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If two vectors $\bf{u}$ and $\bf{v}$ in $\mathbb{R}^2$ are orthogonal to a non-zero vector $\bf{w}$ in $\mathbb{R}^2$, then are $\bf{u}$ and $\bf{v}$ scalar multiples of one another? Prove your claim.

Attempt: From a geometric point of view it seems obvious that they must be scalar multiples of one another but I am having difficulties trying to prove it. My approach was to use the Cauchy-Schwarz Inequality by assuming $|\bf{u}\cdot \bf{v}| < ||\bf{u}|| ||\bf{v}|| $ and somehow reaching a contradiction but I can't seem to obtain one. Maybe I need to try a different approach? It would be great (if possible) if someone can continue using my approach or show that it won't work (Assuming my answer is correct in the first place).

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    $\begingroup$ Hint: $\bf w$ and any nonzero vector orthogonal to it form a basis of ${\mathbb R}^2$. $\endgroup$
    – Robert Israel
    Feb 9, 2012 at 6:38
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    $\begingroup$ Whatever method you use, it must somehow and very intrinsically use the fact that you are working in a 2-dimensional space, since the claim is false in 3 or more dimensions (just take three elements of any orthonormal basis); but I don't see how one could try to leverage the dimension into the Cauchy-Schwarz Inequality, so I would have little hope of being able to solve it along your approach (doesn't mean it can't be done, just that I can't see how I would make it work...) $\endgroup$
    – Arturo Magidin
    Feb 9, 2012 at 6:43

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Without using dimensionality arguments:

Suppose $(a,b)$ and $(c,d)$ are both orthogonal to $(e,f)\ne (0,0)$.

Then, from the definition of orthogonality: $$ ae+bf =0 $$ and $$ ce+df =0 .$$

If $e=0$, we must have $f\ne 0$, which implies $b=d=0$. Thus, $(a,b)=(a,0)$ and $(c,d)=(c,0)$ are scalar multiples of each other.

If $e\ne 0$, then, from the above system $$ a=-\textstyle{ f\over e}\, b \quad \text{and}\quad c=-{f\over e}\,d; $$ whence, $$(a,b)=(\textstyle{- f\over e} \thinspace b, b)=b(\textstyle{-f\over e}, 1)$$and$$(c,d)=(\textstyle{- f\over e} \thinspace d, d)=d(\textstyle{-f\over e},1).$$ This implies that $(a,b)$ and $(c,d)$ are scalar multiples of each other.

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  • $\begingroup$ More simply, if $\bf w$ is orthogonal to $(e,f)$ with $e\ne0$,say, then $\bf w$ is of the form $\alpha({-f\over e},1 ) $. $\endgroup$
    – David Mitra
    Feb 9, 2012 at 15:35
  • $\begingroup$ This is most likely the solution (and probably the only possible one without the use of dimension and basis) the book is after (this section of the book doesn't have answers). Thank you for your contribution. $\endgroup$
    – tcmtan
    Feb 9, 2012 at 22:17
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If either $\mathbf{u}$ or $\mathbf{v}$ are zero, then it is a scalar multiple of the other and you are done. So you may assume that $\mathbf{u}\neq\mathbf{0}$ and $\mathbf{v}\neq\mathbf{0}$.

Note that $\mathbf{u}$ and $\mathbf{w}$ are linearly independent. Hence they span $\mathbb{R}^2$, so we can write $\mathbf{v}=\alpha\mathbf{u}+\beta\mathbf{w}$. So... what is $\langle \mathbf{v},\mathbf{w}\rangle$ going to be, then, and what should it be?

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  • $\begingroup$ Thanks for the hint. So $(\alpha\bf{u}+\beta\bf{w})\cdot\bf{w}=0$ and as a result we have $\beta=0$ as $\bf{w}\neq\bf{0}$ vector. The interesting point is that in the textbook (Anton&Busby Contemporary Linear Algebra) the concept of linear independence, basis and spanning set were not yet introduced in this section of the book and is presented a few chapters later. So I'm assuming there is another way to solve the problem. $\endgroup$
    – tcmtan
    Feb 9, 2012 at 9:21
  • $\begingroup$ @user22678: In that case, you probably want something along the lines of David Mitra's result below, which relies on the precise definition of the inner product in $\mathbb{R}^2$. $\endgroup$
    – Arturo Magidin
    Feb 9, 2012 at 14:23
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    $\begingroup$ @user22678: That is one very good reason why we keep telling posters to include information such as the source of the problem into the question; knowing where it comes from often prevents people (like me) from using material that is not covered yet to solve the problem. $\endgroup$
    – Arturo Magidin
    Feb 9, 2012 at 14:28
  • $\begingroup$ Thanks for the heads up. I will keep the in mind for future posts. $\endgroup$
    – tcmtan
    Feb 9, 2012 at 22:12
  • $\begingroup$ Referring back to your statement where you say you don't see how one could incorporate the dimension into the Cauchy-Schwarz, would you be able to give a somewhat theoretical/pragmatic explanation as to why it seems not possible? $\endgroup$
    – tcmtan
    Feb 9, 2012 at 22:25

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