MAT 232 Chapter 6

1604 days ago by kcrisman

Chapter 6 - Orthogonality

Section 6.1

There are many ways for things to be orthogonal.

vector([1,2,3]).dot_product(vector([-3,0,1])) 
       
0
0
vector([1,2,3])*vector([-3,0,1]) 
       
0
0

For instance, we have some polynomials that are, if you consider the inner product to be $\int_{-1}^1 f(x)g(x)dx$.

for i in range(4): show(expand(legendre_P(i,x))) 
       



for i in range(4): for j in range(i+1,4): f(x) = legendre_P(i,x) g(x) = legendre_P(j,x) html("$$\int_{-1}^1 \left(%s\\right) \left(%s\\right)\; dx = %s$$"%(latex(f(x)),latex(g(x)),integrate(legendre_P(i,x)*legendre_P(j,x),(x,-1,1)))) 
       




















Or you can try some complex vectors.

vector([1+i,2])*vector([1-i,4]) 
       
0
0

The functions $\sin(kx)$ and $\cos(\ell x)$ are also all orthogonal on $[0,2\pi]$.

@interact def _(i=([1..5]),j=([1..5])): html("The integral $$\int_0^{2\pi} \sin(%s x)\sin(%s x)\; dx$$"%(i,j)) html("is $%s$"%latex(integrate(sin(i*x)*sin(j*x),(x,0,2*pi)))) html("The integral $$\int_0^{2\pi} \cos(%s x)\sin(%s x)\; dx$$"%(i,j)) html("is $%s$"%latex(integrate(cos(i*x)*sin(j*x),(x,0,2*pi)))) 
       

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Anyway, now we have a way to talk about what we claimed before - that the row space and null space are "perpendicular"!

M = matrix([[1,2,3],[1,2,3],[1,2,3]]); show(M) 
       
V = M.row_space() W = M.right_kernel() 
       
V;W 
       
Free module of degree 3 and rank 1 over Integer Ring
Echelon basis matrix:
[1 2 3]
Free module of degree 3 and rank 2 over Integer Ring
Echelon basis matrix:
[ 1  1 -1]
[ 0  3 -2]
Free module of degree 3 and rank 1 over Integer Ring
Echelon basis matrix:
[1 2 3]
Free module of degree 3 and rank 2 over Integer Ring
Echelon basis matrix:
[ 1  1 -1]
[ 0  3 -2]

We just multiply every basis vector of the row space with every basis vector of the null space.

for v in V.basis(): for w in W.basis(): html("$%s\cdot %s=%s$"%(latex(v),latex(w),latex(v*w))) 
       




@interact def _(M = input_grid(4, 5, default=[[-1,-2,1,-1,0],[3,6,6,0,3],[3,6,4,1,3],[-2,-4,-5,0,-3]])): M = matrix(M) V = M.right_kernel() W = M.row_space() for v in V.basis(): for w in W.basis(): html("$%s\cdot %s=%s$"%(latex(v),latex(w),latex(v*w))) 
       

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Incidentally, the same is true of the transpose (the column space and kernel of the transpose):

@interact def _(M = input_grid(4, 5, default=[[-1,-2,1,-1,0],[3,6,6,0,3],[3,6,4,1,3],[-2,-4,-5,0,-3]])): M = matrix(M) V = M.left_kernel() W = M.column_space() html("Now we're doing it with the column space and <em>left</em> kernel") for v in V.basis(): for w in W.basis(): html("$%s\cdot %s=%s$"%(latex(v),latex(w),latex(v*w))) 
       

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Warning!  There is a lot we are not saying about these complements - see Sage ticket #7522 for some discussion about this between those interested in undergraduate pedagogical situations and those warning of keeping things completely mathematically correct!

Section 6.2

Let's do some orthogonal projection.

@interact def _(V = matrix([[1,3]]),W = matrix([[-2,3]])): V = vector(V.list()) W = vector(W.list()) ProjVW = (V*W)/(W*W)*W show(plot(ProjVW,color='green',zorder=5)+plot(V,color='green')+plot(W)+line([V.list(),ProjVW.list()],color='black',linestyle='--'),aspect_ratio=1) 
       

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Section 6.3

Well, what about in more than two dimensions?

