Basis – "Math for Non-Geeks"

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In the last two articles we got to know the two concepts generator and linear independence. In this articles we combine the two concepts and introduce the notion of basis of a vector space.

We want to work towards the concept of dimension. Intuitively, we can think of it as the maximum number of linearly independent directions in a space. If we stick to this intuition, we can give the following preliminary definition of dimension: The dimension of a vector space is the maximum number of linearly independent vectors that we can simultaneously choose in a vector space. To find this, we need to find a maximal system of linearly independent vectors.

To see if this preliminary definition makes sense, let's try to apply it in the ${\displaystyle \mathbb {R} ^{3}}$: Clearly, the ${\displaystyle \mathbb {R} ^{3}}$ should be three-dimensional. This means we have to ask ourselves two questions: Do three linearly independent vectors fit into the ${\displaystyle \mathbb {R} ^{3}}$ and is there no fourth vector being linearly independent to any of such three independent vectors? Let us now try this out: We take any first vector, e.g. ${\displaystyle v_{1}:=(2,-1,0)^{T}}$. Now we want to find a linearly independent vector, i.e. a vector that is not a multiple of ${\displaystyle v_{1}}$ or does not lie in ${\displaystyle \operatorname {span} (\{v_{1}\})}$. An example for this is ${\displaystyle v_{2}:=(2,1,0)^{T}}$.

If we want to follow the intuition that ${\displaystyle \mathbb {R} ^{3}}$ is three-dimensional, we still have to find a third vector that is linearly independent to the system ${\displaystyle \{v_{1},v_{2}\}}$. Now ${\displaystyle v_{1}}$ and ${\displaystyle v_{2}}$ span a plane in which the last component is always zero. Thus ${\displaystyle v_{3}=(0,0,1)^{T}}$ is a vector linearly independent of ${\displaystyle \{v_{1},v_{2}\}}$.

Now the question is whether with these three vectors the maximum number of linearly independent vectors has already been reached. To answer this, let us first consider the vector ${\displaystyle (1,0,1)^{T}}$ as an example. We want to check whether we can add this vector to ${\displaystyle v_{1},v_{2}}$ and ${\displaystyle v_{3}}$ and still obtain a system of linearly independent vectors. First we note that

${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}}={\frac {1}{4}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}+{\frac {1}{4}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}}$

So we can write the above vector as

${\displaystyle {\begin{pmatrix}1\\0\\1\end{pmatrix}}={\frac {1}{4}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}+{\frac {1}{4}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}+1\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}}$

So ${\displaystyle (1,0,1)^{T}\in \operatorname {span} (\{v_{1},v_{2},v_{3}\})}$. Now let us ask the same question for any vector ${\displaystyle (a,b,c)^{T}\in \mathbb {R} ^{3}}$ with ${\displaystyle a,b,c\in \mathbb {R} }$. For this we first get

${\displaystyle {\begin{aligned}{\begin{pmatrix}1\\0\\0\end{pmatrix}}&={\frac {1}{4}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}+{\frac {1}{4}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}\\[0.5em]{\begin{pmatrix}0\\1\\0\end{pmatrix}}&={\frac {1}{2}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}-{\frac {1}{2}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}\end{aligned}}}$

With this consideration we obtain for ${\displaystyle (a,b,c)^{T}}$ the representation

${\displaystyle {\begin{aligned}{\begin{pmatrix}a\\b\\c\end{pmatrix}}&=a\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+b\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}+c\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}\\[0.5em]&=a\cdot \left({\frac {1}{4}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}+{\frac {1}{4}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}\right)+b\cdot \left({\frac {1}{2}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}-{\frac {1}{2}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}\right)+c\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}\\[0.5em]&={\frac {a+2b}{4}}\cdot {\begin{pmatrix}2\\1\\0\end{pmatrix}}+{\frac {a-2b}{4}}\cdot {\begin{pmatrix}2\\-1\\0\end{pmatrix}}+c\cdot {\begin{pmatrix}0\\0\\1\end{pmatrix}}\end{aligned}}}$

