Math question: orthogonal and orthonormal

[KnightWhoSaysNi]

I'm studying a linear algebra course through correspondence. The disadvantage I have with this is that I'm a strongly a visual learner. The professor is not there in person to help me understand the logic of how certain mathematical concepts work and my textbook doesn't explain them that well.

I'd like to ask, how do you geometrically represent visually (either 2D or 3D) something that is 'orthogonal' and 'orthonormal'? I can understand how to do the math by following simple steps, but I'd like to what sort of logic is taking place behind the equations. The best way for me to do that is to see what is visually going on.

Jason

[Benjamin Franklin]

Originally posted by Nightshade
I'm studying a linear algebra course through correspondence. The disadvantage I have with this is that I'm a strongly a visual learner. The professor is not there in person to help me understand the logic of how certain mathematical concepts work and my textbook doesn't explain them that well.

I'd like to ask, how do you geometrically represent visually (either 2D or 3D) something that is 'orthogonal' and 'orthonormal'? I can understand how to do the math by following simple steps, but I'd like to what sort of logic is taking place behind the equations. The best way for me to do that is to see what is visually going on.

Jason

Two vectors being orthogonal to each other simply means that they are 90 degrees to each other. Orthornomal vectors are just orthogonal vectors of unit length.


I find it strange that any course on linear algebra would not explain how to represent orthogonality geometrically. Is the correspondence course reputable ?


BF

[Santas little helper]

What BF says !

Just to make sure there is no misunderstanding it is meaningless to say
that a single vector is orthonormal.You always need at least two and it means
that each one of them has unit length and that they form a right angle with
each other.Sometimes a collection of several vectors ( ie at least 2 ) is referred
to as ortonormal like in the term ortonormal basis .The usual basis of R^2 or
R^3 or ( R^n for that matter ) has this property.Once again it means that each one has
unit length and any pair of them form a right angle.

[enigma555]

What everyone else said.

If you stand a pencil up on its end on a desk, the pencil is orthogonal to the desk.

[jfryejr]

If you stand a pencil up on its end on a desk, the pencil is orthogonal to the desk.

And if the pencil and the desk are each unit length, they are orthonormal...

[pmurray]

In N dimensional space.

The dot product of a pair of unit vectors will be equal to the cosine of the angle petween them. The dot product of a pair of orthonormal vectors will be zero, which is the cosine of a right angle.

[ohwilleke]

The great virtue of an algebraic approach, as as opposed to the geometric approach is that it isn't limited to two or three dimensions. On the other hand, dot product and cross product are more intuitive geometrically.

[NialScorva]

Originally posted by jfryejr
And if the pencil and the desk are each unit length, they are orthonormal...

I prefer to just define the desk and pencil to be the unit length rather than hope they are the unit length.

[Ted Hoffman]

How does orthogonality apply while studying physics? I mean I encounter it all the time when manifolds are being described.

When its stated that space is orthogonal to time (for arguments sake) what does it mean? that they are 90 degreed from each other? So what? Why is it important to note that two things are orthogonal to each other especially when describing topological space?

While we are at it, could someone tell me what an isomorphism is?

[enigma555]

Originally posted by Jacob Aliet
When its stated that space is orthogonal to time (for arguments sake) what does it mean? that they are 90 degreed from each other? So what? Why is it important to note that two things are orthogonal to each other especially when describing topological space?


Orthogonal vectors do not 'intrude' on each other. Continuing your exampe, if space were not orthogonal to time - if you had a non-orthogonal basis - , then it would be possible to move through space and time by travelling along one of the unit vectors. Since they are orthogonal, that means that they are totally seperate entities.

hope that's right... :cool:

[Feather]

Originally posted by Jacob Aliet
How does orthogonality apply while studying physics? I mean I encounter it all the time when manifolds are being described.

When its stated that space is orthogonal to time (for arguments sake) what does it mean? that they are 90 degreed from each other? So what? Why is it important to note that two things are orthogonal to each other especially when describing topological space?

