Question

Could someone provide a source/tell himself to me some essential parts about dealing with matrices and vectors? All kinds of tips and hints that you should remember; fist-rules/formulas/key concepts etc. By the way I don't have to do any calculations that would require using cartesian form or a calculator. Thanks in advance ! : ]

Answer 1

That's a big question. What is it that you need this for -- are you cramming for an exam or just looking to refresh what you once learned about matrices? They are useful because they make it simpler to solve a bunch of equations at once. For anything more just go to Google and type in "main concepts of matrices" or something like that and you'll have plenty of great sites.

Answer 2

It's been a while, and I've only used matrices for finding the cross productwith the determinant. So here's what I know for sure.

"Right hand rule" - good for cross product, if you need to know the direction of the resulting vector!
1. Put your vectors together - tail to tail.
2. Take right hand, make it flat.
3. Point your right hand in the direction of the FIRST vector of the cross product.
4. Orient your hand so you're able to curl your fingers normally towards the SECOND vector.
5. The resulting vector will extend in the direction that your thumb extends.

Scalars - in general.
They're boring. A scalar is a quantity. No direction. (It can still be positive, 0, or negative).

Vectors - in general.
Vectors are, very basically, a direction and a magnitude. Beyond that, all the complication is things you can do with vectors.

And naming. There are quite a few ways to show that a variable is actually a vector. In my experience, it depends on the field of study, or user's personal preference. Some people type the variable in bold. Some people put an arrow over it, some a line, and there might be more.

Then there's a way of showing the guts of the vector - it's dimensions. If I'm guessing right, each dimension will be a scalar. I guess you use column vectors? I've only used row vectors. Yet, either way, you know what they look like. (I guess they work special together, and are used for different feels)
http://www.physicsforums.com/showthread.php?t=590812

Another way to show a vector will require an explanation of unit vectors, and multiplying vectors with scalars. And I think that this way of showing a vector is weird, because it involves other vectors. Anyway.

First, multiplying scalar (like 5) with vector (like <1, 2, 3>).
(5) <1, 2, 3> = <(5)(1), (5)(2), (5)(3)> = <5, 10, 15>
The result is a vector that changes in magnitude proportional to the scalar.
Also, it's in the same direction as the vector, because each of its dimensions increased proportionally.
Unit vectors are vectors whose overall magnitude equals 1. Thus, they all have direction (can be any direction). When multiplying with a scalar, like earlier, a unit vector will - like earlier - keep it's direction, and change it's magnitude proportionally. Handy whenever you want to involve a vector's direction, but not it's length.

Here's the other way to express a vector. Well, a vector has dimensions, like
<a, b>. Say they correspond to x, and y directions respectively. Then it's "a" in the x direction. So it's really a scalar, "a", in the positive x direction. And that x direction can be pointed to with a vector (since they do pointing). That vector, it's good to note, extends no (zero)(0) units in the y direction - it's all x, and x doesn't go in the y direction. Just doesn't. So this vector describing the x direction is <some positive number, 0>. Multiply it with some scalar, and you have a vector of some length in the x direction. Make that "some positive number" 1, (then you have a unit vector, and) multiply it with a scalar and you have a vector along the x dimension with a length of the scalar! So
<a,b> = (a) <1, 0> + (b) <0, 1> (vector dimensions are variables)
And, to become familiar with <1,0> and <0, 1> we'll name them.
<1,0> = vector "i"
<0,1> = vector "j"
So, really, <a, b> = (a) i + (b) j

And vectors of 3 dimensions or less can be visualized, where each scalar dimension has a part in the creation of the vector. If you have a vector <a, b, c> in the x, y, z dimensions respectively, then the vector extends "a" units in the x dimension, "b" units in the y dimension, and "c" units in the z dimension. So you pick an arbitrary point in the dimensions to start drawing the vector from - <0, 0, 0> seems good for now. And then you draw your line to the one point in all dimensions that meets the definition - "a" units along x, "b" units along y, and such - and that point will mark the.. not tail end.. Draw an arrow there.

By the way, any vector of all dimensions being 0 is the zero vector, denoted as a "0" symbolized as a vector (you know - maybe arrow, maybe line, maybe bold).

Also, adding and subtracting vectors is just adding and subtracting like dimensions.
<3, 4> + <1, 2> = <3+1, 4+2> = <4, 6>
<3, 4> - <1, 2> = <3-1, 4-2> = <2, 2>

You can take your direction-and-magnitude vector and pull out the magnitude. ||'s around a vector mean that we're talking about its length.
||<a, b, c>|| = square route (a^2 + b^2 + c^2)

You can take your direction-and-magnitude vector and get the unit vector - same direction and a value of 1.
<a, b, c> / ||<a, b, c>|| = unit vector
Each dimension reduced proportionately by it's own length, together = unit vector. Reminiscent of 5 / 5 = 1, 7/7 = 1.


Dot Product -
The result is a scalar.
Two ways to get the same answer for (u dot v)
<u1, u2, u3> dot <v1, v2, v3>:
1. (u1)(v1) + (u2)(v2) + (u3)(v3)
2. (||u||)(||v||) cos(angle between them)

It's important to note that you can solve for the angle between the vectors if you don't know it.
((u1)(v1) + (u2)(v2) + (u3)(v3)) = (||u||)(||v||) cos(angle between them)
therefore...
cos-1( ((u1)(v1) + (u2)(v2) + (u3)(v3)) / (||u||)(||v||) ) = (angle between them)

Good to know:
If dot product is 0, at least one vector is the 0 vector, or vectors are orthogonal (at right angles with each other - perpendicular if put together)

Projection: shown via drawings

Answer 3

Projections use vectors

Projection of u onto v
\[proj_{v}u\]


Projection of v onto u
\[proj_{u}v\]



Projection magnitude of u onto v:
= (u dot v) / ||v||
scalar / scalar = scalar
from
(u dot v) = ||u|| ||v|| cos(angle between)
and substituting (cos(angle between)) = adjacent/ hypotenuse = proj_{v}u / ||u||
so
(u dot v) = ||u|| / ||u|| * ||v|| * proj_{v}u = ||v|| * proj_{v}u
so
proj_{v}u = (u dot v) \ ||v||



Projection vector of u onto v:
= ( (u dot v) / ||v||^2 ) (v)
scalar / scalar * vector = vector
= proj_{v}u length in direction of v (captured by unit vector)
= ( (u dot v) \ ||v|| ) (v) / ||v||
scalar * vector / scalar = vector
"(u dot v) / ||v||)" is proj_{v}u length, copiec from above
= ( (u dot v) / ||v||^2 ) (v)