- #1

Carla1985

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I am trying to understand some examples given to me by my supervisor but am struggling with some bits. The part I don't understand is: if the equation

$$ax+b\lambda=\bar{a}x-\bar{d}y$$

holds for any $x,y\in V$, an open neighbourhood of the origin, and $\lambda$ is a mapping from $V$ to $R^2$, then the coefficients of $x$ must be equal, so $a=\bar{a}$. I don't see how I can state this for definite without yet knowing what $\lambda$ is. Can someone please explain why this is the case? Thanks.

Also if I have

$$ax+bx\lambda +c\lambda+d=\bar{a}x+\bar{b}xy+\bar{c}y$$

can I apply the same principles?

Thanks in advance

Carla