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

• (a) Apply the Gram-Schmidt procedure to the vectors v1 = (1,1,0)t, v2 = (1,0,1)t, v3 = (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 u1 = v1, c1 = u1*u1 = 2. Next we get u2 = v2 - (1/c1)(v2*u1)u1 = (1,0,1)t - (1/2)(1)(1,1,0)t = (1/2)(1,-1,2)t, c2 = u2*u2 = 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 u2 = (1,-1,2)t and c2 = 6 will be easier since we thereby avoid some fractions. Finally, we get u3 = v3 - (1/c2)(v3*u2)u2 - (1/c1)(v3*u1)u1 = (0,1,1)t - (1/6)(1)(1,-1,2)t - (1/2)(1)(1,1,0). Thus 6u3 = (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 u3 if we want; then c3 = 3. u1 = (1,1,0)t, u2 = (1,-1,2)t, u3 = (-1,1,1)t is still an orthogonal basis. To get an orthogonal basis, just divide each ui by the square root of ci.
• (b) Let A be the 2x2 real symmetric matrix whose top row is (2,1). Find a basis of R2 orthogonal with respect to the bilinear form v*w defined as vtAw.
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 Wperp (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 Wperp with x*x not zero (since A is positive definite, that means any nonzero vector x = (a, b) in Wperp). But Aw = (2,1)t, so we want any nonzero solution to xt(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/c1)(1,0)t and (1/c2)(1,-2)t, where c1 = 21/2 and c2 = 61/2.
1. Let Pn be the vector space of all real polynomials of the form a0 + a1x + ... + anxn. Given f(x) and g(x) in Pn, define the symmetric bilinear form f*g by integrating f(x)g(x) from -1 to 1. Find an orthonormal basis for Pn for n = 1, 2 and 3.
Solution: A basis for Pn is given by 1, x, x2, ..., xn. 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 u1 = 1; then 1*1 = c1 = 2. Next, u2 = x - (1/c1)(x*1)1 = x, since 1*x = 0. Also, c2 = x*x = 2/3. Next, u3 = x2 - (1/c2)(x2*x)x - (1/c1)(x2*1)1 = x2 - 0x - (1/2)(2/3)1 = x2 - (1/3); to avoid fractions, we'll use u3 = 3x2 - 1, instead. Now c3 = 8/5. And finally u4 = x3 - (1/c3)(x3*(3x2 - 1))(3x2 - 1) - (1/c2)(x3*x)x - (1/c1)(x3*1)1 = x3 - 0x2 - (3/5)x - 0 = x3 - (3/5)x. As usual, we can take 5x3 - 3x, instead of x3 - (3/5)x; then c4 = 8/7. Normalize by dividing ui by the square root of ci. Normalizing u1 through un+1 gives an orthonormal basis of Pn. (Thus we don't need to start from scratch to do P3, after having done P2. We just need to do the next vector.)