Thus ${\displaystyle (a,b,c)^{T}\in \operatorname {span} (\{v_{1},v_{2},v_{3}\})}$. Since this vector was arbitrarily chosen, every vector from ${\displaystyle \mathbb {R} ^{3}}$ is representable as a linear combination of the linearly independent vectors ${\displaystyle v_{1},v_{2}}$ and ${\displaystyle v_{3}}$. Thus ${\displaystyle \{v_{1},v_{2},v_{3}\}}$ is a generator of ${\displaystyle \mathbb {R} ^{3}}$. Therefore, we cannot add another vector to ${\displaystyle v_{1},v_{2}}$ and ${\displaystyle v_{3}}$, so the system remains linearly independent, since every other vector from ${\displaystyle \mathbb {R} ^{3}}$ can be represented as a linear combination of ${\displaystyle v_{1},v_{2}}$ and ${\displaystyle v_{3}}$. In other words, ${\displaystyle v_{1},v_{2}}$ and ${\displaystyle v_{3}}$ form a maximal system of linearly independent vectors.

In summary, we proceeded as follows to find a maximal system of linearly independent vectors: We start with a vector that is not the zero vector. That means we should not consider the null vector space here. Then we proceed step by step: Once we have found linearly independent vectors ${\displaystyle v_{1},...,v_{n}}$, we form the span of these vectors. If this is the entire vector space, we have found a generator and are done. We choose a vector ${\displaystyle v_{n+1}}$ that is not in ${\displaystyle \operatorname {span} (\{v_{1},...,v_{n}\})}$. This contributes a new direction and the system ${\displaystyle v_{1},...,v_{n+1}}$ is linearly independent again. Then we do the same again until we find a generator. We thus obtain the characterisation that a maximal system of linearly independent vectors is a generator of linearly independent vectors.

So far we have started with a system of linearly independent vectors (which is not yet a generator) and extended it until it became a maximal system of linearly independent vectors. Now we want to investigate what happens when we reverse the direction. That is, we start with a generator (which is not linearly independent) and reduce it until we find a minimum (i.e., linearly independent) generator.

Let us consider

${\displaystyle E=\left\{{\begin{pmatrix}2\\1\\1\end{pmatrix}},{\begin{pmatrix}-1\\0\\-1\end{pmatrix}},{\begin{pmatrix}2\\1\\2\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\1\end{pmatrix}}\right\}}$

First, we establish that ${\displaystyle E}$ is a generator of ${\displaystyle \mathbb {R} ^{3}}$ by following calculation: For a vector ${\displaystyle (\alpha ,\beta ,\gamma )^{T}\in \mathbb {R} ^{3}}$ with ${\displaystyle \alpha ,\beta ,\gamma \in \mathbb {R} }$ we have that

${\displaystyle {\begin{pmatrix}\alpha \\\beta \\\gamma \end{pmatrix}}=(\alpha -\beta -\gamma )\cdot {\begin{pmatrix}2\\1\\1\end{pmatrix}}+\beta \cdot {\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+(\beta +\gamma -\alpha )\cdot {\begin{pmatrix}2\\1\\2\end{pmatrix}}+\beta \cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}+\alpha \cdot {\begin{pmatrix}1\\0\\1\end{pmatrix}}}$

So we can represent any vector ${\displaystyle (\alpha ,\beta ,\gamma )^{T}\in \mathbb {R} ^{3}}$ as a linear combination of vectors from the generator.

Now we ask ourselves whether we can reduce the above generator without losing the property of it being a generator. The vector ${\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\displaystyle E}$ and obtain a new generator

${\displaystyle E_{1}=\left\{{\begin{pmatrix}2\\1\\1\end{pmatrix}},{\begin{pmatrix}2\\1\\2\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\1\end{pmatrix}}\right\}}$

Can we reduce the size of this generator, as well? Yes, since

${\displaystyle {\begin{pmatrix}2\\1\\1\end{pmatrix}}={\begin{pmatrix}1\\0\\1\end{pmatrix}}+{\begin{pmatrix}1\\1\\0\end{pmatrix}}}$

So ${\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\displaystyle (1,0,1)^{T}}$ and ${\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator

${\displaystyle E_{2}=\left\{{\begin{pmatrix}2\\1\\2\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\1\end{pmatrix}}\right\}}$

Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\displaystyle \lambda ,\mu \in \mathbb {R} }$, so that

