Math 817: ``All about (Gram)Schmidt'' Handout Solutions

 (a) Apply the GramSchmidt procedure to the vectors
v_{1} = (1,1,0)^{t},
v_{2} = (1,0,1)^{t},
v_{3} = (0,1,1)^{t},
using the form given by v*w, where * denotes
the usual dot product.
Solution: The procedure I gave in class first produces
an orthogonal set of vectors. We can toss out any zero vectors
as we go, but the last thing we do is
normalize the nonzero vectors that we get.
So first we take u_{1} = v_{1},
c_{1} = u_{1}*u_{1} = 2.
Next we get
u_{2} = v_{2} 
(1/c_{1})(v_{2}*u_{1})u_{1}
= (1,0,1)^{t}  (1/2)(1)(1,1,0)^{t}
= (1/2)(1,1,2)^{t},
c_{2} = u_{2}*u_{2} = 6/4 = 3/2.
Since right now all we want is to get
orthogonal vectors, we can even replace
(1/2)(1,1,2)^{t} by (1,1,2)^{t}.
Using u_{2} = (1,1,2)^{t}
and c_{2} = 6 will be easier since we thereby avoid
some fractions. Finally, we get
u_{3} = v_{3} 
(1/c_{2})(v_{3}*u_{2})u_{2} 
(1/c_{1})(v_{3}*u_{1})u_{1}
= (0,1,1)^{t}  (1/6)(1)(1,1,2)^{t}  (1/2)(1)(1,1,0).
Thus 6u_{3} = (0,6,6)^{t}  (1,1,2)^{t}  (3,3,0)
= (4,4,4)^{t}. We can even take (1,1,1)^{t}
to be u_{3} if we want; then c_{3} = 3.
u_{1} = (1,1,0)^{t},
u_{2} = (1,1,2)^{t},
u_{3} = (1,1,1)^{t} is still an orthogonal
basis. To get an orthogonal basis, just divide each u_{i}
by the square root of c_{i}.
 (b) Let A be the 2x2 real symmetric matrix whose top row
is (2,1). Find a basis of R^{2} orthogonal with respect to
the bilinear form v*w defined as v^{t}Aw.
Solution: We'll use the method from Friday's class that
gives a basis orthogonal with respect to any real symmetric bilinear form.
Start with any vector w for which w*w is not zero. That's easy here;
since the eigenvalues of A are 3 and 1, we know A is positive definite,
so w*w > 0 for any nonzero vector w. So let's take w = (1,0)^{t}.
Now we just need to find an orthogonal basis for the
space W^{perp} (there's no perp symbol in html,
so I'll just use the "word" perp) orthogonal to w. So we pick any element
x of W^{perp} with x*x not zero (since A is positive definite,
that means any nonzero vector x = (a, b) in W^{perp}). But
Aw = (2,1)^{t}, so we want any nonzero solution to
x^{t}(2,1)^{t} = 0, which is just 2a + b1 = 0.
So let's take a = 1, b = 2. Thus our orthogonal basis is
(1,0)^{t}, (1,2)^{t}. To get an orthonormal basis, divide
each vector by the square root of the value of the form
evaluated on the vector against itself:\.
We get (1/c_{1})(1,0)^{t} and
(1/c_{2})(1,2)^{t}, where
c_{1} = 2^{1/2} and
c_{2} = 6^{1/2}.
 Let P_{n} be the vector space of all real polynomials
of the form a_{0} + a_{1}x + ... + a_{n}x^{n}.
Given f(x) and g(x) in P_{n}, define the symmetric bilinear form
f*g by integrating f(x)g(x) from 1 to 1.
Find an orthonormal basis for P_{n} for n = 1, 2 and 3.
Solution: A basis for P_{n} is given by
1, x, x^{2}, ..., x^{n}. Now we find an
orthogonal basis, which in the end we normalize to get an
orthonormal basis. (The resulting
polynomials, up to scalar multiples,
are called Legendre polynomials. They also come up in
differential equations. You can google them to find out more.)
So we take u_{1} = 1; then 1*1 = c_{1} = 2.
Next, u_{2} = x  (1/c_{1})(x*1)1 = x,
since 1*x = 0. Also, c_{2} = x*x = 2/3.
Next, u_{3} = x^{2}
 (1/c_{2})(x^{2}*x)x
 (1/c_{1})(x^{2}*1)1 =
x^{2}  0x  (1/2)(2/3)1 = x^{2}  (1/3);
to avoid fractions, we'll use u_{3} = 3x^{2}  1,
instead. Now c_{3} = 8/5.
And finally u_{4} = x^{3}
 (1/c_{3})(x^{3}*(3x^{2}  1))(3x^{2}  1)
 (1/c_{2})(x^{3}*x)x
 (1/c_{1})(x^{3}*1)1 =
x^{3}  0x^{2}  (3/5)x  0 =
x^{3}  (3/5)x. As usual, we can take 5x^{3}  3x,
instead of x^{3}  (3/5)x; then c_{4} = 8/7.
Normalize by dividing u_{i} by the square root of
c_{i}. Normalizing u_{1} through u_{n+1}
gives an orthonormal basis of P_{n}. (Thus we don't need to start from scratch to do P_{3}, after having done P_{2}.
We just need to do the next vector.)