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Thread: When will be this surjective or injective?

  1. #1
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    When will be this surjective or injective?

    Let $\displaystyle \bold{A}$ be an $\displaystyle m \times n$ matrix and $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ :$\displaystyle \bold{x} \rightarrow \bold{A} \cdot \bold{x}$ What is a necessary and sufficient condition for f to be injective. And surjective?
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    Re: When will be this surjective or injective?

    Quote Originally Posted by gotmejerry View Post
    Let $\displaystyle \bold{A}$ be an $\displaystyle m \times n$ matrix and $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ :$\displaystyle \bold{x} \rightarrow \bold{A} \cdot \bold{x}$ What is a necessary and sufficient condition for f to be injective. And surjective?
    I'ii give you the answer, and you will try to explain me why is that...

    $\displaystyle f$ injective iff $\displaystyle rank(A)=n$

    $\displaystyle f$ surjective iff $\displaystyle rank(A)=m$
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  3. #3
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    Re: When will be this surjective or injective?

    Can you give me a bit more hint where to start?
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  4. #4
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    Re: When will be this surjective or injective?

    For injectivity I did. It is injectien for every f(x)=f(y) ---> x=y.

    $\displaystyle \begin{bmatrix} {a_{11}}\\{a_{21}}\\ \vdots\\{a_{m1}\end{bmatrix}(x_1 -y_1)+\begin{bmatrix} {a_{12}}\\{a_{22}}\\ \vdots\\{a_{m2}\end{bmatrix}(x_2-y_2)+\hdots + \begin{bmatrix} {a_{1n}}\\{a_{2n}}\\ \vdots\\{a_{mn}\end{bmatrix}(x_n-y_n)=0} $ So if the rank is n it can only be zero if $\displaystyle \bold{x}=\bold{y}$

    What should i do with the surjective part?
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