${\displaystyle {\begin{pmatrix}1\\0\\1\end{pmatrix}}=\lambda {\begin{pmatrix}1\\1\\0\end{pmatrix}}+\mu {\begin{pmatrix}2\\1\\2\end{pmatrix}}}$

Then ${\displaystyle \lambda =-\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\displaystyle \mu ={\frac {1}{2}}}$ must be true. Thus we obtain the contradiction

${\displaystyle {\begin{pmatrix}1\\0\\1\end{pmatrix}}=-{\frac {1}{2}}{\begin{pmatrix}1\\1\\0\end{pmatrix}}+{\frac {1}{2}}{\begin{pmatrix}2\\1\\2\end{pmatrix}}={\begin{pmatrix}0.5\\0\\1\end{pmatrix}}}$

So ${\displaystyle (1,0,1)^{T}}$ is not in ${\displaystyle \operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator).

In summary, we proceeded as follows: We started with a generator of vectors ${\displaystyle v_{1},\dots ,v_{n}}$ and reduced it according to the following algorithm: If ${\displaystyle v_{1},\dots ,v_{n}}$ is a system of linearly independent vectors, we cannot remove any vector without losing the property of having a generator. That means we are done and we have found a minimal generator. In the converse case, we find a vector ${\displaystyle v_{i}}$ with ${\displaystyle i\in \{1,\dots ,n\}}$ that is in ${\displaystyle \operatorname {span} (\{v_{1},\dots ,v_{i-1},v_{i+1},\dots ,v_{n}\})}$. This vector can be omitted and we obtain a new generator consisting of ${\displaystyle (n-1)}$ vectors. With this generator, we do the same steps again until we have found a system of linearly independent vectors.

We thus obtain the characterisation that a minimal generator is a generator consisting of linearly independent vectors.

We have found the characterisations of linearly independent generators as minimal generators and as maximal linearly independent subsets. Thus the property of being a linearly independent generator is a special property of both linearly independent sets and generators. A set with both properties is called a basis in Linear Algebra.

Since bases are generators, every vector has a representation as a linear combination of basis vectors. This representation is unambiguous because bases are linearly independent. So we found another way of describing bases:

.

The vector space ${\displaystyle K^{n}}$ of the ${\displaystyle n}$ tuples over the field ${\displaystyle K}$ has a so-called standard basis or canonical basis

${\displaystyle B=\lbrace \underbrace {(1,0,\ldots ,0)^{T}} _{n{\text{ components}}},\,\underbrace {(0,1,\ldots ,0)^{T}} _{n{\text{ components}}},\,\cdots ,\,\underbrace {(0,0,\ldots ,1)^{T}} _{n{\text{ components}}}\rbrace }$

For instance, ${\displaystyle \mathbb {R} ^{3}}$ has the canonical basis ${\displaystyle B_{3}=\lbrace (1,0,0)^{T},\,(0,1,0)^{T},\,(0,0,1)^{T}\rbrace }$ and the ${\displaystyle \mathbb {R} ^{2}}$ the canonical basis ${\displaystyle B_{2}=\lbrace (1,0)^{T},\,(0,1)^{T}\rbrace }$.

The set of complex numbers ${\displaystyle \mathbb {C} }$ is a vector space over ${\displaystyle \mathbb {R} }$, with multiplication by real numbers:

${\displaystyle \cdot \,\colon \mathbb {R} \times \mathbb {C} \to \mathbb {C} ;(\lambda ,z)\mapsto \lambda \cdot z}$

Then ${\displaystyle \mathbb {C} }$ has as an ${\displaystyle \mathbb {R} }$-vector space the basis ${\displaystyle \lbrace 1,\mathrm {i} \rbrace }$, because for ${\displaystyle z\in \mathbb {C} }$ we have unique ${\displaystyle \lambda ,\mu \in \mathbb {R} }$ with ${\displaystyle z=\lambda +\mathrm {i} \mu =\lambda \cdot 1+\mu \cdot \mathrm {i} }$.

If we consider ${\displaystyle \mathbb {C} }$ as a ${\displaystyle \mathbb {C} }$-vector space, ${\displaystyle \{1,i\}}$ is no longer a basis of ${\displaystyle \mathbb {C} }$, since.

