The notion of basis

Vector calculus in plane and space: Bases, Coordinates and Equations

The notion of basis

In order to carry out actual calculations it is useful to describe vectors using numbers. The concepts we require for this are basis and coordinate.

Basis of plane and space

A basis (plural: bases) of the plane consists of two vectors #\vec{e}_1# and #\vec{e}_2#, which are not on the same line through #\vec{0}#. Each vector #\vec{v}# of the plane can now be written in its own unique way as a linear combination of the vectors #\vec{e}_1# , #\vec{e}_2#:
\[
\vec{v}= v_1 \cdot \vec{e}_1 + v_2 \cdot \vec{e}_2
\]
for certain numbers #v_1# and #v_2# unique to #\vec{v}#. The numbers #v_1# , #v_2# are called the coordinates of the vector #\vec{v}# relative to the basis.
A basis of the space consists of three vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#, which are not in the same plane as #\vec{0}#. Each vector #\vec{v}# can be written as a linear combination of these three vectors:
\[
\vec{v} = v_1 \cdot\vec{e}_1 + v_2 \cdot\vec{e}_2 + v_3 \cdot\vec{e}_3
\]
where the scalars #v_1#, #v_2#, and #v_3# are uniquely determined. The vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# form a basis of the space, and the numbers #v_1#, #v_2#, and #v_3#, are called the coordinates of the vector #\vec{v}# with respect to the basis.
The vectors #\vec{e}_1=\rv{1,0}# and #\vec{e}_2=\rv{0,1}# form the basis of #\mathbb{R}^2#, which we call the standard basis of #\mathbb{R}^2#.
The vectors #\vec{e}_1=\rv{1,0,0}#, #\vec{e}_2=\rv{0,1,0}#, and #\vec{e}_3=\rv{0,0,1}#, form the basis of #\mathbb{R}^3#, which we call the standard basis of #\mathbb{R}^3#.

The coordinates in the plane #\mathbb{R}^2# and in the space #\mathbb{R}^3#, which have been previously defined, are the coordinates as defined here for the standard basis:\[\rv{v_1,v_2}=v_1\cdot\rv{1,0}+v_2\cdot\rv{0,1}=v_1\cdot\vec{e}_1+v_2\cdot\vec{e}_2\] for #\mathbb{R}^2# and similarly, but with an additional term #v_3\cdot\vec{e}_3#, for #\rv{v_1,v_2,v_3}# in #\mathbb{R}^3#.

We will now check why the coordinates are uniquely determined by a basis of the plane: say #v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2=w_1\cdot\vec{e}_1+w_2\cdot\vec{e}_2#. What follows, after subtracting #v_2\vec{e}_2-w_1\cdot\vec{e}_1# on both sides:\[\left(v_1-w_1\right)\cdot \vec{e}_1 = \left(w_2-v_2\right)\cdot \vec{e}_2\] Because #\vec{e}_1# and #\vec{e}_2# are not on the same line as #\vec{0}#, we must conclude with the Criteria for two vectors on a line through the origin that #v_1-w_1=0# and #w_2-v_2=0#, in other words #v_1=w_1# and #v_2=w_2#. This shows that the coordinates #v_1# and #v_2# are unique.

The same goes for the space: say \[v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3=w_1\cdot\vec{e}_1+w_2\cdot\vec{e}_2+w_3\cdot\vec{e}_3\] This can be rewritten as \[\left(v_1-w_1\right)\cdot \vec{e}_1 + \left(v_2-w_2\right)\cdot \vec{e}_2+ \left(v_3-w_3\right)\cdot\vec{e}_3=0\] If #v_1-w_1\ne0#, then after multiplying both sides with the the scalar #\frac{1}{v_1-w_1}#, we see that #\vec{e}_1# is a linear combination of #\vec{e}_2# and #\vec{e}_3#; this means that #\vec{e}_1# is in the plane through #\vec{0}#, #\vec{e}_2#, and #\vec{e}_3#, a contradiction to the assumption. What should apply is: #v_1=w_1#. With the same reasoning for the indices #2# and #3#, rather than #1#, we deduce that #v_2=w_2# and #v_3=w_3#. This shows that the coordinates #v_1#, #v_2#, and #v_3# are unique.

OrthormalityLater we will discuss, next to length, the notion of perpendicularity. The vectors of a basis need not have length #1# and need not be perpendicular. If these requirements are met, the basis will be called orthonormal.

We next discuss how to recognize a basis.

Two characteristics of a basis

We fix an origin of the plane and the space.

Two vectors in the plane, respectively three vectors in the space, form a basis if and only if each vector of the plane, respectively the space, is a linear combination of these.
Two vectors in the plane, respectively three vectors in the space, form a basis if and only if the only way to write the origin as a linear combination of these vectors, is when all scalars are equal to #0#.

We only give the proof in the case of the space. The proof for the plane is similar. The origin is indicated by #\vec{0}#.

1. Let #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# be a basis of the space, and #\vec{v}# a random vector. We show that #\vec{v}# is a linear combination of the basis vectors. We look at the line # \ell# with parametric representation #\vec{v}+\lambda\cdot \vec{e}_1# and the plane #V# through the three vectors #\vec{0}#, #\vec{e}_2#, and #\vec{e}_3#. The plane #V# has the parametric representation \[ \mu\cdot\vec{e}_2+\nu\cdot\vec{e}_3\]

The line #\ell# is not parallel to #V# because the direction vector #\vec{e}_1# of #\ell# is not the direction vector of #V#: if it were, then #\vec{e}_1# would be a linear combination of the direction vectors #\vec{e}_2# and #\vec{e}_3# of #V#, which contradicts that the basis vectors are not on the same plane as #\vec{0}#.

