More on Special Symmetric Matrices
We now want to see how we can use eigenvalues and eigenvectors to classify different type of quadratic equations as circles, ellipses and hyperpolas.
We show already now that a circle centered at the origion is of the form
.
You should also know that an ellipse centered at the origion is of the form 
. This equation crosses the x axis a plus or minus a and the y axis at plus or minus b.
Thirdly the equation of an the equation of a hyperbola centered at the orgion has two different forms depending on whether it is crosses the x axis or crosses the y axis.
1. 
This one crosses the x axis at plus or minus a Also you can use the lines bx+ay=0 and bx-ay to aid in drawing the graph of the equation.
2. 
. This one corsses the y asis at plus or minus b. Also you have the same lines of interest at in 1.
The third conic section is y=ax^2 or x = py^2. This is the equation of a parabola. They are only of interest when we have both a rotation and translation. I will give one example of this type of equation. You should also note that if you look at the equation
this is again a circle.
Let us now translate each eqution into a matrix times a vector.
1. The circle 
can be written as 
= 1. Notice that the matrix here has two positive and equal eigenvalue.
2. The ellipse 
can be written as 
= 1. Here again the matrix has two positive but unequal eigenvalues.
3. We will only consider on of the two cases of a hyperbola
can be written as 
= 1. This time the matrix
has one positive and one negative eigenvalue.
4.. The parabola 
can be written as
= y
Here the matrix has one eigenvalue of zero. The p can be either positive on negative.
We will now plot a few examples of conic sections.
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implicitplot(x^2+y^2=4,x=-2..2,y=-2..2); |
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implicitplot(x^2/4 + y^2/16=1, x=-2..2,y=-4..4,scaling=constrained); |
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implicitplot(x^2/4 -y^2/9=1, x=-5..5,y=-5..5,scaling=constrained); |
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implicitplot(y=3*x^2, x=-1..1,y=0..3); |
We now would like to use the ideas of diagonalizing a symmetric matrix to be able to determine what type of conic section we are dealing with when we look at the following type of quadratic forms
We will first look at the pure rotation which will be of the from
For this equation we will only be interested in non-zero eigenvalues. The quadratic from
can be written as 
= d. Since the matrix
is symmetric it is similar to a diagonal matrix with the eigenvalues on the diagonal and the eigenvalues can be made orthonormal. All this will tell us the type of conic section and the anlge through which we should rotate to get the matrix in standard form.
Examples of Conics with cross product terms
Example 1
The first conic to be considered is the following.
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implicitplot(3*x^2 + 2*x*y + 3*y^2=8, x=-3..3,y=-3..3,scaling=constrained); |
| > |
EA:=Eigenvectors(A,output='list'); |
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q1:=Normalize(EA[1][3][1],2); |
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q2:=Normalize(EA[2][3][1],2); |
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pp1:=simplify(U^%T.D1.U); |
Our equation in the rotated coordinates is then given by 
.
= 
The above is the translation matrix times the unknowns x and y. The equation in u and v is a result of a rotation of 45 degrees. The cos of the angle is 
as is the sine of the angle.
Example 2
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B:=<<5,sqrt(3)>|<sqrt(3),7>>; |
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EB:=Eigenvectors(B,output='list'); |
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q1:=Normalize(EB[1][3][1],2); |
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q2:=Normalize(EB[2][3][1],2); |
Our equation in rotated form becomes 
| > |
implicitplot(5*x^2 + 2*x*y*sqrt(3) + 7*y^2 = 16, x=-3..3,y=-3..3,scaling=constrained); |
The rotations matrix is then 
= 
for this problem the rotation then is given by
Here the angle of rotation can be found by setting the cos(θ)= 1/2 and the sin(θ)=sqrt(3)/2.
Example 3
Consider now the quadratic from 
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implicitplot(x^2 -6*x*y + y^2 = 8, x=-5..5,y=-5..5,scaling=constrained); |
We now want to find the rotational matrix and make use this is a hyperbola that has been rotated.
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EC:=Eigenvectors(C,output='list'); |
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q1:=Normalize(EC[1][3][1],2); |
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q2:=Normalize(EC[2][3][1],2); |
The order of the q1 and q2 vectors are selected so that the negative sign is in the first row second column. This is due to the way the rotation matrix is defined throughout math courses.
Our conic section in standard form in the rotated coordinates is then
or 
Again, the angle of rotation to get the equation into standard form is 45 degrees.
Example 5
Let us look at another rotated hyperbola.
We will plot the quadratic form before we try to find the rotational matrix and the rotated from of the equation.
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implicitplot(5*x^2 - 6*sqrt(3)*x*y -y^2 =16,x=-3..3,y=-3..3,scaling=constrained); |
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G:=<<5,3*sqrt(3)>|<3*sqrt(3),-1>>; |
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EG:=Eigenvectors(G,output='list'); |
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q1:=Normalize(EG[1][3][1],2); |
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q2:=Normalize(EG[2][3][1],2); |
This tells us that the equation is a hyperbola 
Notice that for the ellipse we had two eigenvalues and they were positive. What was really important was that they were both of the same sign. In the current case we have two eigenvalues of opposite sign. This will always be the case for hyperbolas. Another way to put it is that the product of the eigenvalues is positive for an ellipse or circle and the product of the eigenvalues in the case of a hyperbola is negative.
Example 6
The following example is not contained in your text. We have not discuss the parabola. The reason for this is that we do not always get a parabola in the case when the product of the eigenvalues is zero. We can get a parabola or we can get two straight lines. We will give two examples that are very closely related but will give two straight lines first and then a rotated parbola in the second case.
Consider the two equations
and
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implicitplot(x^2+2*x*y+y^2 = sqrt(2)*(x+y),x=-3..3,y=-3..3,scaling=constrained); |
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implicitplot(x^2+2*x*y+y^2 = sqrt(2)*(y-x),x=-3..3,y=-3..3,scaling=constrained); |
We have same matrix representation of the quadratic form since they do not change. What changed was the next term. We need to take a close look at how to write a vector matrix equation of each of these equations.
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Ta:=<-sqrt(2),sqrt(2)>; |
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EH:=Eigenvectors(H,output='list'); |
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q1:=Normalize(EH[1][3][1],2); |
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q2:=Normalize(EH[2][3][1],2); |
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U^%T.Lambda.U=simplify(T^%T.Q.U); |
| > |
U^%T.Lambda.U=Ta^%T.Q.U; |
The first equation is the equation of two straight lines while the second equation is parabola.
Analysis of Examples
The standard way of writing a quadratic from is the following.
and the matrix is of the form 
. This is a symmetric matrix which can always be made orthogonally similar to a diagonal matrix with the eigenvalues on the diagonal. The determinate of the matrix is then 
and it is equal to the product of the eigenvalues 
(why?).
By our discussion of the relationship of the product of the eigenvalues to the conice section we have a second why of classifying conic sections.
1. If
then the eigenvalues are of one sign and the conic is either a circle or a ellipse
2. If 
then the eigenvalues are of opposite sign and the conic is a hyperpola.
3. If 
then one of eigenvalues is zero and we have either two straight lines or a parabola.
