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Suppose $V=\{v_1,v_2,...,v_n\}$ and $W=\{w_1,w_2,...,w_m\}$ are linearly independent sets of vectors in $\mathbb{R}^n$ such that $v_i$ is orthogonal to $w_j$ for all $i$ and $j$.

Prove that their union is also linearly independent.

2 Answers 2

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Let $w =\sum_i a_i w_i$, $v=\sum_j b_j v_j$ and suppose $w+v = 0$. Then $w^T(w+v) = \|w\|^2 = 0$ shows that $w=0$ and hence $v=0$. Since the $w_i, v_j$ are linearly independent, it follows that $a_i = 0$ and $b_j = 0$.

Note: Since $w_i \bot v_j$ for all $i,j$ we have $w^T v = 0$ Since $w^T(w+v) = w^T w + w^T v = \|w\|^2$.

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    thank your for answer, but I do not undersand why the transpose of w times(w+v) equals to the square of determinant of w? – 2017-02-03
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    You have $w^T w = \|w\|^2$, this is the definition of the Euclidean norm. The fact that $w^T v = 0$ follows by expanding the summations and using the fact that $w_i$ is orthogonal to $v_j$ for all $i,j$. – 2017-02-03
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    Why the downvote? – 2017-02-04
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    Don't know, I voted you up – 2017-02-06
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    @angryTomy: Thanks! I believe I know who the downvoter was, nothing to do with the question, seems to be a personal thing. My very own personal serial downvoter :-). – 2017-02-06
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let X=a1w1+a2w2+...+amwm,Y=b1v1+b2v2+...+bnvn. Because vi is orthogonal to wj for all i and j, so XY=0

Suppose X+Y=0, then X and Y are both zero since XY=0,because X and Y are both linearly independent, so all ai and all bi are zero, hence the union is linearly independent.

Is my answer correct?