5
$\begingroup$

The definition of inner product is $\langle u,v\rangle=u_1\bar{v_1}+\cdots+u_n\bar{v_n}$. Two vectors $u,v \in V$ are said to be orthogonal if $\langle u,v\rangle=0$, where $V$ is complex vector space. But why is it true? A lot of books just present this fact without giving any proof. Why such an expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}=0$ will lead to the fact that $u$ and $v$ are orthogonal? Is it an intuition behind the expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}$ that can explain orthogonality?

I read some proof about it but they don't seem correct. For example, some said that $\langle u,v\rangle=|u||v|\cos\theta$ and thus $\theta=90^\circ$. But why are they equivalent?

Some use the Pythagoras theorem, $$\begin{array}{rl} |u+v|^2 &=\langle u+v,u+v\rangle\\ &=|u|^2+2\langle u,v\rangle+|v|^2 \end{array}$$ So dot product $= 0$ implies orthogonality, but this only works for real vector space. So why exactly inner product $= 0$ implies orthogonality?

$\endgroup$
  • 3
    $\begingroup$ Depends on the context. For general inner product spaces this is a definition; however, abstract definitions in mathematics usually don't come out of nowhere. In this situation the abstract definition of an inner product grows out of the dot product of vectors in $\mathbb{R}^3$, and the notion of orthogonality grows out of the notion of two vectors being perpendicular. So it is good to check that the abstract definition, when applied to the concrete scenario of $\mathbb{R}^3$, coincides with the basic intuition. $\endgroup$ – Willie Wong Aug 24 '16 at 13:43
  • $\begingroup$ Definition of orthogonal is that the inner product is zero. By definition. HOWEVER. If we define $\mathbb R^n$ to be a vector space with geometric rays (or planes) represented by vectors. Then it is a provable theorem that geometrically the rays (or planes) are perpendicular if and only if the vectors are orthogonal. (And you prove it with the pythogorean theorem). $\endgroup$ – fleablood Aug 24 '16 at 15:20
4
$\begingroup$

Just to clarify the excellent other answers and Willie Wong's illucid comment:

Orthongonal := inner product equals zero--- a linear algebra term about vectors (applies to all vector spaces); is a definition.

Perpendicular := intersect at a right angle--- a geometry term about lines, planes or higher dimensional hyper-planes (applies to Euclidean planes and spaces); is a definition.

$\mathbb R^n$ representing Euclidean $n$-space and vectors in $\mathbb R^n$ representing Euclidean lines or planes or hyper planes; is an interpretation.

In $\mathbb R^n$, orthogonal if and only if perpendicular; is a theorem (which is provable by the Pythagorean Theorem)

(-- and which is frequently not considered an important theorem and ignored, as many [most?] texts do not considered classic geometry to be important; instead all geometry is only interpreted in linear algebra terms, in which "perpendicular" is not used.)

(In a way, asking why orthogonal/perpendicular means inner product is $0$, is like asking why $\{(t, t(x)+a|x \in \mathbb R^n\}$ is a line. It's a result of interpreting classical geometry into analytical terms, and then deciding the classical interpretation is no longer pertinent and, from then on, only using the analytic interpretation as the very basis and definition of geometry instead.)

$\endgroup$
2
$\begingroup$

As Willie Wong mentioned in his comment, the definition we take in general vector spaces is that two elements of the vector space are orthogonal if their inner product is zero. However, in $\Bbb R^2$, we can see motivation for this definition in terms of the Euclidean dot product. Suppose $a, b\in \Bbb R^2$ and $a\cdot b = 0$. Then, $$ a_1b_1 + a_2b_2 = 0 \iff a_1/b_2 = -a_2/b_1. $$ This is a manifestation of the fact that if two lines have slopes that are negative reciprocals of one another, the two lines are perpendicular in $\Bbb R^2$. Hence, two lines are perpendicular in the sense of Euclidean geometry provided that their dot product is zero. This is motivation for the general definition of orthogonality in more abstract vector spaces.

$\endgroup$
  • $\begingroup$ I like this way of understanding the inner product geometrically. But how would be the equivalent of your explication in 3 dimensions? Here we do not have slopes... $\endgroup$ – Patricio Apr 8 '18 at 19:53
  • $\begingroup$ I am conscious that you said that this is just a motivatuion for more dimensions, but I think that this should work in 3 dimensions as well...I am trying to get the 3 dimension equivalent of slopes and see how $a_1b_1+a_2b_2+a_3b_3=0$ stands for orthogonality, but until now I have not obtained any result. $\endgroup$ – Patricio Apr 9 '18 at 7:21
2
$\begingroup$

After some research I found another way to prove this, which in my opinion is the most elementary one:

Suppose you have two vectors $v,u\in\mathbb{R}^n$, and you want to determine their "angle". Supposing that their angle is 90 degrees, we could apply Pythagoras in order to calculate the area of $|u-v|^2$. This yields,

$||v||^2+||u||^2=||u-v||^2$,

which is equivalent to

$v_1u_1+...+v_nu_n=0$.

In the two dimensional case we could have argued that $u$ and $v$ has the slopes $t_1$ and $t_2$ respectively, and by applying the same argument you arrive to the condition that $t_1t_2=-1$ (this follows if you take $u=(1,t_1)$ and $v=(1,t_2)$). The last case was already mentioned by AOritz. Actually, I am still trying to figure out how the "slope" argument could be extended to three dimensions but I cannot see this.

$\endgroup$
0
$\begingroup$

The geometrical interpretation of the inner product have made usage of some of the terminology of geometry in vector spaces. I guess that the concept of orthogonality is one of them, and as the dot product defines orthogonality when it is equal to zero, then orthogonality in a vector space is defined as zero inner product.

Read what this wikipedia article says about it: https://en.wikipedia.org/wiki/Inner_product_space (in orthogonality section)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.