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The Statement

Show that if a set of vectors is linearly independent, then any vector in the span of that set has a unique representation as a linear combination of these vectors.

My Proof

I'm going for a proof by contrapositive.

Let $S=(v_{1},v_{2},\dots,v_{n})$ be our set of vectors.

Let $x_{0}$ be a vector in the span of S. If it does not have a unique representation, it can be written in atleast two distinct ways:

$x_{0}= a_1v_{1}+a_{2}v_{2}+\dots a_{n}v_{n}$

$x_{0}= b_{1}v_1+b_2v_2+\dots+b_nv_n$

$1)$For atleast one $b_j$ for $1\leq j\leq n$ in those linear combinations it will be the case that $b_j \neq a_j$

$0 = x_0-x_0=(a_1-b_1)v_1+(a_2-b_2)v_2+\cdots+(a_n-b_n)v_n$

By fact 1 that means atleast one of these coefficients is not zero, which means S is linearly dependent.

My Question

I'm pretty happy with my solution(which I think is correct). I was wondering if for learning sake anyone could provide a direct proof(or a proof via another method if a direct is impossible)?

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This proof works fine. Here's a "direct" one.

Suppose that $$ \alpha_1 v_1+\dotsb+\alpha_n v_n=\beta_1 v_1+\dotsb+\beta_n v_n $$ Then $$ (\alpha_1-\beta_1)v_1+\dotsb+(\alpha_n-\beta_n) v_n=\vec 0 $$ But $\{v_1,\dotsc,v_n\}$ is linearly independent so $\alpha_j-\beta_j=0$ for each $1\leq j\leq n$. Hence $\alpha_j=\beta_j$ for $1\leq j\leq n$.

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A direct method is of course possible, and it is pretty straightforward.

Let $x \in L(\{v_1, \ldots, v_n\})$. Then, there are $a_1, \ldots, a_n \in \mathbb K$, such that $x = \sum_i a_i v_i$. Let $x = \sum_i b_i v_i$ be another representation of $x$ in those vectors. We have:

$$\sum_i (a_i - b_i)v_i = 0$$

As the $v_i$'s are linearly independent, it follows: $a_i - b_i = 0$, $\forall \ i$. Which shows that the $b_i$'s are exactly the $a_i$'s.

This demonstrates the uniqueness of the $a_i$'s. Essentially, this is the same as your argument, but working by contradiction wasn't necessary.

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Your proof is great. A direct proof is very similar. Suppose we have $S=(v_1,\dots,v_n)$, a linearly independent set of vectors. Then suppose we have two ways to write $x_o\in \text{sp}(S)$,

$$a_1v_1+\dots+a_nv_n=x_o=b_1v_1+\dots+b_nv_n$$

$$\implies (a_1-b_1)v_1+\dots+(a_n-b_n)v_n=0$$

because the set is linearly independent, this means

$$a_i-b_i=0\implies a_i=b_i\text{ ($0\leq i\leq n$)}$$

So our way of writing $x_o$ is unique.

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