# Show that a vector that is orthogonal to every other vector is the zero vector

I have the following question, and I'd like to get some tips on how to write the proof. I know why it is, but I'm still not so great at writing it mathematically.

If $u$ is a vector in $\mathbb{R}^n$ that is orthogonal to every vector in $\mathbb{R}^n$, then $u$ must be the zero vector. Why?

I'm starting off like this, but I don't know if it's the right way to do it, or if it is and I just don't know how to continue.

\begin{align} (\exists u\in\mathbb{R}^n)(\forall v\in\mathbb{R}^n)[\text{u is orthogonal to v}]&\iff u\cdot v=0\\ &\iff ? \end{align}

From here, instinctively I want to divide both sides by $v$, but I don't know if there is such a thing as dividing a dot product.

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Let $e_i$ be the standard basis vectors of $\mathbb{R}^n$. What does the equation $u\cdot e_i=0$ tell us? –  Jared May 23 '13 at 15:48
Your title has the implication reversed from the body question. Please fix it or confirm and I will. –  Ross Millikan May 23 '13 at 15:50

The dot product $\cdot$ is an unusual type of multiplication. It takes two vectors in and produces a scalar out. Imagine a mommy elephant and a daddy elephant giving birth to a giraffe.

There's no such thing as division in this context. You need to do the proof by looking at components of the vectors. $u=(u_1,u_2,\ldots, u_n), v=(v_1, v_2,\ldots, v_n)$, and $$u\cdot v=u_1v_1+u_2v_2+\cdots+u_nv_n$$

Then, following Jared's hint, you can try specific values for $v$ and learn things about the components of $u$.

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LOL, I am going to use your ZOO terminology in my lin algebra class :) –  imranfat May 23 '13 at 15:51
Could you explain to me the action of giraffes on elephants? :-) –  Sammy Black May 23 '13 at 16:13

Hint: what would it mean for a vector to be orthogonal to itself?

If the problem says "every other vector" as in the title (but not the quoted text) you could instead consider what it means for a given vector $\vec{u}$ to be orthogonal to multiples of itself, e.g. to $-\vec{u}$.

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Of course, for $-\vec u$ to be distinct from $\vec u$, we'd have to assume that $\vec u\ne\vec 0$. –  Cameron Buie May 23 '13 at 17:59
@Cameron True, but if $\vec{u} = \vec{0}$ then we're done. –  Trevor Wilson May 23 '13 at 18:02

Perhaps this is slightly off topic but it's worth explaining slightly more intuitively why you can't divide by a vector.

The key idea is to think about information (or more precisely numbers of constraints/equations). A dot product takes two vectors, both with $n$ numbers in them, and combines them in a particular way (related to projecting then onto each other) to give you one number. So $2n\to 1$.

Lots of information is clearly not being used in this function. There must be many different inputs giving the same output so it's hard to invert the process. But that's not unfamiliar, multiplication (or even addition) does something similar, taking $2\to 1$. Indeed $4\times 1=2\times 2$.

The idea of division normally works by saying "Okay, so suppose I know one of the inputs as well as the output. Can I figure out the other input?" The answer is yes. Schematically, you're now given two pieces of information at the end, so $2\to 2$ pieces of information. Then you have enough to figure out what you started with! (So long as you weren't just given a zero for the input, in the case of multiplication. It works perfectly for addition.)

But what about the dot product? If someone told you what one of the vectors was as well as the dot product, that's $n$ extra pieces of information. Therefore you now get $2n\to n+1$. But unless $n=1$ (normal multiplication) you still don't have enough data! The problem is everything in the vector you seek which is orthogonal to the input vector you were told is thrown away. You have a gap of $2n-(n+1)=n-1$ in your data. You have improved by just $1$ on your original ignorance.

