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I want to prove the linear map $L : \mathbb R^n \to \mathbb R^n$ is onto if and only if $L$ is one to one.

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    Not only in *asking* is where success resides but also in showing some effort, giving some reasons/background of the problem, etc.2012-07-30

3 Answers 3

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The statement is true not only for any linear operator on $\mathbb{R^n}$, but for any linear map between finite dimensional vector spaces of equal dimension. This follows from the rank-nullity theorem, which states that given vector spaces $V$ and $W$, and a linear map $T: V \to W$,

$\dim T(V) + \dim\ker T= \dim V$

Fact: $T$ is one to one $\Longleftrightarrow$ $\ker T = \{0\}$

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That is false. Consider the map $L:\mathbb{R}^n\to\mathbb{R}^n$ defined by $L(a_1,\ldots,a_n)=(a_1,\ldots,a_n+1)$ This is both onto and one-to-one, but not linear.


You probably meant to ask about the fact that, given a linear map $L:\mathbb{R}^n\to\mathbb{R}^n$, $L$ is onto if and only if it is one-to-one. Note that the statements $A\iff B\text{ and }C$ and $A\implies (B\iff C)$ are different.

In this discussion, we are using $A$ = "$L$ is linear", $B$ = "$L$ is one-to-one", and $C$ = "$L$ is onto"

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    Yes exactly, I edited my question. Sorry.2012-07-29
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This is false. The trivial map $L : \mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $L(v) = 0$ is linear, but not one to one or onto.


For the new phrasing of the question. Suppose $L$ is linear. If $L$ is injective, then $\text{ker}(L) = \{0\}$. By the first isomorphism theorem, $im(L) \simeq \mathbb{R}^n / \{0\}$. This is a $n$-dimensional vector space. The only $n$ dimensional vector subspace of a $n$-dimensional vector space is the entire vector space. So $L$ is surjective.

Suppose that $L$ is surjective. Then again by the first isomorphism theorem, $\mathbb{R}^n / \text{ker}(L) \simeq \mathbb{R}^n$. So $\text{ker}(L)$ is a zero dimensional vector subspace. Hence $\text{ker}(L) = \{0\}$. $L$ is injective.