Does the matrix[latex]\left[ \begin{array}{cc} x&y \end{array} \right][/latex][latex]\left[ \begin{array}{ccc} a&b&c\\d&e&f\\k&m&n \end{array} \right][/latex][latex]\left[ \begin{array}{c} x\\y\\1 \end{array} \right][/latex] has a name?

I believe it opens to become [latex]{a{x^2}+e{y^2}+(b+d){xy}+(k+c){x}+(f+m){y}+n}[/latex]

Does the matrix[latex]\left[ \begin{array}{cc} x&y \end{array} \right][/latex] [latex]\left[ \begin{array}{ccc} a&b&c\\d&e&f\\k&m&n \end{array} \right][/latex] [latex]\left[ \begin{array}{c} x\\y\\1 \end{array} \right][/latex] has a name?

There are three matrices there. Three. THREE.

Edited November 2, 201114 yr by the tree

I usually call them Jim, John, and Terry. Terry is recently divorced though and doesn't feel like talking much anymore.

I usually call them Jim, John, and Terry. Terry is recently divorced though and doesn't feel like talking much anymore.

 

Quick someone get to Mississippi's lab - he has been sniffing the wrong chemicals again

There are three matrices there. Three. THREE.

 

Sorry. It should be matrices. So it doesn't, collectively, have a name? I can see that it represents the equation of a curve.

It doesn't really represent anything. Are those supposed to be multiplied together? Are we talking dot multiplication? Even that only makes sense if there are the same amount of rows on the left as columns on the right of the dot.

 

edit Okay, it occurs to me that you might have meant

[math]\left[ \begin{array}{ccc} x&y&1 \end{array} \right] \cdot \left[ \begin{array}{ccc} a&b&c\\d&e&f\\k&m&n \end{array} \right] \cdot \left[ \begin{array}{c} x\\y\\1 \end{array} \right] [/math]

 

Which does indeed expand to the expression in the OP. In which case, no name, just a very obscure and typo ridden way of writing the general form of a two variable polynomial of degree two.

 

It doesn't represent an equation. It's an expression. Equations have equals in them.

Edited November 3, 201114 yr by the tree

Set equal to zero and with certain coefficients you will obtain an ellipse - other forms will give hyperbola or no real plots

I think it covers all the conic sections, although I really can't see any merit to writing them like that.

I think it covers all the conic sections, although I really can't see any merit to writing them like that.

 

yes of course. should have thought - if the coefficient for the y^2 and xyare zero it will be a parabola. But agree it is a very strange way of writing it

It turns out this is actually a thing, and you can sort of get something from it. Still no name for it though.

It doesn't really represent anything. Are those supposed to be multiplied together? Are we talking dot multiplication? Even that only makes sense if there are the same amount of rows on the left as columns on the right of the dot.

 

edit Okay, it occurs to me that you might have meant

[math]\left[ \begin{array}{ccc} x&y&1 \end{array} \right] \cdot \left[ \begin{array}{ccc} a&b&c\\d&e&f\\k&m&n \end{array} \right] \cdot \left[ \begin{array}{c} x\\y\\1 \end{array} \right] [/math]

 

Which does indeed expand to the expression in the OP. In which case, no name, just a very obscure and typo ridden way of writing the general form of a two variable polynomial of degree two.

 

It doesn't represent an equation. It's an expression. Equations have equals in them.

 

Yes I did. Sorry I missed out an element. So basically it can be used to represent various conics...but I see that the normal way is much simpler and less fussier. I actually done a differentiation operation of it by expanding it but I don't think it has much use anyway.

If you take a look at the wikipedia article I linked, you'll see that by taking determinants of the matrix in the middle and of a minor of that matrix, you can see what kind of conic you are looking at.

We can do the same by determining the eccentricity of the conic using the conventional method...but the matrix method conveys it much easily

Yes I did. Sorry I missed out an element. So basically it can be used to represent various conics...but I see that the normal way is much simpler and less fussier. I actually done a differentiation operation of it by expanding it but I don't think it has much use anyway.

 

What you have is the restriction of a quadratic form on 3-space to the two-dimensional subspace determined by the variables x and y. This is a rather unusual notion.

 

More commonly one studies quadratic forms of some given dimension which are determined by a symmetric matrix, often positive-definite, and almost always non-degenerate (determinant not 0). There is a large literature on quadratic forms -- see for instance the books by O'Meara or Lam.

What you have is the restriction of a quadratic form on 3-space to the two-dimensional subspace determined by the variables x and y. This is a rather unusual notion.

 

Quadratic forms are expressions of space?

Quadratic forms are expressions of space?

 

No. Quadratic forms are functions defined on vectors from an n-dimensional vector space.

 

I have no idea what an "expression of space" could possibly mean.

Edited November 6, 201114 yr by DrRocket