${\displaystyle 0=\mathrm {i} \cdot 1+(-1)\cdot \mathrm {i} }$

For ${\displaystyle \mathbb {C} }$ as a ${\displaystyle \mathbb {C} }$-vector space we have the one-element basis ${\displaystyle \{1\}}$. As ${\displaystyle \mathbb {R} }$-vector space the complex numbers have a two-element basis (dimension is ${\displaystyle 2}$) and as ${\displaystyle \mathbb {C} }$-vector space a one-element basis (dimension is ${\displaystyle 1}$). So be cautious: it is important over which field we take a fixed vector space!

The vector space ${\displaystyle \mathbb {R} [x]}$ of the polynomials with coefficients from ${\displaystyle \mathbb {R} }$ has a basis with infinitely many elements. An example for a basis are the powers of ${\displaystyle x}$:

${\displaystyle B=\{1,x,x^{2},x^{3},...\}}$

This is a generator, because for a polynomial ${\displaystyle f}$ of degree ${\displaystyle n}$ we have a representation

${\displaystyle f=\lambda _{n}x^{n}+\dots \lambda _{1}x+\lambda _{0}}$

Where ${\displaystyle \lambda _{i}\in \mathbb {R} }$ for all ${\displaystyle 0\leq i\leq n}$. Thus every polynomial is a finite linear combination of elements from ${\displaystyle B}$. Consequently, ${\displaystyle B}$ is a generator.

For linear independence we consider the following: With ${\displaystyle \lambda _{i}\in \mathbb {R} }$ let:

${\displaystyle 0=\lambda _{n}x^{n}+\dots \lambda _{1}x+\lambda _{0}}$

We can also write the zero polynomial as ${\displaystyle 0=0\cdot x^{n}+\dots +0\cdot x+0}$. A comparison of coefficients yields that

${\displaystyle \lambda _{0}=\dots =\lambda _{n}=0}$

So ${\displaystyle B}$ is a Basis.

We will show by using the plane ${\displaystyle \mathbb {R} ^{2}}$ that the basis of a vector space is not unique. Let us look at the (canonical) basis for ${\displaystyle \mathbb {R} ^{2}}$ consisting of the unit vectors:

${\displaystyle B_{1}=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}}$

These vectors obviously form a generator:

${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}=a\cdot {\begin{pmatrix}1\\0\end{pmatrix}}+b\cdot {\begin{pmatrix}0\\1\end{pmatrix}}}$

They are also linearly independent because it is impossible to find a linear combination of zero with non-trivial coefficients. Thus ${\displaystyle B_{1}}$ is a basis. However, for the plane there are also a lot of other bases. An example is the following set:

${\displaystyle B_{2}=\left\{{\begin{pmatrix}4\\-2\end{pmatrix}},{\begin{pmatrix}-2\\4\end{pmatrix}}\right\}}$

We can generate all vectors with these two vectors:

${\displaystyle {\begin{pmatrix}a\\b\end{pmatrix}}=\left({\frac {a}{3}}+{\frac {b}{6}}\right)\cdot {\begin{pmatrix}4\\-2\end{pmatrix}}+\left({\frac {b}{3}}+{\frac {a}{6}}\right)\cdot {\begin{pmatrix}-2\\4\end{pmatrix}}}$

These vectors are linearly independent because one vector is not a multiple of the other vector (two vectors are linearly dependent if one vector is a multiple of the other vector). Thus ${\displaystyle B_{2}}$ is also a basis. These two examples show that the basis for ${\displaystyle \mathbb {R} ^{2}}$ is not unique. And one can indeed find a lot of other bases, for instance by stretching and rotating.

In general, for every ${\displaystyle \mathbb {R} }$-vector space with a basis, we can construct a second, different basis: Consider the two-dimensional vector space ${\displaystyle V}$ with a basis ${\displaystyle B_{1}=\{b_{1},b_{2}\}}$. Then ${\displaystyle B_{2}=\{b_{1},b_{2}-b_{1}\}}$ is also a basis of ${\displaystyle V}$. The same argument of "substituting the last vector" also works in higher dimensions. We first show that ${\displaystyle B_{2}}$ is a generator and then that the basis vectors are linearly independent.

