# Prove a set S of vectors is linearly independent

I saw the following question on an exam and my teacher told me my proof is wrong because "these are not the same scalars".

Let $$v_1, v_2, ..., v_k, w$$ be different vectors in a linear space over $$\mathbb{R}$$.

Suppose $$w \notin Sp\{v_1 - w, v_2 - w, \ldots, v_k -w\}$$.

Prove that if the set $$\{v_1 - w, v_2 - w, \ldots, v_k - w\}$$ is linearly independent, then the set $$\{v_1, \ldots, v_k\}$$ is also linearly independent.

Can you verify my proof?

Pf:

Suppose $$\{v_1, \ldots, v_k\}$$ is linearly dependent. Therefore, there exist $$\lambda_1, \ldots \lambda_k$$ such that $$\{\lambda_1, \ldots, \lambda_k\} \neq \{0\}$$ and

$$\lambda_1 v1 + \ldots + \lambda_k v_k = 0$$

$$\{v_1 - w, \ldots, v_k - w\}$$ is linearly independent and therefore

* $$\lambda_1(v_1 - w) + \ldots + \lambda_k(v_k - w) \neq 0$$

** Assume $$\lambda_1(v_1 - w) + \ldots + \lambda_k(v_k - w) = u$$.

Therefore, $$\lambda_1v_1 - \lambda_1w + \ldots + \lambda_kv_k - \lambda_k w = u$$.

$$\lambda_1v_1 + \ldots + \lambda_kvk - (\lambda_1 + \ldots + \lambda_k)w = u \Rightarrow \lambda_1v_1 + \ldots + \lambda_kv_k - u = (\lambda_1 + \ldots \lambda_k)w$$.

We will look at 2 cases:

If $$\lambda_1 + \ldots + \lambda_k \neq 0$$, then

$$w = \frac{\lambda_1v1}{\lambda_1 + \ldots + \lambda_k} + \ldots + \frac{\lambda_kv_k}{\lambda_1 + \ldots + \lambda_k} - \frac{u}{\lambda_1 + \ldots + \lambda_k}$$.

$$u \in Sp\{v_1 - w, \ldots, v_k - w \}$$ and therefore the expression above is an element of $$Sp\{v_1 - w, \ldots, v_k - w\}$$ and therefore $$w \in Sp\{v_1 - w, \ldots, v_k - w\}$$ - contradiction.

Assume $$\lambda_1 + \ldots + \lambda_k = 0$$, therefore, $$\lambda_1v_1 + \ldots \lambda_kv_k - u = 0 \Rightarrow \lambda_1v_1 + \lambda_kv_k = u$$, and therefore, $$u = 0$$, but according to (*) and (**), $$u \neq 0$$ - contradiction.

Therefore, the set is linearly independent.

• I don't understand this step: "$u \in Sp\{v_1 - w, \ldots, v_k - w \}$ and therefore the expression above is an element of $Sp\{v_1 - w, \ldots, v_k - w\}$ and therefore $w \in Sp\{v_1 - w, \ldots, v_k - w\}$ - contradiction.". The expression above that is $w = \frac{\lambda_1v1}{\lambda_1 + \ldots + \lambda_k} + \ldots + \frac{\lambda_kv_k}{\lambda_1 + \ldots + \lambda_k} - \frac{u}{\lambda_1 + \ldots + \lambda_k}$, and even though the last term is in $Sp\{v_1 - w, \ldots, v_k - w\}$, the remaining sum is in $Sp\{v_1, \ldots, v_k\}$
– user700480
Mar 13, 2023 at 9:48
• But the remaining sum is equal to 0 Mar 13, 2023 at 9:53
• Ah, you just have a long-winded way to say that, if $\lambda_1 v_1+\ldots+\lambda_k v_k=0$ then $\lambda_1 (v_1-w)+\ldots+\lambda_k (v_k-w)=-(\lambda_1+\ldots+\lambda_k)w$. Then, $\lambda_1+\ldots+\lambda_k=0$ implies that $v_1-w, \ldots, v_k-w$ are linearly dependent, but $\lambda_1+\ldots+\lambda_k\ne 0$ implies that $w$ is in the span of $\lambda_1-w,\ldots,\lambda_k-w$ - contradiction both ways! In that case, I would say that the proof looks good but is maybe slightly confusingly written and your teacher may have misunderstood it.
– user700480
Mar 13, 2023 at 10:27
• I agree that your proof looks good. And when he says "these are not the same scalars", I think he got confused in the way you used the $\lambda_i$'s both to say something related to $v_1$, ..., $v_k$ being LD and $v_1 - w$,...., $v_k-w$ being LI. To make it easier to understand, try saying something like: these $\lambda_i$ which are such that they are not all zero are also such that $\lambda_1 (v_1-w) + ... + \lambda_k (v_k - w) \neq 0$. And then you say that this happens because if it were equal to zero, then since they are LI, it would mean that all $\lambda_i=0$ Mar 13, 2023 at 11:16

First, if $$w=0$$ the theorem is trivial, so assume $$w\neq 0$$

Second, for linear dependence of the vectors $$\{v_1, ...v_k\}$$ it is sufficient that in:

$$\beta_1 v_1+...+\beta_k v_k=0$$

At least on of the $$\beta_i$$ is different from 0.

$$(\exists j)(1\leq j\leq k)(\beta_j\neq0)$$

Assume wlog $$j=1$$ (you can always rearrange a finite set).

Thus: $$v_1=-\frac{1}{\beta_1}(\beta_2 v_2+...+\beta_k v_k)$$ $$v_1\neq 0\implies (\exists j)((2\leq j\leq k)\land \beta_j\neq 0)$$

Now use the hypothesis:

$$\alpha_1(v_1-w)+...+\alpha_k(v_k-w)=0\implies (\forall i)(\alpha_i=0)$$

But this in turn implies:

$$\alpha_1(-\frac{1}{\beta_1}(\beta_2 v_2+...+\beta_k v_k)-w)+\alpha_2(v_2-w)...+\alpha_k(v_k-w)=0$$

And can be rewritten as:

$$(-\frac{1}{\beta_1}\beta_2\alpha_1+\alpha2)v_2+...+(-\frac{1}{\beta_1}\beta_k\alpha_1+\alpha_k)v_k-(\alpha_1+...+\alpha_k)w=0$$

Again assume wlog $$\beta_2\neq 0$$. Then

$$\alpha_2=\frac{\beta_2}{\beta_1}\alpha_1$$