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My Linear Algebra textbook offers the following theorem:

Let $T: R^n \to R^m$ be a linear transformation. Then $T$ is one-to-one if and only if the equation $T(\vec{x})$ = $\vec{0}$ has only the trivial solution.

I don't understand why this is true.

From discrete mathematics, I know that a mapping/function $T: A \to B$ is one-to-one if, for $a \in A$ and $b \in B$, we have that $f(a) = f(b)$ implies that $a = b$. In other words, no two elements in the domain map to the same element in the codomain.

In terms of vectors, that means that no two vectors $\vec{x}$ in the domain $R^n$ map to the same vector $A\vec{x}$ in $R^m$. Or in other words, $A\vec{x} = \vec{b}$ has at most one solution for all $\vec{x} \in R^n$.

In the theorem they give, why are they only talking about $\vec{b} = \vec{0}$, the zero vector?

What if $T(\vec{x}) = \vec{b}$ has infinitely many solutions for $\vec{b} \ne \vec{0}$, violating the condition for one-to-oneness?

I don't see why only examining the special case of $T(\vec{x}) = \vec{0}$ can allow us to conclude that the entire mapping is one to one.

Edit: or is that not what they're saying? Is it because of the wording "if and only if"? Does the theorem actually translate to:

  • if $T$ is one to one, then $T(\vec{x}) = \vec{0}$ has only the trivial solution

  • if $T(\vec{x}) = \vec{0}$ only has the trivial solution, then $T(\vec{x})$ is one to one

I can see both of these being true.

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    $\begingroup$ If $T(\vec x_1)=\vec b=T(\vec x_2)$ then $T(\vec x_1-\vec x_2)=0$. $\endgroup$
    – lulu
    Sep 14, 2017 at 14:04
  • $\begingroup$ And vice versa. $\endgroup$
    – Bernard
    Sep 14, 2017 at 14:06
  • $\begingroup$ @lulu Would love a bit more clarification. I wrote that down, and here's what I got. Correct me if this is wrong: suppose $T(\vec{x_1}) = \vec{b}$ and $T(\vec{x_2}) = \vec{b}$. Then by linearity, $T(\vec{x_1} - \vec{x_2}) = T(\vec{x_1}) - T(\vec{x_2}) = \vec{b} - \vec{b} = \vec{0}$. So we have that $T(\vec{x_1} - \vec{x_2}) = \vec{0}$. Such a mapping is only one to one if $\vec{x_1} = \vec{x_2}$. This would in turn imply that $T(\vec{x_1} - \vec{x_2}) = T(\vec{0}) = \vec{0})$, or in other words, $T(\vec{x}) = \vec{0} $ only has the trivial solution. $\endgroup$ Sep 14, 2017 at 14:10
  • $\begingroup$ @Bernard And if what I did is correct, how would I do the reverse? Re: your comment of "vice versa". Thanks! $\endgroup$ Sep 14, 2017 at 14:12
  • $\begingroup$ I just meant it's an equivalence, not just an implication. $\endgroup$
    – Bernard
    Sep 14, 2017 at 14:13

2 Answers 2

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Suppose $T$ is not one-to-one. Then there exist $a$ and $b$ with $a \neq b$ such that $T(a) = T(b)$. But then by linearity we have that $T(a-b) = 0$, and so there is a nontrivial solution to $T(x) = 0$. (Aside: it is the linearity that allows us to make broad statements and consider only the point $0$).

By the contrapositive of what we've just shown, we now have that $T(x)$ having no nontrivial solutions implies that $T$ is one-to-one.

Conversely, suppose that $T$ is one-to-one. Then as $T(0) = 0$, there can be no more solutions to $T(x) = 0$. So $T$ being one-to-one implies that there are no nontrivial solutions to $T(x) = 0$.

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Special thanks to @lulu and @Bernard in the comments. Gonna write up an answer.

The theorem is equivalent to:

  • if $T$ is one to one, then $T(\vec{x}) = \vec{0}$ has only the trivial solution

  • if $T(\vec{x}) = \vec{0}$ only has the trivial solution, then $T(\vec{x})$ is one to one

We need to prove both these cases to prove the theorem. In both cases, we'll assume $T$ is linear.


1) if $T$ is one to one, then $T(\vec{x}) = \vec{0}$ has only the trivial solution

We proceed with a direct proof. Assume that $T(\vec{x})$ is one to one. Then for $\vec{a}, \vec{b} \in R^n$, we have that $T(\vec{a}) = T(\vec{b}) \implies \vec{a} = \vec{b}$, by definition of one to one. Now, by linearity, we have that $T(\vec{a} - \vec{b}) = T(\vec{a}) - T(\vec{b}).$ Furthermore, since $\vec{a} = \vec{b}$ and $T(\vec{a}) = T(\vec{b})$, we can conclude that $T(\vec{a} - \vec{b}) = T(\vec{0}) = T(\vec{a}) - T(\vec{b}) = \vec{0}$. In other words, $T(\vec{0}) = \vec{0}$. This means $T(\vec{x}) = \vec{0}$ only has the trivial solution.

1) if $T(\vec{x}) = \vec{0}$ only has the trivial solution, then $T$ is one to one

We proceed with a direct proof. Assume that $T(\vec{x}) = \vec{0}$ only has the trivial solution. Then $\vec{x} = \vec{0}$. Let's take vectors $\vec{a}, {b} \in R^n$. Because $T(\vec{x})$ is linear, we have that $T(\vec{a} - \vec{b}) = T(\vec{a}) - T(\vec{b}) = \vec{0}.$ Rearranging, $T(\vec{a}) = T(\vec{b}).$ We now need to prove that this implies that $\vec{a} = \vec{b}$. we do so by noticing that since $T(\vec{x}) = \vec{0} \implies \vec{x} = \vec{0}$, then $T(\vec{a} - \vec{b}) = \vec{0} \implies \vec{a} - \vec{b} = \vec{0}$. Solving, we have that $\vec{a} = \vec{b}$, as desired.

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