### Vector spaces: Spans

### Independence

In *Reduction with vectors* we have seen an example of a subspace spanned by three vectors that can be spanned by two vectors. We now discuss how to find from a given set of vectors an efficient set of vectors that spans the same subspace as the given set. In addition to the theorem *Standard operations with spanning sets* and the *Exchange Theorem*, the concepts of *dependence* and *independence *play a role here.

(In)dependent set

A sequence of vectors #\vec{u}_1,\ldots ,\vec{u}_n# of a vector space is called (**linearly**) **dependent **if at least one of the vectors is dependent on the others, and (**linearly**) **independent** if none of the vectors is a linear combination of the others.

If one of the vectors of the above sequence is a linear combination of some of the others, it is said to be **(linearly) dependent** on these others. If not, it is said to be **(linearly) independent** of the others.

A **non-trivial relation** between the vectors #\vec{u}_1,\ldots ,\vec{u}_n# is an equality

\[

\lambda_1 \cdot\vec{u}_1 + \lambda_2\cdot \vec{u}_2 + \cdots + \lambda_n\cdot \vec{u}_n =\vec{0},

\] in which at least one of the numbers #\lambda_1, \lambda_2, \ldots ,\lambda_n# is distinct from #0#.

To determine whether a given set of vectors is linearly independent, we often use the following criterion:

Dependence criterion

A set of vectors #\vec{u}_1,\ldots ,\vec{u}_n# of a vector space is dependent if and only if there is a non-trivial relation between the vectors #\vec{u}_1,\ldots ,\vec{u}_n#.

With this criterion, we can test (in)dependence of the set of vectors #\vec{u}_1,\ldots ,\vec{u}_n# by solving the system of linear equations \[\lambda_1\cdot\vec{u}_1 + \lambda_2\cdot \vec{u}_2 + \cdots + \lambda_n \cdot\vec{u}_n =\vec{0}\] with unknowns #\lambda_1#, #\lambda_2, \ldots ,\lambda_n#. If the only solution is #\lambda_1 =0, \lambda_2=0, \ldots , \lambda_n=0#, then the set is independent. If there is another solution, then the set is dependent.

Thinning

Each set #\vec{u}_1,\ldots,\vec{u}_m# in a vector space #V# that does not solely consist of zero vectors, can be thinned to an independent set with the same span #U=\linspan{\vec{u}_1,\ldots,\vec{u}_m }#:

- If #\vec{u}_1,\ldots \vec{u}_m# is an independent set, then we are ready.
- If not, then one of the vectors is dependent on the others. Remove this vector, so we are left with #m-1# vectors that still span #U#, and go back to the previous step.

In this way we find an independent set spanning #U# after at most #m# removals.

The set found that spans #U# will *later* turn out to be minimal, in the sense that no smaller number of spanning vectors for #U# can be found.

\[

\lambda_1 \cdot\vec{e}_1 + \cdots + \lambda_n\cdot \vec{e}_n =\rv{0,\ldots ,0}

\] with unknowns #\lambda_1,\ldots, \lambda_n#. After writing out the sum in coordinates, it comes down to #\rv{\lambda_1 ,\ldots , \lambda_n }=\rv{0,\ldots ,0}#. This system has only one solution: #\lambda_1 =\cdots =\lambda_n=0#. According to the

*Dependence criterion,*this proves that the set of vectors #\vec{e}_1 ,\vec{e}_2, \ldots,\vec{e}_n# is independent.

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