# Matrix in Rotation

1. May 5, 2005

### laminatedevildoll

Find the matrix of a relflection at y=-4x.

I found out that b1 = [1,-4] in the direction y = -4x
I found out that b2 = [-4,-1] orthogonal to the line

Therefore,

rho(b1) = b1 - 4b2
= -4b1 - b2

Will the matrix be

|1 -4 |
|-4 -1|

Find the matrix of rotation, if the angle of rotation is 45 degrees, the axis of rotation points in the direction of [1,2,2] and if the rotation id counterclockwise. Also repeat the same problem, if the rotation is clockwise.

I started off saying that
I found an orthogonal vector to [1,2,2]
b1=[ -4,1,1]

b3= [1/(sqrt5), 1/(sqrt5), 1/(sqrt5)]

b2 = -b1 * b3

Am I on the right track?

2. May 6, 2005

### kleinwolf

Well a reflection is an isometry (it does not change the length of the vector)....hence the matrix should have a determinant d=1 or d=-1...which is not the case for your matrix....d=-17....in fact what you wrote is the change of basis :

let [x y] be a vector...then in the basis made of the axis of reflection b1 and it's orthogonal, then [x y]=a*b1+b*b2=[a-4b -4a-b]=M[a b] with $$M=\left(\begin{array}{cc} 1 & -4\\-4 & -1\end{array}\right)$$

The reflection of [x y] is clearly obtained in the [b1 b2] basis by changing the sign of the b2 component :

R[x y]=a*b1-b*b2=[a+4b -4a+b]=N[a b] $$N=\left(\begin{array}{cc} 1 & 4\\-4 & 1\end{array}\right)$$

Then you see that R[x y]=N*inv(M)*[x y] where inv(M) is the inverse matrix of M.

3. May 7, 2005

### mathwonk

here is alittle summary of elementary linear algebra, with some remarks on reflections and rotations.

BASIC FACTS:

1) basic algebraic theory: a linear homogeneous system with s equations and n variables has infinitely many non - zero solutions, provided the number n of variables is greater than the number s of equations.

refined algebraic theory: since the reduced system has the same solutions as the original system, in fact there are infinitely many non - zero solutions if the number n of variables is greater than the number r of non zero equations in the reduced system.

since r may be smaller than s, this is more precise information, but some calculation is necessary to obtain it.

2) basic geometric theory: the solutions of a homogeneous system of s equations in n variables forms a linear subspace of R^n, having dimension at least n - s, and containing the origin.

refined geometric theory: the solutions of a homogeneous system of s equations in n variables forms a linear subspace of R^n, passing through the origin, and having dimension exactly n - r, where r is the number of non zero equations in the reduced system, i.e. r is the "rank".

[although s may be greater than n, r is never greater than n.]

for non homogeneous equations, there may be no solutions, but if there is at least one solution p, then the full set of solutions is a linear set, also of dimension exactly n-r, passing through p, and parallel to the solution space of the corresponding homogenous system.

3) linear map point of view: A system AX of s linear equations in n variables may be viewed as a linear "map" from R^n to R^s, and the solutions of the homogeneous system AX=0, is the subspace of vectors in R^n which are mapped to zero in R^s.

The map my be viewed as projecting all of R^n perpendicularly onto the "row space" R(A) and then including the r dimensional row space isomorphically onto the r dimensional image space C(A) in R^s.

Thus the solutions of the homogenous system AX=0 form the n-r dimensional linear space N(A) in R^n perpendicular to the row space, and if b happens to lie on the column space in R^s, then the solution set of the non homogeneous system AX=b, is the n-r dimensional set parallel to N(A) and passing through the point on R(A) which corresponds to the point b in R^n.

MORE REFINED RESULTS:

1) The determinant of a linear map from R^n to itself, is the oriented volume of the block which is the image of the standard unit block, by the map. The map is invertible if and only if the determinant is not zero.

2) If A is the square matrix for a map from R^n to itself, and P is any invertible matrix, then P^(-1)AP represents the matrix of the same map with respect to the new coordinate system, having the columns of P as unit vectors. for any invertible P, det(A) = det(P^(-1)AP).

3) If there is an invertible matrix P such that P^(-1)AP = B is diagonal, then A is said to be diagonalizable and the entries c1,...,cn on the diagonal of B are called eigenvalues of A. The columns v1,...,vn of P are called eigenvectors. This simply means that v1,...,vn is a basis of R^n with AVj = cjvj, for every j.

4) the eigenvalues of A are the roots of the characteristic polynomial p(t) = det(A-tI), a polynomial of degree n. for any invertible P, det(A-tI) = det(P^(-1)AP-tI).

5) If A is any matrix, with roots c1,...,cn of its characteristic polynomial p(t), then the constant term of p(t) = the product of the cj = the determinant of A; and the sum of the cj equals the coefficient of t^(n-1) in p(t) = the sum of the entries on the main diagonal of A = the "trace" of A.

In particular if A is diagonalizable with diagonal version B, the entries on the diagonals of A and B, although different, have the same sum. Hence we can determine something about the eigenvalues of A directly from A. For instance if A is 2 by 2, and has a fixed non zero vector, and the entries on the diagonal of A are a,b then the eigenvalues of A are 1 and a+b-1.

6) A matrix A which is symmetric, is always diagonalizable, and by a matrix P whose columns are mutually orthogonal.

7) Every length preserving, orientation preserving linear map of R^(2n) is a product of rotations in a family of n perpendicular planes. Every such map of R^(2n+1) has a fixed axis, and on the 2n dimensional space orthogonal to that axis, is a product of n rotations in a family of n perpendicular planes.

e.g. in R^3 every matrix A with det +1, and with transpose equal its inverse is a rotation.

In R^2 every orientation reversing isometry is a reflection. IN R^3, every symmetric, orientation reversing isometry except -I is a reflection.