# Let $\{v_1,v_2,\cdots,v_n\}$ is a basis for vector space $V$. Let $w \in V$, prove that $W =\{v_1+w,v_2+w,\cdots,v_n+w\}$ is a basis for $V$

The complete question is:

Let $$\{v_1,v_2,\cdots,v_n\}$$ is a basis for vector space $$V$$. Let $$w \in V$$, prove that $$W =\{v_1+w,v_2+w,\cdots,v_n+w\}$$ is a basis for $$V$$ iff $$w \neq a_1v_1 + a_2v_2 + \cdots + a_nv_n$$, where $$a_1 + a_2 + \cdots + a_n = -1$$.

The proof is divided into two parts. For the first part, we will prove that if $$W$$ is a basis for $$V$$ then $$w \neq a_1v_1 + a_2v_2 + \cdots + a_nv_n$$, where $$a_1 + a_2 + \cdots + a_n = -1$$ by contrapositive.
Assume $$w = a_1v_1 + a_2v_2 + \cdots + a_nv_n$$, where $$a_1 + a_2 + \cdots + a_n = -1$$. Thus,

\begin{align} (-1)(-w) & = a_1v_1 + a_2v_2 + \cdots + a_nv_n \\ (a_1+a_2+\cdots+a_n)(-w) & = a_1v_1 + a_2v_2 + \cdots + a_nv_n \\ 0 & = a_1(v_1+w) + a_2(v_1+w) + \cdots + a_n(v_1+w) \end{align}

We see that $$W$$ is not linearly independent because $$a_1 + a_2 + \cdots + a_n = -1$$. Hence, $$W$$ is not a basis for $$V$$.

Is this correct? Also, how to prove the second part (converse)?

• See this answer to a rather similar issue. Sep 12, 2020 at 7:38
• Thank you @JeanMarie Sep 12, 2020 at 7:53

Your half of the proof is correct. For the other direction: suppose $$W$$ is not a basis for $$V$$. Then there exist $$c_1, \ldots, c_n$$ not all zero such that \begin{align} c_1 (v_1 + w) + \cdots + c_n(v_n + w)&=0\\ c_1 v_1 + \cdots + c_n v_n &= -(c_1 + \cdots + c_n) w\\ - \frac{c_1}{c_1 + \cdots + c_n} v_1 - \cdots - \frac{c_n}{c_1 + \cdots + c_n} v_n &= w \end{align} Can you conclude from here?

• Yes, thank you very much Sep 12, 2020 at 7:52
• What happens if $c_1 + \dotsm + c_n = 0$? Sep 12, 2020 at 10:01
• @J.-E.Pin If so, then in the second equation, we have $c_1 v_1 + \cdots + c_n v_n = 0$, which is a contradiction since the $v_i$ form a basis. Sep 12, 2020 at 20:44

We wish to show that $$v_k+w$$ is a basis iff $$-w \notin \operatorname{aff} \{ v_k \}$$ where $$\operatorname{aff} V$$ is the affine hull of $$V$$ (that is all points of the form $$\sum_k \lambda_k v_k$$ where $$\sum_k \lambda_k = 1$$).

Suppose $$-w \in \operatorname{aff} \{ v_k \}$$ then there are $$\lambda_k$$ summing to one such that $$\sum_k \lambda_k (v_k+w) = 0$$ which contradicts $$v_k+w$$ being a basis.

For the other direction, suppose $$-w \notin \operatorname{aff} \{ v_k \}$$ and suppose $$\sum_k \alpha_k (v_k+w) = 0$$. If $$\sum_k \alpha_k = 0$$ this gives $$\sum_k \alpha_k v_k = 0$$ which in turn gives $$\alpha_k = 0$$. Otherwise, let $$\lambda_k = {\alpha_k \over \sum_j \alpha_j }$$ and note that the $$\lambda_k$$ sum to one, hence $$\sum_k \lambda_k (v_k+w) =0$$ (or $$-w = \sum_k \lambda_k v_k$$) which is a contradiction. Hence $$\alpha_k = 0$$ and so $$v+w_k$$ are linearly independent.

another approach: collect the abstract vectors in 'hypervectors'
$$\mathbf {V} :=\bigg[\begin{array}{c|c|c|c|c} v_1 & v_2 &\cdots & v_{n}\end{array}\bigg]$$
$$\mathbf {W} :=\bigg[\begin{array}{c|c|c|c|c} v_1 +w & v_2+w &\cdots & v_{n}+w\end{array}\bigg]=\mathbf V + w\mathbf 1^T$$

$$w= \mathbf V \mathbf a\implies \mathbf W= \mathbf V + \big(\mathbf V\mathbf a\big)\mathbf 1^T = \mathbf V\big(I_n +\mathbf a \mathbf 1^T\big)$$
$$\text{rank}\Big(\mathbf W\Big) = \text{rank}\Big(\mathbf V\big(I_n +\mathbf a \mathbf 1^T\big)\Big)\leq \text{rank}\Big(\mathbf {V}\Big)$$
since the RHS consists entirely of linearly independent vectors, equality holds iff $$\det\big(I_n +\mathbf a \mathbf 1^T\big)\neq 0$$.

Finally, apply the matrix determinant lemma for rank one updates:
$$\det\big(I_n +\mathbf a \mathbf 1^T\big)= \det\big(I_n\big)\cdot\big(1 +\mathbf 1^TI_n^{-1}\mathbf a\big) = 1 \cdot \big(1 +\sum_{i=1}^n a_i\big)$$

may find easier to understand. Let $$V$$ be the matrix of the given

basis. Then $$Vv_{1}=e_{1},..,Vv_{n}=e_{n}$$ and let $$w=Vc$$. Then the

condition $$-w\notin affV$$ is equivalent to $$-c\,\notin aff\left\{e_{1},...,e_{n} \right\}$$ which is equivalent to

$$0\,\notin aff\left\{e_{1}+c,...,e_{n}+c \right\}$$

. This can be easily proved that it is

equivalent to $$e_{1}+c,...,e_{n}+c$$ being linearly independent and this

(multiplying with matrix $$V$$) gives that $$v_{1}+w,,,,,,v_{n}+w$$ are

linearly independent and hence a basis of $$V$$.!!