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I know that if $u\cdot v = 0$ then by definition, $u_1v_1 + u_2v_2 + \cdots + u_nv_n = 0$

I also know that if $u$ and $v$ are linearly independent and a matrix $A$ has $u$ and $v$ as its successive column vectors then the equation $Ax = 0$ will have only the trivial solution.

However I can't think of anything else that might help me answer this question.

mathstack
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  • Just to clarify, if you mean that the matrix system $[u,v]x=0$ has a unique solution, you are correct. But if you meant that the matrix has $u,v$ as two of its column vectors but also has some other columns, the statement is not true; all the columns must be independent, not just the two. Sorry if you knew that already, but I wanted to clarify in case some other beginner might get confused. – Zach Effman May 13 '15 at 00:00

2 Answers2

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Assume that $$c_1u + c_2v = 0$$ for some scalars $c_1, c_2$. Then $$0 = \langle u, c_1u + c_2v\rangle = \langle u,c_1u\rangle + \langle u,c_2v\rangle = c_1\langle u,u\rangle + c_2\langle u,v\rangle = c_1\langle u,u\rangle,$$ since $\langle u,v\rangle=0$. Since $u\ne0$, it follows that $c_1=0$. From there we see immediately that $c_2=0$.

Math1000
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You can use the fact that two vectors are linearly dependent iff one is a scalar multiple of the other:

For the sake of contradiction suppose $u$ is nonzero and $u=cv$ (i.e. the vectors are dependent). Then $0=u\cdot v=c(u_1\cdot u_1+\dots+u_n\cdot u_n)=c(u_1^2+\dots+u_n^2)$, but this can only be $0$ if each $u_i^2=0$ (assuming we are working over the reals so that $x^2\ge0$ with equality iff $x=0$). Hence $u=0$, but this is a contradiction. Therefore, the two vectors $u$ and $v$ are in fact linearly independent.

Peter Woolfitt
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