A vector space is a mathematical structure formed by a set of vectors, which can be added up and multiplied by a scalar. Will work on vector spaces, and we will operate with vectors and will define the concept of basis.
On the plane, two vectors $$\vec{u}$$ and $$\vec{v}$$ form a basis if they are linearly independent, since any vector $$\vec{w}$$ can be expressed as a linear combination of these two vectors.
The basis formed by $$\vec{u}$$ and $$\vec{v}$$ is represented like $$B=\{\vec{u}, \vec{v}\}$$.
Given any basis $$B=\{\vec{u}, \vec{v}\}$$, $$$\vec{w}= \lambda\vec{u}+\mu\vec{v}$$$
This expression is unique, or, in other words, $$\lambda$$ and $$\mu$$ are uniquely determined.
The coordinates of $$\vec{w}$$ in the basis $$B$$ are $$\lambda$$ and $$\mu$$. We can say that $$\vec{w}=(\lambda,\mu)$$ in the base $$B$$.
From the infinite number of basis that we can find among the vectors of the plane there is one that is especially simple: it is the one that is formed by two vectors $$\vec{i}$$ and $$\vec{j}$$ perpendicular to each other and with module $$1$$. This basis is named the canonical basis of the plane.
Remember that two vectors are perpendicular when they form an angle of $$90^\circ$$.

The vector $$\vec{v}=(2,3)$$ expressed in the canonical basis $$B=\{\vec{i}, \vec{j}\}$$ is $$\vec{v}=2\vec{i}+3\vec{j}$$.

Do the following vectors form a basis in the plane?

$$\vec{u}=(1,1)$$, $$\vec{v}=(3,3)$$. Com que $$\dfrac{1}{3}= \dfrac{1}{3}$$ Therefore, they are l.d. (linearly dependent) vectors, so they cannot form a base.

$$\vec{u}=(1,2)$$, $$\vec{v}=(2,3)$$. Com que $$\dfrac{1}{2}\neq \dfrac{2}{3}$$ They are l.i. (linearly independent) vectors, therefore they form a basis in the plane.
 $$\vec{u}=(4,2)$$, $$\vec{v}=(2,1)$$. Com que $$2=\dfrac{4}{2}= \dfrac{2}{1}=2$$ Therefore, they are l.d. (linearly dependent) vectors, so they cannot form a base.


To express the vector $$\vec{w}=(4,5)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$ where $$\vec{u}=(1,1)$$ and $$\vec{v}=(2,3)$$.
We want to find $$\lambda$$ and $$\mu$$ such that: $$$ (4,5)=\lambda(1,1)+\mu(2,3)= (\lambda,\lambda)+(2\mu,3\mu)=(\lambda+2\mu,\lambda+3\mu)$$$ therefore, $$$ \left. \begin{array}{lr} 4=\lambda+2\mu & (a) \\ 5=\lambda +3\mu & (b) \end{array} \right\} \Rightarrow \ \text{ subtracting } \ (a)(b) \Rightarrow 1=\mu \Rightarrow \lambda=42\mu=42=2$$$
The vector $$\vec{w}=(4,5)$$ will be $$(2,1)$$ in the basis $$B=\{\vec{u},\vec{v}\}$$.