While we are at it, could someone tell me what an isomorphism is?


http://mathworld.wolfram.com/ is a good place to start for mathematics and physics information.

An isomorphism is commonly described as a transformation or mapping that leaves some property or set of properties unchanged or a set of relations that describes a symmetry within a given set of related "things." Perhaps more clear is http://mathworld.wolfram.com/Isomorphism.html :).

[Ted Hoffman]

Originally posted by Nightshade
I'm studying a linear algebra course through correspondence.
Jason
Hi Jason, I would like to know how you got to start the course. If you could PM me the details, link etc, I'd be grateful.

I need to understand these things. Everytime I am studying physics, everytime .... these things come up.

[KnightWhoSaysNi]

So I take a pencil and stand it on its end on the desk and call it "orthogonal" since they are 90° to each other. But if I say the pencil is 20cm, and the spot where the pencil touches the desk is 30 cm from the edge, then that is "orthonormal" right? Is that what you mean by 'unit length'?

I'm trying to get my head around the "Gram-Schmidt" and "normalizing" a vector. Again, I understand the mathematical steps, but the logic and what happens in a ''visual" sense still eludes me. All I seem to be learning is how to plug in numbers but can't fathom why I'm doing it.

Jason

[Ted Hoffman]

Originally posted by enigma555
Orthogonal vectors do not 'intrude' on each other. Continuing your exampe, if space were not orthogonal to time - if you had a non-orthogonal basis - , then it would be possible to move through space and time by travelling along one of the unit vectors. Since they are orthogonal, that means that they are totally seperate entities.

hope that's right... :cool:
So that means that when 2 things are orthogonal, they have a common origin or they are simply placed at right angles to each other?

[nermal]

Originally posted by Jacob Aliet
So that means that when 2 things are orthogonal, they have a common origin or they are simply placed at right angles to each other?
Since the properties of a vector are independent of its location, yes.

What it means is simply that they are linearly independent. That is, if two vectors ( and let's include functions for sake of generality) are orthogonal, then neither function can be expressed as a linear combination of any of the other function to which it is orthogonal.

IE: v is orthogonal to t and z therefore there exists no combination:
c1t+c2z=v

This is very useful in physics because exploiting this quality eliminates all but the influencing terms in differential equations such as the wave equation automatically, leaving a simple solution such as:
w=c1sin[x]+c2cos[x]
Since sin and cos are orthogonal functions, w is a unique solution if the boundary conditions are defined.

Ed

[KnightWhoSaysNi]

Originally posted by nermal
Since the properties of a vector are independent of its location, yes.

What it means is simply that they are linearly independent. That is, if two vectors ( and let's include functions for sake of generality) are orthogonal, then neither function can be expressed as a linear combination of any of the other function to which it is orthogonal.

IE: v is orthogonal to t and z therefore there exists no combination:
c1t+c2z=v

This is very useful in physics because exploiting this quality eliminates all but the influencing terms in differential equations such as the wave equation automatically, leaving a simple solution such as:
w=c1sin[x]+c2cos[x]
Since sin and cos are orthogonal functions, w is a unique solution if the boundary conditions are defined.

Ed

Thanks Ed, this clears thing up a little for me. :) I'm a really damn slow learner with math, but I guess that's what will make me a great teacher with it. :)

Also, a quick question. If say two planes are not orthogonal to eacher other, then is it still linearly independent? Or is it only linearly dependent if say it was like 2 sheets of paper layed flat on top of each other?

Jason

[Undercurrent]

Originally posted by nermal
What it means is simply that they are linearly independent. That is, if two vectors ( and let's include functions for sake of generality) are orthogonal, then neither function can be expressed as a linear combination of any of the other function to which it is orthogonal.

This isn't true. There are plenty of linearly independent sets of vectors that are not orthogonal. {(1, 0), (1, 1)} for example.

I believe that it is true that an orthogonal set of vectors not containing the zero vector is linearly independent, the converse is not true.

