Unique representations in a vector space

Suppose I have a vector space V over a field k, and a finite basis \mathcal B=\{v_1,v_2,\dots,v_n\}. Then each vector has a unique representation of the form c_1 v_1+c_2 v_2+\cdots+c_nv_n, where the c_i come from the field k.

Clearly it has at least one, since by definition of being a basis, \mbox{span}(\mathcal B)=V. This means that every vector can be written as a linear combination of vectors in \mathcal B.

Suppose v\in V has two such representations:

v=b_1v_1+b_2v_2+\cdots+b_nv_n=c_1v_1+c_2v_2+\cdots+c_nv_n

Subracting the two representations gives us that

(b_1-c_1)v_1+(b_2-c_2)v_2+\cdots+(b_n-c_n)v_n=0

But, since a basis is also a linearly independent set, this means that for each i=1,2,\dots,n, b_i-c_i=0, so b_i=c_i. But then we don’t have two different representations. So there is exactly one.

In fact, this is true of vector spaces without finite bases too, but we did not develop any of the necessary tools to talk about infinity yet, so I’m going to restrict our attention to vector spaces with finite bases.

This result is an important fact and will be used over and over again, many times without mention. We’ll just say something like “take a basis, and write v in that basis. The fact that we can and we can uniquely is what we just showed.

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One Response to Unique representations in a vector space

  1. Pingback: The Tower Rule « Andy Soffer

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