Let $v_1, … , v_n$ be independent vectors in a vector space V, and let $w ∈ V$. Show that the span of $v_1 + w, … , v_n + w$ has dimension n − 1 or n. Show that both values can be realized.
We know that span of $v_1, … , v_n$ has dimension $n$ since there are $n$ independent vectors. If we add $w$ to all vectors, it can make them stay independent, or make 2 of these dependent. Which leads to dimension n and n-1. But, how do I prove that?
Let $v_1, … , v_n$ be independent vectors. Show that the span of $v_1 + w, … , v_n + w$ has dimension n − 1 or n
abstract-algebralinear algebra
Best Answer
If someone is interested, there is another solution that does not require using matrix algebra or systems of equations. The case where $n=1$ is trivial, so let us consider $n\geq 2$.
Let $v_1,\dots,v_n$ be linearly independent. Consider $W = span(v_1+w,\dots,v_n+w)$, the generated space of $v_1+w,\dots,v_n+w$. Now, subtracting $v_1+w$ from all the vectors $v_j+w$ we get that $v_j-v_1\in W$ for all $2\leq j\leq n$. But it is not hard to prove* that these vectors are linearly independent; and that there are $n-1$ of them. Therefore the dimension of $W$ must be at least $n-1$ since it contains $n-1$ linearly independent vectors.
*Let us prove that, in fact, the vectors $v_j-v_1$ for $2\leq j\leq n$ are linearly independent. Consider the sum $$ \sum_{j=2}^n\alpha_j(v_j-v_1) = -Sv_1 + \sum_{j=2}^n\alpha_jv_j=0, $$ where $S = \sum_{j=2}^n\alpha_j$. Then, because of linear independence of the $v_j$, it is necessary that $\alpha_j =0$ for all $2\leq j\leq n$, hence proving that the vectors are linearly independent.
Now proving that dimension $n-1$ is attainable, just pick $w = -v_1$ and the vectors we get are precisely $0,v_2-v_1, v_3-v_1,\dots v_n -v_1$, since we just proved that $v_j-v_1$ for $2\leq j\leq n$ are linearly independent vectors, and $0$ is automatically linearly dependent then these vectors span a space of dimension $n-1$.