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 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?
 A: 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 lineary independent. Consider $W = span(v_1+w,\dots,v_n+w)$, the generated space of $v_1+w,\dots,v_n+w$. Now, substracting $v_1+w$ 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 this vectors are linearly independent; and 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 vectos $v_j-v_1$ for $2\leq j\leq n$ are linealry 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.
A: If $w$ is linearly independent from $v_i$s (in case $\dim (V)>n$), then we have $${\sum_{i=1}^na_i(v_i+w)=0\iff\\\sum_{i=1}^na_iv_i+w\cdot \sum_{i=1}^na_i=0\iff 
\\a_i=0,\sum a_i=0\iff \\a_i=0}$$which means that the dimension is still $n$.
If $w$ is linearly dependent to $v_i$s, since $v_i$s are $w$ linearly independent, we can find unique $c_1,c_2,\cdots , c_n$ such that $$w=\sum_i c_iv_i$$and the equality $\sum_{i=1}^na_i(v_i+w)=0$ reduces to $$\sum_i (a_i+S\cdot c_i)v_i=0\iff a_i+S\cdot c_i=0\iff A\cdot \underline a=0$$where $S=\sum a_i$, $\underline a=[a_1,a_2,\cdots ,a_n]$ and $$A=\begin{bmatrix}1+c_1&c_2&c_3&\cdots &c_n\\c_1&1+c_2&c_3&\cdots &c_n
\\\vdots\\c_1&c_2&c_3&\cdots &1+c_n
\end{bmatrix}=I+\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}\begin{bmatrix}c_1&c_2&\cdots&c_n\end{bmatrix}$$ It is not hard to see that the matrix $\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}\begin{bmatrix}c_1&c_2&\cdots&c_n\end{bmatrix}$ has $n-1$ eigenvalues equal to $0$ and an eigenvalue equal to $\sum c_i$, hence $A$ has at least $n-1$ eigenvalues equal to $1$ which leads to a subspace with dimension at least equal to $n-1$. In brief

Conclusion
If $w$ can be expressed as a linear combination of $\{v_i\}$ with coefficient whose sum equals $-1$, the dimension becomes $n-1$, otherwise the dimension is $n$.

