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Let {$v_1$, $v_2$,..., $v_r$} be basis of V and {$w_1$, $w_2$,..., $w_m$} be basis of W. Denote S = {$v_1$, $v_2$,..., $v_r$, $w_1$, $w_2$,..., $w_m$}. Show that S is linearly independent.

Suppose I assume that V∩W = $\{0\},$ how do I go about showing that S is linearly independent, meaning $a_1v_1+a_2v_2+...+a_rv_r+b_1w_1+b_2w_2+...+b_mw_m = 0$?

In another case if I assume instead that dim(V∩W) ≥ 1, how do I show that S is also linearly independent?

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  • $\begingroup$ For arbitrary spaces $V$ and $W$, the statement at the top is not true. The sentence is missing something. I'm guessing the missing bit of information is given by your assumptions below, but it is not clear to me what the question in its whole is. $\endgroup$ – Niki Di Giano Oct 17 '18 at 17:21
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    $\begingroup$ This is not true in general . For example if $V=W$ and if you take two different bases, the union is obviously linearly dependent $\endgroup$ – Basanta R Pahari Oct 17 '18 at 17:22
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The statement requires some adjustments, otherwise it makes no sense at all.

Let $V$ and $W$ be subspaces of the vector space $Z$ and suppose that $\{v_1,\dots,v_r\}$ is a basis of $V$ and $\{w_1,\dots,w_m\}$ is a basis of $W$. If $V\cap W=\{0\}$, then $\{v_1,\dots,v_r,w_1,\dots,w_m\}$ is linearly independent.

How to prove it? Suppose $a_1v_1+\dots+a_rv_r+b_1w_1+\dots+b_mw_m=0$. Set $$ z=a_1v_1+\dots+a_rv_r=-(b_1w_1+\dots+b_mw_m) $$ Then $z\in V\cap W$.

Can you finish?

By the way, the set $\{v_1,\dots,v_r,w_1,\dots,w_m\}$ is linearly independent if and only if $V\cap W=\{0\}$.

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