Let ${\displaystyle c\in V}$ be any vector and ${\displaystyle c=\lambda _{1}b_{1}+\lambda _{2}b_{2}}$ a linear combination of it using elements of the basis ${\displaystyle B_{1}}$. Then a linear combination of ${\displaystyle c}$ can also be found via vectors from ${\displaystyle B_{2}}$:

${\displaystyle c=(\lambda _{1}+\lambda _{2})\cdot b_{1}+\lambda _{2}\cdot (b_{2}-b_{1})}$

Thus ${\displaystyle B_{2}}$ is a generator because the vector ${\displaystyle c}$ was arbitrarily chosen.

We show that the basis vectors are linearly independent via proof by contradiction. To do this, we prove that if ${\displaystyle B_{2}}$ is linearly dependent, then ${\displaystyle B_{1}}$ must also be linearly dependent. By contraposition, it thus follows that ${\displaystyle B_{2}}$ is linearly independent if ${\displaystyle B_{1}}$ is linearly independent (which it must be the case it is a basis). If ${\displaystyle B_{2}}$ is linearly dependent, there is a representation of zero:

${\displaystyle 0=\lambda _{1}\cdot b_{1}+\lambda _{2}\cdot (b_{2}-b_{1})}$

Here ${\displaystyle \lambda _{1}\neq 0}$ or ${\displaystyle \lambda _{2}\neq 0}$. Now we also find a representation of the zero with the basis ${\displaystyle B_{1}}$:

${\displaystyle 0=(\lambda _{1}-\lambda _{2})\cdot b_{1}+\lambda _{2}\cdot b_{2}}$

We still have to show that one of the coefficients ${\displaystyle \lambda _{1}-\lambda _{2}}$ or ${\displaystyle \lambda _{2}}$ is not equal to zero. As a premise we have ${\displaystyle \lambda _{1}\neq 0}$ or ${\displaystyle \lambda _{2}\neq 0}$. The case ${\displaystyle \lambda _{2}\neq 0}$ leads directly to the non-trivial representation of the zero, since this factor also appears in the second equation.

If ${\displaystyle \lambda _{1}\neq 0}$ holds, we have to distinguish again. If ${\displaystyle \lambda _{1}\neq 0}$ and ${\displaystyle \lambda _{1}=\lambda _{2}}$, then ${\displaystyle \lambda _{2}\neq 0}$ and hence one of the coefficients is not zero. If ${\displaystyle \lambda _{1}\neq 0}$ and ${\displaystyle \lambda _{1}\neq \lambda _{2}}$ hold, then ${\displaystyle \lambda _{1}-\lambda _{2}\neq 0}$ and hence the first coefficient is non-zero.

It follows that one of the coefficients is always non-zero. Thus the vectors of the basis ${\displaystyle B_{1}}$ are also linearly dependent. It follows (by contradiction) that ${\displaystyle B_{2}}$ is linearly independent if ${\displaystyle B_{1}}$ is linearly independent. ${\displaystyle B_{2}}$ is thus a new basis constructed from the first basis.

This principle can also be applied to larger bases and shows: The basis of a vector space is (usually) not unique. A vector space with dimension equal to or larger than 2 has several bases.

We have not yet answered the question of whether there is a basis for every n vector space at all. Attention, this is not self-evident, as bases may be infinitely large! Nevertheless, you can rejoice, because the answer is: Yes, every vector space has (at least) one basis.

Of course, we still have to justify this answer. For the case of finitely generated vector spaces, i.e. all vector spaces that have a finite generator, we will prove this in a moment. For infinitely generated vector spaces, i.e. vector spaces that do not have a finite generating system, the proof is much more complicated and uses the axiom of choice.

Now we know that every vector space has a basis, but how can you find a basis for a given vector space? For finitely generated vector spaces, the proof of the theorem on the existence of a basis gives you a procedure for constructing a basis in finitely many steps (it is not applicable for infinitely generated vector spaces). According to the basis completion theorem, we can proceed as follows:

1. Find a finite generator ${\displaystyle E}$ of the vector space.
2. Check whether ${\displaystyle E}$ is linearly independent.
   • If yes: We are done and the generator is a basis.
   • If no: Find a smaller generator ${\displaystyle E'\subsetneq E}$ of the vector space and repeat step 2.

