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Show that if S = {$v_1$, ... , $v_n$} is a basis for a vector space V then each vector v $\in$ V can be expressed as v = $k_1v_1$ + $k_2v_2$ + ... + $k_nv_n$ (where $k_i \in R$ for i = 1, ... , n) in exactly one way.

Since the vectors in S form a basis, $v_1$, ... , $v_n$ are all independent and span the vector space.

This means that for any vector v $\in$ V, v can be reached with a linear combination of $v_1$, ... , $v_n$.

So v = $k_1v_1$ + $k_2v_2$ + ... + $k_nv_n$.

However, I do not know how to prove it can be expressed in exactly one way. Geometrically speaking, I understand that every $v_i$ points in a different direction, so to reach a new vector there is only one linear combination of the basis vectors that will reach it.

I think that it could be shown that if you assume that there are multiple linear combinations to reach a vector then it implies that $v_1$, ... , $v_n$ are not all linearly independent, but I am not sure how to do this.

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  • $\begingroup$ If $\sum{\alpha_i v_i} = v = \sum{\beta_i v_i}$, then what can you say about the relation between $\alpha_i$ and $\beta_i$ using the definition of linear independence? $\endgroup$ – Andrew May 11 '18 at 14:50
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You're on the right track. Here's how to start implementing your idea:

Suppose you have two ways to write $v$ as a linear combination of basis elements, say, as both \begin{align*} v &= k_1 v_1 + \cdots + k_n v_n \\ v &= \ell_1 v_1 + \cdots + \ell_n v_n . \end{align*} Then, subtracting the two equations gives $$0 = (\ell_1 - k_1) v_1 + \cdots + (\ell_n - k_n) v_n .$$ Now, what does linear independence tell us about a linear combination of basis elements that is equal to zero?

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    $\begingroup$ The definition of linear independence states that $k_1v_1 + ... + k_nv_n$ = 0 if, and only if every $k_i$ is equal to zero. So this means that every $(l_i-k_i)$ must be equal to zero, which means every $k_i$ and $l_i$ are equal which shows there is exactly one linear combination that can represent v? $\endgroup$ – Jackbid May 11 '18 at 15:43
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If there are two ways, $v=k_1v_1+...k_nv_n=l_1v_1+...l_nv_n$.

Hence $(k_1-l_1)v_1+...(k_n-l_n)v_n=0$, which contradicts with the independence of $v_1,...v_n$.

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