Basis and coordinates

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, $$\left. \begin{array}{lr} 4=\lambda+2\mu & (a) \\ 5=\lambda +3\mu & (b) \end{array} \right\} \Rightarrow \ \text{ subtracting } \ (a)-(b) \Rightarrow 1=\mu \Rightarrow \lambda=4-2\mu=4-2=2$$

The vector $\vec{w}=(4,5)$ will be $(2,1)$ in the basis $B=\{\vec{u},\vec{v}\}$.