If $n$ vectors are linearly independent, is there only one way to write a vector as a linear combination of those vectors?

I know the converse is true, because if you can write 0 in two ways you can keep adding 0 to get an infinite number of linear combinations that sum to the same thing.

(Sorry, it's been years since I've had linear algebra.)

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Technically it's at most one; for example, a collection consisting of a single non-zero vector is linearly independent, but if the vector space $V$ we're working in has dimension $>1$, there will be vectors that are not linear combinations of that vector (i.e. scalar multiples of it). If $n=\dim(V)$, then you are correct that it is actually exactly one.

If $v_1,\ldots,v_n$ are linearly independent, then by definition $$\sum_{i=1}^nc_iv_i=0\implies c_i=0\text{ for all }i,$$ so if $$w=\sum_{i=1}^nc_iv_i=\sum_{i=1}^nd_iv_i$$ then $$0=w-w=\sum_{i=1}^n(c_i-d_i)v_i,$$ hence $c_i-d_i=0$ for all $i$, hence $c_i=d_i$ for all $i$. Thus, if the $v_i$ are linearly independent, there is at most one way of writing a given vector as a linear combination of them.

Conversely, if there is at most one way of writing a given vector as a linear combination of the $v_i$, that is if $$\sum_{i=1}^nc_iv_i=\sum_{i=1}^nd_iv_i\implies c_i=d_i\text{ for all }i,$$ then if $$\sum_{i=1}^nc_iv_i=0,$$ we have $$\sum_{i=1}^nc_iv_i=\sum_{i=1}^n0v_i\implies c_i=0\text{ for all }i,$$ so the $v_i$ are linearly independent.

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Thanks, good to know! Good catch on the "at most one way"... – badatmath Dec 1 '11 at 6:11

Yes. Any vector that can be expressed as a linear combination of linearly independent vectors can be expressed in one and only one way.

To see this, suppose that $\mathbf{v}$ can be expressed as a linear combination of $\mathbf{v}_1,\ldots,\mathbf{v}_n$, with scalars $\alpha_i$ and $\beta_i$: $$\mathbf{v} = \alpha_1\mathbf{v}_1+\cdots + \alpha_n\mathbf{v}_n = \beta_1\mathbf{v}_1+\cdots + \beta_n\mathbf{v}_n.$$ Then: \begin{align*} \mathbf{0} &= \mathbf{v}-\mathbf{v} \\ &=\bigl( \alpha_1\mathbf{v}_1+\cdots + \alpha_n\mathbf{v}_n\bigr)-\bigl( \beta_1\mathbf{v}_1+\cdots + \beta_n\mathbf{v}_n\bigr)\\ &= (\alpha_1-\beta_1)\mathbf{v}_1 + \cdots + (\alpha_n-\beta_n)\mathbf{v}_n. \end{align*} Since $\mathbf{v}_1,\ldots,\mathbf{v}_n$ are linearly independent, this means $\alpha_1-\beta_1=\alpha_2-\beta_2=\cdots=\alpha_n-\beta_n = 0$, so $\alpha_1=\beta_1,\ldots,\alpha_n=\beta_n$. That is: the two expressions are actually identical.

One way to remember this is:

A set $S$ of vectors of $V$ spans $V$ if and only if every vector of $V$ can be written as a linear combination of vectors in $S$ in at least one way. A set $I$ of vectors of $V$ is linearly independent if and only if every vector of $V$ can be written as a linear combination of vectors in $I$ in at most one way. A set $B$ of vectors of $V$ is a basis if and only if every vector of $V$ can be written as a linear combination of vectors in $V$ in exactly one way.

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Thank you, I feel enlightened :D – badatmath Dec 1 '11 at 6:10