@interact def _(V = matrix([[1,3,2]]),W = matrix([[-2,3,2]])): V = vector(V.list()) W = vector(W.list()) ProjVW = (V*W)/(W*W)*W show(plot(ProjVW,color='green',zorder=5)+plot(V,color='green')+plot(W)+line([V.list(),ProjVW.list()],color='black',linestyle='--'),aspect_ratio=1) 
       

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Note that the default setting here doesn't have the $W1$ and $W2$ being orthogonal!  You can't use the formula without doing that first.

var('u,v') @interact def _(V = matrix([[1,3,2]]),W1 = matrix([[-2,3,2]]),W2 = matrix([[-2,1,1]])): V = vector(V.list()) W1 = vector(W1.list()) W2 = vector(W2.list()) if W1*W2 != 0: print "you don't have orthogonal vectors" else: ProjVW = (V*W1)/(W1*W1)*W1+(V*W2)/(W2*W2)*W2 show(plot(ProjVW,color='green',zorder=5)+plot(V,color='blue',thickness=2)+parametric_plot3d(u*W1+v*W2, (u,-3,3), (v,-3,3),color='green',opacity=.7)+line([V.list(),ProjVW.list()],color='black',linestyle='--',thickness=5),aspect_ratio=1) 
       

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Of course, we can do all this for the polynomial or other inner products...

w1 = x w2 = 3/2*x^2-1/2 v = 1+x+x^2 
       
F = integrate(w1*v,x,-1,1)/integrate(w1*w1,x,-1,1)*w1+integrate(w2*v,x,-1,1)/integrate(w2*w2,x,-1,1)*w2 F 
       
x^2 + x - 1/3
x^2 + x - 1/3

Here, $F$ is the projection and $v-F$ is the residual.

integrate((x^2+x-1/3)*(4/3),x,-1,1) 
       
0
0
integrate((v-F)*F,x,-1,1) 
       
0
0

Inner product is zero.  Yeah.  And the distance from $v$ to this subspace is...

sqrt(integrate((v-F)^2,x,-1,1)) 
       
4/3*sqrt(2)
4/3*sqrt(2)

Huh.

Section 6.4

So, how does one exactly get this orthogonal basis?  The answer is the Gram-Schmidt process!  

#A = matrix([[1,1,0,1,0],[1,0,1,1,0]]) #A = matrix([[1,1,0,1,0],[1,0,1,1,0],[0,1,1,1,0]]) A = matrix([[1,1,0,1,0],[1,0,1,1,0],[0,1,1,1,0],[5,0,0,0,1]]) 
       
show(A.gram_schmidt()[0]) 
       

Let's see what this looks like for three general vectors.

var('u,v') @interact def _(U1 = matrix([[1,3,2]]),U2 = matrix([[-2,3,2]]),U3 = matrix([[-2,1,1]])): U1 = vector(U1.list()) U2 = vector(U2.list()) U3 = vector(U3.list()) M = matrix([U1,U2,U3]) if M.det()==0: print "you don't have a basis!" else: M = M.gram_schmidt()[0] G = Graphics() G += plot(U1,thickness=2)+plot(U2,thickness=2)+plot(U3,thickness=2) G += parametric_plot3d(u*U1+v*U2, (u,-1,1), (v,-1,1),color='green',opacity=.6) V1,V2,V3 = M.rows() G += plot(V1,thickness=2,color='red')+plot(V2,thickness=2,color='red')+plot(V3,thickness=2,color='red') G += line([V2.list(),U2.list()],color='black',linestyle='--',thickness=2) G += line([V3.list(),U3.list()],color='black',linestyle='--',thickness=2) show(G,aspect_ratio=1) 
       

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We can use a similar method to get a set of orthogonal polynomials - very useful in applications.

legendre_P? 
       

File: /usr/local/sage-5.6-linux-64bit-ubuntu_8.04.4_lts-x86_64-Linux/local/lib/python2.7/site-packages/sage/functions/orthogonal_polys.py

Type: <type ‘function’>

Definition: legendre_P(n, x)

Docstring:

Returns the Legendre polynomial of the first kind for integers \(n > -1\).

REFERENCE:

  • AS 22.5.35 page 779.

