When will be this surjective or injective?

**gotmejerry** · November 27th 2011, 05:49 AM

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

**Also sprach Zarathustra** · November 27th 2011, 06:54 AM

Originally Posted by gotmejerry

Let $\bold{A}$ be an $m \times n$ matrix and $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ : $\bold{x} \rightarrow \bold{A} \cdot \bold{x}$ 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...

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

$f$ surjective iff $rank(A)=m$

**gotmejerry** · November 27th 2011, 08:04 AM

Can you give me a bit more hint where to start?

**gotmejerry** · November 27th 2011, 10:20 AM

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

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

What should i do with the surjective part?

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