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 $V$

Question

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 $a_1 + a_2 + \cdots + a_n = -1$.

Here is my answer:

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)?

Answer 1

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?

Answer 2

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.

Answer 3

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:
$\det\big(I_n +\mathbf a \mathbf 1^T\big)= \det\big(I_n\big)\cdot\big(1 +\mathbf 1^TI_n^{-1}\mathbf a\big) = 1 \cdot \big(1 +\sum_{i=1}^n a_i\big)$

Answer 4

A slightly different from a previous answer approach which the readers

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 $V$.!!