Matematik och Datavetenskap, Chalmers Tekniska Högskola och Göteborgs Universitet

ALA-B, 2000, studio 6.1

Home, w 1, w 2, w 3, w 4, w 5, w 6, w 7. Matlab: analysis, linear algebra, facit.
 

obs!! innehållet i denna sida är ännu inte slutligt fastställt!! obs!!

Linear algebra in R^n, part 4.

1. Projection onto subspace V=R(A),  Ch 45.31-45.33.   Let the column vectors a,b,c in R^4 and the matrix A be as in the previous exercise:

a=[1; 2; 3; 4], b=[4; 3; 2; 1], c=2*a-3*b, A=[a b c],  B=[a b]

Remember that  u=[1;1;1;0]  does not belong to V=R(A).   We will compute the projection  Pu  of  u  onto  V.   Remember that  a,b, but not a,b,c, are linearly independent, so a,b is a basis for V and  V=R(A)=span(a,b,c)=R(B)=span(a,b).   Pu  belongs to V, so it is a linear combination of a,b,  i.e.,

Pu=Bx=x1a+x2b.

We will determine the coefficients x1, x2, so that the residual  u-Pu  is orthogonal to  V.  This means that

(u-Pu,v)=0  for all  v in V.

Since a,b is a basis for V, it is sufficient to take  v=a and v=b here:

(u-Pu,a)=0               or equivalently                  (Pu,a)=(u,a)
(u-Pu,b)=0                                                      (Pu,b)=(u,b)

If we insert  Pu=x1a+x2b here we get a system of two equations:

x1(a,a) + x2(b,a) = (u,a)
x1(a,b) + x2(b,b) = (u,b)

Insert the given vectors a,b,u and solve for the coefficients  x1, x2.   Hint:  the matrix of the system is
C=[a'*a  a'*b;  b'*a  b'*b].
What is the vector of the right hand side?

Form the vector v=Pu and check that  w=u-Pu is orthogonal to a, b, c, and Pu  as required.    This illustrates the proof of Theorem 45.10.

We have an  orthogonal decomposition of the space Rn so that every vector u can be written
u = Pu + (u-Pu) = v + w,
where v=Pu belongs to V and w=u-Pu is orthogonal to V.   Show that
|| u ||2 = || Pu ||2 + || u-Pu ||2 .                   (Pythagoras' theorem)

Hint:  expand the square  || v+w ||2 = (v+w,v+w)  and use orthogonality.

Check that this equality holds for the vectors v, w that you computed earlier.

2.  The least squares method, Ch 45.42.   We recall that the vector u does not belong to R(B), so that the equation Bx=u has no solution.   When it is impossible to make the residual Bx-u equal to 0, we try instead to minimize the norm of the residual:

find x in R2 such that  || Bx-u ||2 is minimized.

The is called "the least squares method".   We know that the distance   || Bx-u ||  is minimal when Bx is the projection of u onto V=R(B)=span(a,b), which is exactly what we computed in exercise 1.  So we have just computed the least squares solution x.    Now we will do it again in a different way.

Pu is determined by the equation

(u-Pu,v)=0  for all v in V=R(B)

Here we have Pu=Bx and v=By for some x,y in R2.  Hence, we want to find x in Rsuch that

(u-Bx,By)=0  for all y in R2.

Here  (u-Bx,By)=(BTu-BTBx,y)  so that (BTu-BTBx,y)=0 for all y in R2.  This implies that

(*)   BTBx=BTu           (the normal equations)

Here we used the fact that (w,y)=0 for all y, implies w=0.  Prove this!   Hint:  we can take y=w because the equality should hold for all y.

Prove that BTB is symmetric.

Set up the normal equations with the given matrices B, u and solve the equations.     Compare with what you got in exercise 1.   In matlab the backslash operator:  x=B\u  gives the least squares solution if the equation Bx=u has no solution.

3.  The tank reactor.    The Arrhenius rate law is  k=k0 exp(-E/(RT)).   The task is to determine the rate constant k0 and the activation energy E from the given information.   Form the logarithm of this equation to get at linear relation between y=log(k) and x=1/T, y=b+ax.  Use the given information about the reaction rate to generate several (5 or 10) data points   Ti, ki.   (Remember that T is the absolute temperature [K].)  Then set up a linear system of equations and solve them by the least squares method.    The idea is that we determine the coefficients a, b by finding a straight line y=b+ax that fits the given data points as well as possible, i.e., find a,b so that the norm of the residual b+axi-yi is minimized.

Hint:   B=[ones(5,1) -1./T].   The result should be somewhere around  k0=1e8, E=8e4.

/stig



Last modified: Fri Oct 22 17:39:46 MET DST 1999