But if you get another $k$ input vectors, and are told both them and their dot products with your mystery vector, you get $(k+1)n\to k(n+1)$. Clearly when $n=k$ you get a balance! Therefore, trying out dot products with $n$ different fixed vectors ought to be enough information (typically) to reverse the process.

So what is this magical new division? Well, you take $n$ vectors of $n$ entries, dot them all with the unknown vector, and collect the $n$ results. If you check it out, this is exactly matrix multiplication by an $n\times n$ matrix formed by taking rows of $A$ to be the vectors chosen. That is, you are given $A u=d$ where $d$ is the vector of dot products. So the division is actually just inverting the matrix $A$ to get $u=A^{-1} d$. Magic! And just as multiplication fails to be invertible when you times by zero, this fails if $A$ is not invertible; which is equivalent (as you might know) to the test input vectors all being linearly independent.

Since in this problem you know that all vectors dotted with $u$ give zero, you can choose any linearly independent set of test vectors and you'll find $u=A^{-1} \pmatrix{0 \\ 0 \\\vdots \\ 0}$ which vanishes. Hence $u$ is the zero vector.

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vadem123's answer is perfectly fine, but I do not like such coordinate-based approaches, and the way how you compute your scalar product and the specific vectors $e_i$ you chose are heavily coordinate-based, in that vein I like Trevor Wilson's answer more, but let me add some more. If you have some orthonormal basis (which could always be choosen by the Gram-Schmidt process) $(u_1, \ldots, u_n)$, then for every $v$ we have $$v = \langle v, u_1 \rangle v + \ldots + \langle v, u_n \rangle v$$ and so for your specific $u \in U$ we have $$u = \langle u, u_1 \rangle u + \ldots + \langle u, u_n \rangle u = 0 \cdot u + \ldots + 0 \cdot u = 0.$$ Another less coordinate-based argument, if you consider $U = \{ \alpha u : \alpha \in K \}$, i.e. the linear space generated by $u$, and look at the orthogonal complement $U^T := \{ v \in V : \langle v, w \rangle = 0 \mbox{ for all } w \in U \}$, then by basic linear algebra we have $\dim V = \dim U + \dim U^T$. But by your assumption we have $U^T = V$, and so $\dim U = 0$, which implies $U = \{ 0 \}$ and therefore $u = 0$.

For an intuitive interpretation, partly inspired by Sharkos answer, let me add some interpretations of the dot product (not a proof in some cases, just an intuitive way of thinking).

i) the dot product is related to length. Specifically by $||u||^2 = u\cdot u$, so in this case $u \cdot u = 0$, which implies by definiteness $u = 0$.

ii) the dot product is related to angle, but by the cosine of it (so no angle equals one, and perpendicular, i.e. maximal distant in terms of angle, equals zero), but the zero vector is the only vector where no angle with any other vector could be meaningfully assigned,

iii) (more abstract) by the Riesz-Fischer representation theorem every linear functional could be unique described by the dot product with some vector, so if a vector represents the zero functional, it must be the zero vector itself by uniqueness (and because this identification is an isomorphism)

iv) for each $u \ne 0$ the equation $u \cdot x = 0$ describes a hyperplane, i.e. a set of dimension $n-1$, so if this set is $V$ itself we must have $u = 0$,

v) the dot product could be interpreted in terms of projection. More specifically, if I have a vector $u$ and want to project it on some other vector $v$ it is done by the formulae $$\frac{u\cdot v}{v\cdot v} v$$ for $v \ne 0$, if I have a vector which is the zero vector projected on every other non-zero vector then it must be zero itself

vi) the dot product could be interpreted as (auto-)correlation or ''how much of some vector is contained in some other vector'' (this becomes more obvious in general function spaces, i.e. infinite dimensional vector spaces, for example the fourier series expansion measures in some way how much of $\sin(nx), \cos(nx)$ oscillations are in a given function), this interpretation is related to v), if I look at every vector how much of another vector is contained in that vector, and this is zero, then the vector itself must be zero.

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