EXAMPLES:

sage: P.<t> = QQ[]
sage: legendre_P(2,t)
3/2*t^2 - 1/2
sage: legendre_P(3, 1.1)
1.67750000000000
sage: legendre_P(3, 1 + I)
7/2*I - 13/2
sage: legendre_P(3, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
[-179  242]
[-484  547]
sage: legendre_P(3, GF(11)(5))
8

File: /usr/local/sage-5.6-linux-64bit-ubuntu_8.04.4_lts-x86_64-Linux/local/lib/python2.7/site-packages/sage/functions/orthogonal_polys.py

Type: <type ‘function’>

Definition: legendre_P(n, x)

Docstring:

Returns the Legendre polynomial of the first kind for integers \(n > -1\).

REFERENCE:

  • AS 22.5.35 page 779.

EXAMPLES:

sage: P.<t> = QQ[]
sage: legendre_P(2,t)
3/2*t^2 - 1/2
sage: legendre_P(3, 1.1)
1.67750000000000
sage: legendre_P(3, 1 + I)
7/2*I - 13/2
sage: legendre_P(3, MatrixSpace(ZZ, 2)([1, 2, -4, 7]))
[-179  242]
[-484  547]
sage: legendre_P(3, GF(11)(5))
8

Section 6.5

Finally, let's fit some linear models.

var('a,b') model(x) = a+b*x Data = [(1,0),(2,1),(4,2),(5,3)] find_fit(Data,model) 
       
[a == -0.6000000000034891, b == 0.6999999999993456]
[a == -0.6000000000034891, b == 0.6999999999993456]
M = find_fit(Data,model) points(Data)+plot(model.subs(a=M[0].rhs(),b=M[1].rhs()),(x,0,5),color='black') 
       

Here's some real-life data about men's and women's record times in the 100-meter free in swimming.

var('a,b,c') model(gender,year) = a+b*year+c*gender Data = [(1,1905,65.8),(1,1908,65.6),(1,1910,62.8),(1,1912,61.6),(1,1918,61.4),(1,1920,60.4),(1,1922,58.6),(1,1924,57.4),(1,1934,56.8),(1,1935,56.6),(1,1936,56.4),(1,1944,55.9),(1,1947,55.8),(1,1948,55.4),(1,1955,54.8),(1,1957,54.6),(1,1961,53.6),(1,1964,52.9),(1,1967,52.6),(1,1968,52.2),(1,1970,51.9),(1,1972,51.22),(1,1975,50.59),(1,1976,49.44),(1,1981,49.36),(1,1985,49.24),(1,1986,48.74),(1,1988,48.42),(1,1994,48.21),(1,2000,48.18),(1,2000,47.84),(0,1908,95),(0,1910,86.6),(0,1911,84.6),(0,1912,78.8),(0,1915,76.2),(0,1920,73.6),(0,1923,72.8),(0,1924,72.2),(0,1926,70),(0,1929,69.4),(0,1930,68),(0,1931,66.6),(0,1933,66),(0,1934,65.4),(0,1936,64.6),(0,1956,62),(0,1958,61.2),(0,1960,60.2),(0,1962,59.5),(0,1964,58.9),(0,1972,58.5),(0,1973,57.54),(0,1974,56.96),(0,1976,55.65),(0,1978,55.41),(0,1980,54.79),(0,1986,54.73),(0,1992,54.48),(0,1994,54.01),(0,2000,53.77),(0,2004,53.52)] 
       
M = find_fit(Data,model); M 
       
[a == 555.7167834308991, b == -0.2514636815413249, c ==
-9.797961450991004]
[a == 555.7167834308991, b == -0.2514636815413249, c == -9.797961450991004]
model = model.subs(a=M[0].rhs(),b=M[1].rhs(),c=M[2].rhs()); model 
       
(gender, year) |--> -9.797961450991004*gender -
0.2514636815413249*year + 555.7167834308991
(gender, year) |--> -9.797961450991004*gender - 0.2514636815413249*year + 555.7167834308991
points([d[1:] for d in Data if d[0]==1])+points([d[1:] for d in Data if d[0]==0],color='red')+plot(model(gender=1),(year,1900,2010),color='blue',legend_label="modeling men's times" )+plot(model(gender=0),(year,1900,2010),color='red',legend_label="modeling women's times")