to verify $T\textbf{0} = \textbf{0}$. Verifying property iii is necessary for verifying that $T$ is linear.

 

[Evgeny.Makarov]

Well, but according to the solution provided, $T\textbf{x}$ does equal $\textbf{0}$. Which means $\textbf{x}$ should also be $\textbf{0}$.

No, it does not mean this. For example, $\sin0=0$, but the fact that $\sin x=0$ does not imply that $x=0$.

If $\textbf{x} \ne \textbf{0}$, then it is not possible to verify $T\textbf{0} = \textbf{0}$.

If by this you mean that $\textbf{x}\ne \textbf{0}$ and $T(\mathbf{x})=\mathbf{0}$ do not imply that $T$ is or is not linear, then I agree.

Verifying property iii is necessary for verifying that $T$ is linear.

Verifying that $T$ is linear is not needed because it is given. The flow of a proof should be from assumptions to the final claim, not back to assumptions. It is not required to take some conclusion of an argument and ask yourself whether this conclusion implies the assumptions. It may or may not imply them.

Here is an example of the situation described in the problem. Let $m=n=2$, $T$ be given by $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\mathbf{u}=(1,0)$ and $\mathbf{v}=(1,1)$ (as columns). Then $T(\mathbf{u})=T(\mathbf{v})$, so $T(\mathbf{u})$ and $T(\mathbf{v})$ are linearly dependent even though $\mathbf{u}$ and $\mathbf{v}$ are linearly independent, n particular, $\mathbf{u}-\mathbf{v}\ne\mathbf{0}$. In this situation, $T(\mathbf{u}-\mathbf{v})=\mathbf{0}$. If you have an alleged argument that results in a contradiction, then you can apply it to these objects to find an error, since this situation is consistent.

If you still have objections, you should come up with a precise and detailed proof a contradiction. That is, you need to show that the assumptions of the problem imply some statement $A$ together with its negation $\neg A$. If you have a formal proof, we can examine it.