I'm solving the following problem:

Be $U=\lbrace(x,y) \in \mathbb{R}^2 | x>0, y>0\rbrace$. If $z_1=(x_1,y_1)$ and $z_2=(x_2,y_2)$ are elements of $U$ and $\lambda \in \mathbb{R}$ define: $$z_1+z_2=(x_1 x_2, y_1 y_2), \lambda z_1=(x_{1}^{\lambda}, y_{1}^{\lambda})$$

I showed that $U$ is a vector space with additive identity $(1,1)$.

Now I'm trying to show that $\mathcal{B}=\lbrace (e,1),(1,e) \rbrace$ is a basis for $U$. I already showed that $(e,1),(1,e)$ are linearly independent. Taking into account the new definition of addition and scalar multiplication, it means that $\lambda_1(e,1)+\lambda_2(1,e)=(1,1) \Rightarrow \lambda_1=\lambda_2=0$:





Now I suppose I have to show that $\mathcal{B}=\lbrace (e,1),(1,e) \rbrace$ generates all of $U$. But I can't wrap my mind around it. $z_1+z_2=(x_1 x_2, y_1 y_2)$ and $\lambda z_1=(x_{1}^{\lambda}, y_{1}^{\lambda})$ are clearly positive, since $x_1, x_2, y_1$ and $y_2$ are always positive. But how can I guarantee that $\mathcal{B}=\lbrace (e,1),(1,e) \rbrace$ will generate all of $U$?

Is it enough to say that $e^{\lambda}$, $\lambda \in \mathbb{R}$, yields only positive values, since all linear combinations between any two vectors will end up with $(e^{\lambda_1},e^{\lambda_2})$?


No, it does not suffice to say that $e^x$ yields only positive numbers, or that $\mathcal{B}$ yields only elements in $U$. You need to show that every element in $U$ is a linear combination of the vectors of $\mathcal{B}$ .

Take a general vector in $U$: $z = (x,y)$, and you need to show that there are two scalars $\lambda_1, \lambda_2 \in \mathbb{R}$ so that $\lambda_1(e, 1) + \lambda_2(1, e) = (x, y)$.

According to your rules of addition and scalar multiplication: $$\lambda_1(e, 1) + \lambda_2(1, e) = (e^{\lambda_1}, 1^{\lambda_1}) + (1^{\lambda_2}, e^{\lambda_2}) = (e^{\lambda_1}, e^{\lambda_2})$$

Can you think of values in $\mathbb{R}$ of $\lambda_1, \lambda_2$ that will make $(e^{\lambda_1}, e^{\lambda_2}) = (x, y)$?

  • $\begingroup$ I could use the fact that $e^{ln \, x}=x$ to show that I can generate any real number $x$. $\endgroup$ Jun 27 '17 at 19:43
  • 1
    $\begingroup$ Exactly. And that will show that $\mathcal{B}$ spans $U$, and therefore $\mathcal{B}$ is a basis $\endgroup$
    – AsafHaas
    Jun 27 '17 at 19:48

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