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I came across the following problem that says:

Let $A$ be an $n \times n$ matrix with real entries and suppose that the system $Ax=0$ has the unique solution $x=0$. Then the mapping $T\colon \mathbb R^n\rightarrow \mathbb R^n$ defined by $Tx=Ax$ is
$1.$ a bijection
$2.$ one-one but not onto
$3.$ onto but not one-one
$4.$ neither one-one nor onto.

I am not sure how to progress with it and in particular how to use the information "$Ax=0$ has the unique solution $x=0.$".Can someone point me in the right direction?Thanks in advance for your time.

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Clue: The rows of $A$ must be linearly independent. – Herng Yi Feb 11 '13 at 10:44
Then $A$ is non-singular (invertible). – vonbrand Feb 11 '13 at 11:05
up vote 1 down vote accepted

What is $\ker T$? By definition is $\ker T=\{x\in\mathbb R^n:Tx=0\}=\{x\in\mathbb R^n:Ax=0\}=\{0\}$ by hypothesis.

Now use that:
A linear map $T:V\longrightarrow V$ ($V$ finite dimensional) is

  • one-one iff $\ker T=\{0\}$,
  • one-one iff is onto.
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When $$Ax=0, ~~\text{and if}~~det(A)\neq 0$$ then $$A^{-1}Ax=A^{-1}0=0\Longrightarrow x=0$$ is our unique solution here. This means that the corresponding transformation $T|_A$ is one-one, so two last options are not true

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It's confusing if you use capital letters for both matrices and vectors. – Tara B Feb 11 '13 at 11:06
@TaraB: I am accustomed to using it. Thanks. I'll fix it. – Babak S. Feb 11 '13 at 11:15
Nicely put. You give great "hints" and explanations! +1 – amWhy Feb 11 '13 at 15:00

You already accepted an answer while I was writing this, but I just wanted to give a very elementary answer.

To check whether $T$ is one-to-one:

Suppose $x,y\in V$ with $Tx = Ty$. Then $Ax = Ay$, so $A(x - y) = 0$, so $x - y = 0$ and hence $x=y$. Thus $T$ is one-to-one.

Any one-to-one function from a set to itself is a bijection.

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+1 for noting me that – Babak S. Feb 11 '13 at 11:16

The condition says precisely that the kernel of $T$ is $\{0\}$. For a linear transformation having trivial kernel is equivalent to being injective. For linear transformations between vectors spaces of the same dimension being injective is the same as being surjective is the same as being bijective.

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  1. A bijection: being one-to-one follows easily from the fact that $Ax=0\iff x=0$, and then conclude that the image of the (standard) basis vectors are linearly independent, so $\dim({\rm im\,} T)=n$.
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