We now need an explicit way to get a smaller generator ${\displaystyle E'\subsetneq E}$ from a finite generator ${\displaystyle E=\{e_{1},\dots ,e_{n}\}}$, which is not a minimal generator. Since ${\displaystyle E}$ is not a minimal generator, ${\displaystyle E}$ is linearly dependent. So we find ${\displaystyle \lambda _{1},\dots ,\lambda _{n}\in K}$ so that not all ${\displaystyle \lambda _{i}=0}$ and we have that

${\displaystyle \lambda _{1}e_{1}+\dots +\lambda _{n}e_{n}=0}$

Now we choose an ${\displaystyle i\in \{1,\dots ,n\}}$ with ${\displaystyle \lambda _{i}\neq 0}$ and set

${\displaystyle E':=E\setminus \{e_{i}\}=\{e_{1},\dots ,e_{i-1},e_{i+1},\dots ,e_{n}\}}$

We now want to show that ${\displaystyle E'}$ is also a generator. In the article linear independence of vectors we have proven that

${\displaystyle e_{i}=-{\frac {\lambda _{1}}{\lambda _{i}}}e_{1}-\dots -{\frac {\lambda _{i-1}}{\lambda _{i}}}e_{i-1}-{\frac {\lambda _{i+1}}{\lambda _{i}}}e_{i+1}-\dots -{\frac {\lambda _{n}}{\lambda _{i}}}e_{n}}$

Since ${\displaystyle E}$ is a generator of ${\displaystyle V}$, we find for a vector ${\displaystyle v\in V}$ scalars ${\displaystyle \mu _{1},\dots ,\mu _{n}\in K}$ such that

${\displaystyle {\begin{aligned}v&=\mu _{1}e_{1}+\dots +\mu _{i}e_{i}+\dots +\mu _{n}e_{n}\\&=\mu _{1}e_{1}+\dots +\mu _{i-1}v_{i-1}+\mu _{i}\left(-{\frac {\lambda _{1}}{\lambda _{i}}}e_{1}-\dots -{\frac {\lambda _{i-1}}{\lambda _{i}}}e_{i-1}-{\frac {\lambda _{i+1}}{\lambda _{i}}}e_{i+1}-\dots -{\frac {\lambda _{n}}{\lambda _{i}}}e_{n}\right)+\mu _{i+1}e_{i+1}+\dots +\mu _{n}e_{n}\\&=\left(\mu _{1}-\mu _{i}\cdot {\frac {\lambda _{1}}{\lambda _{i}}}\right)e_{1}+\dots +\left(\mu _{i-1}-\mu _{i}\cdot {\frac {\lambda _{i-1}}{\lambda _{i}}}\right)e_{i-1}+\left(\mu _{i+1}-\mu _{i}\cdot {\frac {\lambda _{i+1}}{\lambda _{i}}}\right)e_{i+1}+\dots +\left(\mu _{n}-\mu _{i}\cdot {\frac {\lambda _{n}}{\lambda _{i}}}\right)e_{n}\end{aligned}}}$

Thus ${\displaystyle v\in \operatorname {span} (E')}$. Since ${\displaystyle v\in V}$ was arbitrarily chosen, it follows that ${\displaystyle E'}$ is a generator. With the proof of the basis theorem, we now get the following procedure for determining a basis:

Alternatively, we can proceed as in the section Motivation via linear independence; that is, we start with a linearly independent set and extend it until it is maximal, i.e., a basis.

This proof gives you another method to determine a basis of a finitely generated vector space in finitely many steps (this method is also only applicable for finitely generated vector spaces):

1. Choose a finite generator and start with the empty set as your first linearly independent set.
2. Try to find a vector from your generator that is not in the span of your previous linearly independent set. If you don't find one, you're done.
3. Add the vector you found to your linearly independent set and go back to Step 2.

In the next chapter we will see that every two bases of the same finitely generated vector space have the same cardinality. We get that every linearly independent set that has as many elements as a basis is automatically already a maximally linearly independent subset. Therefore, we can change step 2 in the above procedure as follows: "Try to find a vector from your vector space that is not in the span of your previous linearly independent set. If you don't find one, you're done." In this variant of the procedure, you do not have to choose a generator in Step 1.

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