[Undercurrent]

Originally posted by Jacob Aliet
So that means that when 2 things are orthogonal, they have a common origin or they are simply placed at right angles to each other?

In general, questions of where the "origin" of a vector is are irrelevant. Two vectors of the same length and direction are considered to be equal. In a lot of vector spaces there is no concept of where the tail of the vector is.

[Undercurrent]

Originally posted by Nightshade
So I take a pencil and stand it on its end on the desk and call it "orthogonal" since they are 90° to each other. But if I say the pencil is 20cm, and the spot where the pencil touches the desk is 30 cm from the edge, then that is "orthonormal" right? Is that what you mean by 'unit length'?

In "real world" examples like that one, you first have to decide what you are going to be measuring with. If you declare the units to be metres, your vectors aren't "unit length". If your declare your units to be "20cm long things" then your pencil would be a unit vector.

I'm trying to get my head around the "Gram-Schmidt" and "normalizing" a vector. Again, I understand the mathematical steps, but the logic and what happens in a ''visual" sense still eludes me. All I seem to be learning is how to plug in numbers but can't fathom why I'm doing it.

Normalation is easier. Suppose you have a vector that is not a unit vector. For example, a vector represented by your 20cm pencil where you're decalred "unit" is 1m. Your pencil is a fifth the length of the unit, so you multiply it by 5 and you get a unit vector pointing in the same direction as the original vector.

To undestand the G-S process, consider a single "basis" vector, call it u, and another vector v. v can always be written as the sum of a vector paralell to u and a vector orthogonal to u. The G-S process just takes each vector in turn, and subtracts the part of the vector that's paralell to all of the previous vectors (the slew of <v1, v2>/<v1, v1> * v1 terms), leaving only the part that's orthogonal.

[nermal]

Originally posted by Undercurrent
This isn't true. There are plenty of linearly independent sets of vectors that are not orthogonal. {(1, 0), (1, 1)} for example.



I never said all linearly independent vectors were orthogonal, just that all orthogonal vectors were linearly independent.

Ed

[nermal]

Originally posted by Nightshade
Also, a quick question. If say two planes are not orthogonal to eacher other, then is it still linearly independent? Or is it only linearly dependent if say it was like 2 sheets of paper layed flat on top of each other?

Jason [/B]

They are linearly dependent if one can be expressed as a function of the other multiplied by a constant coefficient. Therefore, if they are orthogonal, they will always be linearly independent, but not necessarily the converse, as Undercurrent demonstrated using 1,0 and 1,1 (obviously the angle between these is 45 degrees.)
If you put the vectors that make up the planes into matrix form, and row reduce to row reduced echelon form, you will have the identity matrix if the planes are linearly INdependent.

As for Grahm-Schmidt--it simply calculates the projection of one vector onto another vector (ie. a basis vector) and normalizes it.

Ed

[nermal]

Feeling quite stupid. So long as there are only two vectors, planes, operators, or functions in consideration, and unless one is a constant multiple of the other, then they are always linearly independent.
It's when you see 3 or more that you worry about one lying in the space generated by the other two.

Ed

[HRG]

Originally posted by Nightshade
Thanks Ed, this clears thing up a little for me. :) I'm a really damn slow learner with math, but I guess that's what will make me a great teacher with it. :)

Also, a quick question. If say two planes are not orthogonal to eacher other, then is it still linearly independent? Or is it only linearly dependent if say it was like 2 sheets of paper layed flat on top of each other?

Jason

Linear dependency and independency is usually defined only for vectors, or the one-dimensional subspace they generate. With two- or higherdimensional subspaces, you would say that their intersection contains only the zero vector. With planes (2-dim. subspaces) in R3, this is obviously impossible, since they will intersect in a line (if they are different).

A maximal linearly independent set of vectors is called a basis. It generates the total space in the sense that every vector can be written as a linear combination of vectors from the basis. Gram-Schmidt just makes an orthonormal basis out of any basis, once an inner product has be defined on the total space.

Regards,
HRG.