3

My teacher used the following symbol: $\boxed{\overset{\wedge}{=}}$

We had to write down a vector equation, and he said my direction vector $\begin{pmatrix}6\\2\\2\end{pmatrix}$ could be simplified to $\begin{pmatrix}3\\1\\1\end{pmatrix}$ since length doesn't matter as a direction vector.

He wrote in my vector equation:

$$\begin{pmatrix}6\\2\\2\end{pmatrix} \overset{\wedge}{=} \begin{pmatrix}3\\1\\1\end{pmatrix}.$$

Is this notation correct, and if so, what is the name and when is it used?

Michael Hardy
  • 1
rae306
  • 9,742
  • 1
    Was it $\triangleq$? – Batman Aug 29 '14 at 19:55
  • No, I guess not.. – rae306 Aug 29 '14 at 19:56
  • 1
    I don't think the particular one that you mention is valid. I tried image-searching it and got nothing. He may have meant to write $ \cong $ or $ \equiv $; both of them make sense. Besides, why don't you just ask him/her? – Ahaan S. Rungta Aug 29 '14 at 20:00
  • He isn't my teacher anymore. Is the one with the 'triangle' correct notation for such problems? – rae306 Aug 29 '14 at 20:02
  • 1
    $\cong$ is sometimes used for equality up to a non-zero constant, which is what they may have meant. However, the notation you stated is not standard and you should ask your teacher about it. In any case, the notation you use should be defined before you use it anyway (such as using $\cong$ for equality up to a constant). The triangle in $\triangleq$ means that the equality is a definition and is not correct in this case. – Batman Aug 29 '14 at 20:02

3 Answers3

6

This notation is not standard. One can define an equivalence relation on $\mathbb{R}^{n}$ s.t $v\sim u\iff u=\alpha v$ for $0\neq\alpha\in\mathbb{R}$ and then we can write for example $$ \begin{pmatrix}6\\ 2\\ 2 \end{pmatrix}\sim\begin{pmatrix}3\\ 1\\ 1 \end{pmatrix} $$

Regarding the use of the hat symbol and vectors - it is standard that if $0\neq v\in\mathbb{R}^{n}$ then we denote $$ \hat{v}=\frac{v}{\|v\|} $$

in this case $\hat{v}$ spans the same one dimensional subspace as $v$ and satisfies $\|\hat{v}\|=1$

Michael Hardy
  • 1
Belgi
  • 23,150
  • Thanks for the answer! ;-) What should I use? $\cong$ or $\sim$? – rae306 Aug 29 '14 at 20:12
  • @user3761304 - I would say that $\sim$ but even then I would specify what the equivalence relation $\sim$ means. – Belgi Aug 29 '14 at 20:14
  • Alright. Thank you very much! – rae306 Aug 29 '14 at 20:17
  • @user3761304 - you are welcome :) – Belgi Aug 29 '14 at 20:18
  • I suppose that the teacher is combining the hat symbol for unit vectors with the equality symbol to say that two vectors are hat-equal if their hats (unit vectors) both exist and are equal (and possibly also including the case where both do not exist, which in this case means that both vectors are the zero vector). (Incidentally, to define this equivalence relation in the terms in which you defined $\sim$, we must require $\alpha > 0$ rather than merely $\alpha \ne 0$.) – Toby Bartels Jan 22 '16 at 19:52
  • You could generalize this to any function $f$, so that $x \overset{f}= y$ means that $x$ and $y$ are both in the domain of $f$ and $f(x) = f(y)$ (and possibly also allows that neither is in the domain of $f$, or possibly reverts to equality in the case that either is not in the domain of $f$). For example, if $f$ is the squaring function $f(x) = x^2$, then $-3 \overset{f}= 3$, because $(-3)^2 = 3^2$. When $f$ is defined everywhere, then $\overset{f}=$ is an equivalence relation; the general case is at least a partial equivalence relation, depending on how we handle undefined values. – Toby Bartels Jan 22 '16 at 19:57
3

I don't think the particular one that you mention is valid. See here for a full list. He or she may have meant to write $ \cong $ or $ \equiv $; both of them make sense, since they both imply that the two vectors are equivalent and of the same form and meaning.

Ahaan S. Rungta
  • 7,706
  • I think I got it. I should use $\cong$ in this case. What do you mean by 'define'? Should I write down what I mean at the beginning of my exercises? – rae306 Aug 29 '14 at 20:06
  • @user3761304 I did not mention 'defining' anything, but if you are referring to Batman's comment, I would say that you should define notation if it is not completely standard and you're going to use it. For example, you do not have to define '$=$' since everybody knows what that means, but you should either write out the $\equiv$ in words or define it before or after you start. – Ahaan S. Rungta Aug 29 '14 at 20:08
2

It is not conventional and will not be broadly understood without an explanation of what it means, but there's nothing incorrect about it if you explain it before using it.

Here's another example of its use: Suppose $H_1,\ldots,H_n$ are mutually exclusive hypotheses one of which must be true. Their probabilities given some new data $D$ are desired. Then $$ (P(H_1\mid D),\ldots,P(H_n\mid D)) \overset{\wedge}{=} (P(H_1),\ldots,P(H_n)) \cdot (P(D\mid H_1),\ldots,P(D\mid H_n)) $$ where the dot means term-by-term multiplication. After thus finding the equivalence class of the vector on the left, the constant by which all components must be multiplied to get the actual probabilities is the one that makes their sum equal to $1$.

And yet another example: A linear dependence among vectors $\vec{x}_1,\ldots,\vec{x}_m$ is an $m$-tuple of scalars $c_1,\ldots,c_m$, not all $0$, such that $c_1\vec{x}_1+ \cdots+c_m\vec{x}_m=\vec{0}$. But any nonzero scalar multiple of $(c_1,\ldots,c_n)$ works just as well, and expresses exactly the same nature of dependence among $\vec{x}_1,\ldots,\vec{x}_m$. Therefore a linear dependence is really an equivalence class of such tuples. One way of putting it is that the space of linear dependences is a projective space.

Michael Hardy
  • 1