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[hops]

Say I have a and b, which are arbitrary real numbers, can I say that (a+b), (2b-a), and (a-b) can be represented by x1, x2, and x3 which are arbitrary real numbers? If so, how do I prove that? It seems like something that is possible, but I have never split constants before, I have only ever combined them.

[Arx]

You definitely can say that, since there's no way to combine real numbers to produce a non-real result. I think there should be some kind of proof by contradiction if you assume a and b are real but combine to form some arbitrary (x + iy)? I don't think you'd need to prove it, though.

[Arx]

Yeah, I know the reals are a group (which legitimises the assumption), but proving that the reals are a group (which is essentially what Cinder's asking)? Bit more of a mouthful.

[Arx]

Blark, yes, you're right. I'm on vac, that's my excuse.

[hops]

I was asking about dimensionality, which, in this case, was just two like I suspected.

[hops]

a) should be trivial to solve, but my method (find the normal unit vector of E1 dependent on s) gives me ugly values, so I'm assuming I did something wrong. What should I do?

[MagmaMcFry]

E₁ has the equation 2x₁ + 2x₂ - 3x₃ = c. Determine c by using the fact that P lies in E₁, then intersect E₁ and g to get R.

[hops]

Could someone calculate a transformation matrix that reflects a vector with respect to a plane with (1, -2, 2)T as its normal vector for me? The one I got changes the length of the vector for some reasons and I want to see if someone else's answer matches mine.

EDIT: I normed the normal vector of the plane and it gives me another length. What the fuck is going on!?

[MagmaMcFry]

A reflection matrix that reflects along a given normal vector v is given by M = I-2vv'/v'v. As you can easily check, Mv = -v and Mw = w for any w orthogonal to v. Inserting that particular normal vector into the equation results in (1/9)*[[7,4,-4],[4,1,8],[-4,8,1]].

[TheBiggerFish]

Wait, I think I know how to do that one

Uhh...

Glep.  Maybe when I'm not braindead-tired...

You need to get the unit normal before making the reflection matrix though, I think.

[hops]

Okay, norming the normal vector DOES solve my problem, I apparently was an idiot and fucked up when applying the transformation. Also, the English script of my university's notes omitted the part where you have to norm the thing, and for some reasons use a different definition than the German script.

[hops]

Can someone explain to me the relation between the two part of this problem? And what concept it's trying to get me to use? I'm guessing this has something to do with characteristic polynomials but I'm not sure how that applies here.

[TheDarkStar]

IDK why you have a²₁ rather than b₁ in the first part, but otherwise the problem seems to be about using the determinant of a matrix used for find the equation of a polynomial to see if it has no solution, a unique solution, or many solutions.

In other words, there does seem to be a strange disconnect between the parts of a problem.

[Helgoland]

The coordinates of the matrix are the points that would yield the polynomial x^2, no? The computing of the determinant probably gives you some sort of existence and/or uniqueness condition, but that's just a guess. If the three points are not distinct, then your polynomial is definitely not unique, and that determinant is definitely zero. No idea though

[TheDarkStar]

The coordinates of the matrix are the points that would yield the polynomial x^2, no? The computing of the determinant probably gives you some sort of existence and/or uniqueness condition, but that's just a guess. If the three points are not distinct, then your polynomial is definitely not unique, and that determinant is definitely zero. No idea though

...right, this post reminded me of what's going on.

Given n+1 points, you can find the equation (if one exists) for an n-degree polynomial that passes through those points by setting up n+1 linear equations. For each input a_n, you get an equation of the form (c₀a_n⁰ + c₁a_n¹ + ... + c_na_nⁿ = b_n). When you set up the system with matrices, you get:
[ a₀⁰   a₀¹   ...   a₀ⁿ ] [c₀]      [b₀]
[ a₁⁰   a₁¹   ...   a₁ⁿ ] [c₁]      [b₁]
[  .           .          . ] [ . ]  =  [ . ]
[  .           .          . ] [ . ]      [ . ]
[  .           .          . ] [ . ]      [ . ]
[ a_n⁰   a_n¹   ...   a_nⁿ ] [c_n]      [b_n]

For n=2, the first matrix is the same one that your problem shows inside determinant brackets. The value of the determinant should tell you about the existence of solutions. Everything else is just solving a linear system.