There is an intersection point of the line #\ell# with the plane #V#. In terms of the parametric representations of #\ell# and #V#, this means that there are numbers #\lambda#, #\mu#, and #\nu#, so \[ \vec{v}+\lambda\cdot \vec{e}_1=\mu\cdot \vec{e}_2+\nu\cdot\vec{e}_3\] If we subtract the term #\lambda\cdot \vec{e}_1# on both sides of this equality, we find that #\vec{v}# is a linear combination of #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#.

To prove the converse, we assume that #\vec{e}_1# , #\vec{e}_2# , #\vec{e}_3# is not a basis of the space, and we deduce that not every vector is a linear combination of these three vectors. Because the three vectors do not form a basis, they are located in in a plane #U# together with #\vec{0}#. This means that one of the three, say #\vec{e}_1#, is a linear combination of #\vec{e}_2# and #\vec{e}_2#: there are scalars #\lambda# and #\mu#, so \[\vec{e}_1=\lambda \cdot \vec{e}_2 + \mu\cdot \vec{e}_3\] But then each linear combination of #\vec{e}_1#, #\vec{e}_2 #, and #\vec{e}_3# is a vector in the plane #U#: \[v_1\cdot \vec{e}_1+v_2\cdot \vec{e}_2 + v_3\cdot \vec{e}_3=\left( v_1\lambda+v_2\right)\vec{e}_2+\left(v_1\mu+v_3\right)\vec{e}_3\] In particular, a vector that is not in the plane #U# is not a linear combination of #\vec{e}_1#, #\vec{e}_2 #, and #\vec{e}_3#. Thus, it is not true that each vector is a linear combination of #\vec{e}_1# , #\vec{e}_2 # and #\vec{e}_3# is. This proves the reverse implication by contradiction.

2. If #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# are a basis of the space, then the scalars #v_1#, #v_2#, and #v_3# in the expression \[\vec{0}= v_1 \cdot\vec{e}_1 + v_2 \cdot\vec{e}_2 + v_3 \cdot\vec{e}_3\] are uniquely determined, and so equal to #0#. This proves that the only way to write #\vec{0}# as a linear combination of the vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#, is the one with all scalars equal to #0#.

Now we can assume that the only way to write #\vec{0}# as a linear combination of #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#, is the one with all scalars equal to #0#. We want to deduce that #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# are a basis. Suppose they are not. Then these three vectors would be together in a plane with #\vec{0}#. The vectors #\vec{e}_2# and #\vec{e}_3# are not on the same line as #\vec{0}#, because otherwise there would be scalars #u_2# and #u_3#, both not equal to #0#, with #u_2\cdot\vec{e}_2=u_3\cdot\vec{e}_3#. This means that #\vec{0}=u_2\cdot\vec{e}_2-u_3\cdot\vec{e}_3# is a linear combination of #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#, having at least one scalar not equal to #0#, contradicting our assumption. We conclude that #\vec{0}#, #\vec{e}_2#, and #\vec{e}_3# are in the same plane. Call this plane #U#. It has the parametric representation \[\lambda \cdot\vec{e}_2+\mu\cdot\vec{e}_3\] But the three vectors, including #\vec{e}_1#, are together in the same plane as #\vec{0}#. This plane must therefore be equal to #U#, so there are numbers #\lambda# and #\mu# such that\[\vec{e}_1=\lambda \cdot\vec{e}_2+\mu\cdot\vec{e}_3\] This means that #\vec{0}# can be written as \[\vec{0}=\vec{e}_1-\lambda \cdot\vec{e}_2-\mu\cdot\vec{e}_3\] a contradiction to the assumption. Therefore, the three vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# must be a basis.

This proves the theorem.

Do #\left[ -2 , -5 \right]# and #\left[ 8 , 20 \right]# form a basis of #\mathbb{R}^2#?

No

To determine this, we use the second characteristic of a basis. Suppose there are two scalars #\lambda# and #\mu#, so #\vec{0}=\lambda\cdot \left[ -2 , -5 \right]+\mu\cdot \left[ 8 , 20 \right]#. #\lambda# and #\mu# satisfy the system of linear equations \[\eqs{ 8\cdot \mu-2\cdot \lambda&=&0\cr 20\cdot \mu-5\cdot \lambda&=&0}\] whose solutions are: #\lambda= 4 \mu#. In particular, #\lambda= 8\land \mu=2# is a solution.

This shows that #\vec{0}# can be written as the linear combination \[ \lambda \cdot \left[ -2 , -5 \right]+\mu\cdot \left[ 8 , 20 \right]\] wherein #\lambda= 8\land \mu=2# are scalars that are not both equal to #0#. The second characteristic of a basis gives that #\left[ -2 , -5 \right]# and #\left[ 8 , 20 \right]# do not form a basis.

The first characteristic of a basis can also be used to derive this. If #\vec{v}=\rv{v_1,v_2}# is a linear combination #\vec{v}=\lambda\cdot{ \left[ -2 , -5 \right]}+\mu\cdot {\left[ 8 , 20 \right]}# of #\left[ -2 , -5 \right]# and #\left[ 8 , 20 \right]#, with scalars #\lambda # and #\mu#, those for the scalars satisfy the system of linear equations \[\eqs{ 8\cdot \mu-2\cdot \lambda&=&v_1\cr 20\cdot \mu-5\cdot \lambda&=&v_2 }\] This shows that the vector #\vec{v}=\rv{1,0} # cannot be written as a linear combination of #\left[ -2 , -5 \right]# and #\left[ 8 , 20 \right]#. The first characteristic of a basis gives that #\left[ -2 , -5 \right]# and #\left[ 8 , 20 \right]# do not form a basis.

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