Confirming Understanding of an Orthonormal System

  • MHB
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In summary, the conversation discusses the concept of an ON-system, which refers to an orthonormal basis where all vectors have a length of 1 and are perpendicular to each other. It is mentioned that in an ON-system, the dot product between two vectors is simplified to u*v=a_1a_2+b_1b_2+c_1c_2. The conversation also clarifies the meaning of ON, which stands for orthogonality, and how it relates to the dot product formula.
  • #1
Petrus
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Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we don't use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)

Regards
\(\displaystyle |\pi\rangle\)
 
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  • #2
Petrus said:
Hello MHB,
I would like if someone could confirmed if I did understand correct.
In a ON-system we don't use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)

Regards
\(\displaystyle |\pi\rangle\)

What do you mean by an ON-system?

If u and v are perpendicular to each other, their dot product is indeed zero.
 
  • #3
I like Serena said:
What do you mean by an ON-system?

If u and v are perpendicular to each other, their dot product is indeed zero.
ON=Orthogonality
If I have understand correctly Orthogonality means that it's perpendicular.
Then we say that "Let \(\displaystyle B=(u_1,u_2,u_3)\) be a ON base in the room and we assume that \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in this base ten (they all u and v is vector)
\(\displaystyle u*v=a_1a_2+b_1b_2+c_1c_2\)

Regards,
\(\displaystyle |\pi\rangle\)
 
  • #4
Petrus said:
ON=Orthogonality
If I have understand correctly Orthogonality means that it's perpendicular.
Then we say that "Let \(\displaystyle B=(u_1,u_2,u_3)\) be a ON base in the room and we assume that \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in this base ten (they all u and v is vector)
\(\displaystyle u*v=a_1a_2+b_1b_2+c_1c_2\)

Regards,
\(\displaystyle |\pi\rangle\)

Yep. That's true.
 
  • #5
I like Serena said:
Yep. That's true.
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
\(\displaystyle |\pi\rangle\)
 
  • #6
Petrus said:
Hello,
Have I also understand this correct. ON-Base means that all vector have the length 1?

Regards,
\(\displaystyle |\pi\rangle\)

Yes.
ON would mean orthonormal.
Ortho means perpendicular.
Normal means length 1.
 
  • #7
Petrus said:
In a ON-system we don't use \(\displaystyle u*v=|u||v|\cos\theta\) cause In a ON-system the \(\displaystyle \theta=\frac{\pi}{2}\) so \(\displaystyle \cos(\frac{\pi}{2})=0\)
Just to add: If \(\displaystyle u=(a_1,b_1,c_1)\) and \(\displaystyle v=(a_2,b_2,c_2)\) in an orthonormal basis, then \(\displaystyle u\cdot v=a_1a_2+b_1b_2+c_1c_2\) is a corollary of the fact \(\displaystyle x\cdot y=|x||y|\cos\theta\) for any vectors $x$, $y$ with an angle $\theta$ between them. So it's not the case that we don't use \(\displaystyle x\cdot y=|x||y|\cos\theta\); it's just that in the case of orthonormal basis there is a special formula for the dot product.
 

FAQ: Confirming Understanding of an Orthonormal System

What is an orthonormal system?

An orthonormal system is a set of vectors that are both orthogonal (perpendicular) to each other and normalized (have a length of 1). This means that the vectors are independent and form a basis for the vector space.

Why is it important to confirm understanding of an orthonormal system?

Confirming understanding of an orthonormal system is important because it ensures that the vectors are indeed orthogonal and normalized. This is crucial in various mathematical and scientific applications, such as in linear algebra and signal processing.

How do you confirm understanding of an orthonormal system?

To confirm understanding of an orthonormal system, you can check if the vectors are orthogonal to each other by taking their dot product. If the dot product is 0, then the vectors are orthogonal. You can also check if the vectors have a length of 1 by calculating their magnitude. If the magnitude is 1, then the vectors are normalized.

Can an orthonormal system have more than 3 vectors?

Yes, an orthonormal system can have any number of vectors. As long as the vectors are both orthogonal and normalized, they form an orthonormal system, regardless of the number of vectors.

What are the applications of orthonormal systems?

Orthogonal systems have various applications in mathematics and science, such as in linear algebra, signal processing, and quantum mechanics. They are also used in data compression, image processing, and computer graphics.

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