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

Asked Apr 04 '19 at 05:11

Active Apr 04 '19 at 07:56

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This is a follow up to If $v_1,...,v_m$ are linearly independent, then the span $v_1+w,...,v_m+w$ has dimension $\ge m-1$

I want to prove it using $\dim(U_1 + U_2) = \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2); \dim(U_1 \cap U_2) = 0$.
But can't seem to find $\dim(U_1 \cap U_2)$ so that we obtain $\geq m-1$.

edited Apr 04 '19 at 07:56

Bernard

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asked Apr 04 '19 at 05:11

helios321

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span of $(v_i+w)$'s is not the sum of span of $v_i$'s and span of $w$. – Kavi Rama Murthy Apr 04 '19 at 05:18
Oh? I thought $c_1v_1=...+c_mv_m + (c_1+...+c_m)w$ reduces to $v+w$ for $v \in V$ and $w \in W$ which means its the same as the sum? Where I go wrong? – helios321 Apr 04 '19 at 06:49
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But span of $v_i$'s plus span of $w$ contains all vectors of the form $\sum c_iv_i+cw$. Here $c$ need not be equal to $\sum c_i$. – Kavi Rama Murthy Apr 04 '19 at 07:17

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

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