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 R2 such 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