Proof

(related to Lemma: Uniqueness Lemma of a Finite Basis)

Let \(v\) be a vector in a vector space \(V\) with the basis \(B\).

If \(B\) is finite, then we have \(B=\{b_1,\ldots, b_n\}\). Suppose that there are two different representations \[\begin{array}{ccl} v&=&\alpha_1b_1+\ldots+\alpha_nb_n,\\ v&=&\beta_1b_1+\ldots+\beta_nb_n.\\ \end{array}~~~~~~~~~~~~~~~~~( * )\] Then we have \[\begin{array}{ccl} 0&=&(\alpha_1-\beta_1)b_1+\ldots+(\alpha_n-\beta_n)b_n, \end{array}\] Because the representations \( ( * ) \) are different by assumption, there is at least one \(i\) for which \(\alpha_i\neq \beta_i\). But this would mean that \(\{b_1,\ldots, b_n\}\) are linearly dependent, in contradiction to the definition of a basis.


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References

Bibliography

  1. Koecher Max: "Lineare Algebra und analytische Geometrie", Springer-Verlag Berlin Heidelberg New York, 1992, 3rd Volume