If $v_1,...,v_m$ are linearly independent, then the span $v_1+w,...,v_m+w$ has dimension $\ge m-1$

Suppose $v_1,...,v_m$ is linearly independent in $V$ and $w\in V$. Prove that $$ \dim (\operatorname{span}(v_1+w,...,v_m+w)) \ge m-1$$

It's an exercise in the book Linear Algebra Done Right.

I'm wondering if I can write $U_1 =\operatorname{span}(v_1,...,v_m)$ and $U_2=\operatorname{span}(w)$ then write $$ \dim(\operatorname{span}(v_1+w,...,v_m+w)) = \dim(U_1+U_2)$$ Would you please help me with this problem, I really want a rigorous proof, thanks.

Solution 1:

If we subtract the first vector from the rest, we get: $$\mathbf v_2-\mathbf v_1, \mathbf v_3-\mathbf v_1,\dots,\mathbf v_m - \mathbf v_1 \in \operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)$$

So we have:

$$\dim\operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)\geq\dim\operatorname{span}(\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m - \mathbf v_1)$$

Since $$\dim \operatorname{span}(\mathbf v_1,\dots, \mathbf v_m)=\dim\operatorname{span}(\mathbf v_1,\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m-\mathbf v_1)=m$$

We have $$\dim\operatorname{span}(\mathbf v_2-\mathbf v_1,\dots,\mathbf v_m-\mathbf v_1) = m-1$$

$$\dim\operatorname{span}(\mathbf v_1+\mathbf w,...,\mathbf v_m+\mathbf w)\geq m-1$$