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Let $A$ be the set of affine functions denoted by $$ f_{a,b} (x) = a x + b$$ (with $a \in \mathbb{R}$ and $b \in \mathbb{R}$ and $x \in \mathbb{R}$ and $f_{a,b} (x) \in \mathbb{R}$)

We can easily show that $A$ has the structure of a real vector space (internal addition $f_{a,b} + f_{c,d}= f_{a+c, b+d}$ and external multiplication for $\alpha \in \mathbb{R}$, $\alpha f_{a,b} = f_{\alpha a , \alpha b}$).

I would like to further show that $S=\{ f_{1,0} , f_{0,1} \}$ is a basis for $A$.

$S$ generates $A$: take any $f_{a,b} \in A$ . For $\alpha_1$ and $\alpha_2$ real, we have $f_{a,b}(x) = \alpha_1 f_{1,0} (x) + \alpha_2 f_{0,1} (x)$ $\Leftrightarrow a x + b = \alpha_1 x + \alpha_2$. We get $\alpha_1 = a$ and $\alpha_2 = b$.

However I fail to show that $f_{1,0}$ and $f_{0,1}$ linearly independent, for all $x \in \mathbb{R}$

1 Answers 1

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If $$0=\alpha_1 f_{1,0}(x) + \alpha_2 f_{0,1}(x) = \alpha_1 x + \alpha_2$$ is true for all $x$, then what can you say about $\alpha_1$ and $\alpha_2$?

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    that $ \alpha_2 = - \alpha_1 x $ ( and we don't have $\alpha_1 = \alpha_2 = 0 $ this is exactly where I block... If your statement is true then it should imply that $\alpha_1$ and $\alpha_2$ are equal to zero) – 2017-01-24
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    @Sha That's why I put **for all $x$** in bold. Your equation $\alpha_2 = - \alpha_1 x$ needs to hold **for all $x$**. – 2017-01-24
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    for $x =0$ we have $\alpha_2 =0$ and $\alpha_1 = - \alpha_2 / x = 0 / 0$ which is impossible or makes no sense – 2017-01-24
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    so it is not true for all $x$ but how do we conclude ? can we say since it is not true for all $x$ the $f_{1,0}$ and $f_{0,1}$ are linearly independent? – 2017-01-24
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    For $x = 0$ we have that $\alpha_2 = - \alpha_1 x$ impossible/absurd so necessarily $\alpha_1 = \alpha_2 = 0 $ ? Or by contraposition we take $\alpha_1 =0$ and $\alpha_2 \neq 0$ and we get that $\alpha_1 f_{1,0} (x) + \alpha_2 f_{0,1} (x) \neq 0$ – 2017-01-24
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    @Sha Looking at $x=0$ is not enough. For $x=0$, all you can conclude is $\alpha_2=0$. Then you should ask "which $\alpha_1$ will satisfy $0=-\alpha_1 x$ for all $x$?" – 2017-01-24
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    Answer " \alpha_1 = 0 " , that you! Is the following correct ? We have the statement " A: $0 = \alpha_1 f_{1,0} (x) + \alpha_2 f_{0,1} (x) = \alpha_1 x + \alpha_2$ " implies " B: $\alpha_1 = \alpha_2 = 0$". We want to show that If A true then B is true. We can also show that "not B" is false, so i suppose that $\alpha_1 \neq 0$ and $\alpha_2=0$. We have that FOR ALL $x$, $\alpha_1 x + \alpha_2=0 \Leftrightarrow \alpha_1 x =0$ which is not true for ALL x. Hence, Necessarily $\alpha_1 = 0$... Is this better? – 2017-01-25