Open neighbourhoods and equating coefficients of polynomials

[Carla1985]

Hi all,

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

 

[jvicens]

Hi Carla,

It seems like your supervisor is discussing linear transformations and their properties. Let me try to break down the equation for you to provide a better understanding.

Firstly, the equation
$$ax+b\lambda=\bar{a}x-\bar{d}y$$
is representing a linear transformation in the form of an equation. Here, $x$ and $y$ are variables in a vector space $V$ and $\lambda$ is a mapping from $V$ to $R^2$. This means that for any given values of $x$ and $y$ in an open neighbourhood of the origin, the equation must hold true.

Now, for the coefficients of $x$ to be equal, we can compare the terms $ax$ and $\bar{a}x$ on both sides of the equation. In order for these terms to be equal, the coefficients $a$ and $\bar{a}$ must be equal. This is because they are both multiplied by the same variable $x$.

So, even without knowing the exact form of the mapping $\lambda$, we can still say that $a=\bar{a}$ based on the equation given to us. This is a fundamental property of linear transformations.

Moving on to your second question, the same principles can be applied. In this case, we have multiple variables and terms on both sides of the equation. However, the same logic applies. In order for the terms containing $x$ to be equal, the coefficients $a$ and $\bar{a}$ must be equal. Similarly, for the terms containing $y$ to be equal, the coefficients $b$ and $\bar{b}$ must be equal.

I hope this explanation helps you understand the concept better. Let me know if you have any further questions.