(x, y)=(Sx)T

MA351 EXERCISES FOR SECTION 5.5
Problem 1. Consider a matrix S in R n×n . In R n , define the product
a. For matrices S is this an inner product?
b. For matrices S is (x, y) = x · y?
(x, y) = (Sx) T (Sy).
Solution: a. Note that (x, y) = (Sx) T (Sy) = Sx · Sy. Need to check 4 conditions:
1. (x, y) = Sx · Sy = Sy · Sx = (y, x)
2. (x + y, z) = S(x + y) · Sz = (Sx + Sy) · Sz = Sx · Sz + Sy · Sz = (x, z) + (y, z).
3. (cx, y) = S(cx) ˙Sy = c(Sx) · Sy = c(Sx · Sy) = c(x, y).
4. (x, x) = Sx · Sx = Sx 2 ≥ 0. If (x, x) = 0, then Sx = 0, meaning that Sx = 0. In order for
(·, ·) define an inner product, it must satify that the only way (x, x) = 0 is when x = 0. Therefore, the
homogenous system Sx = 0 must have only the (unique) trivial solution x = 0. Namely, Ker(S) = {0},
which tells us that S must be invertible.
b. Note that (x, y) = (Sx) T Sy = x T S T Sy. It is required that x T (S T S)y = (x, y) = x T y for all x, y.
This is the case when S T S = In, that is, S is orthogonal.
Problem 2. In R n×n , consider the inner product
(A, B) = Trace(A T B).
a. Find a formula for this inner product in R n×1 , i.e., when A and B are vectors in R n .
b. Find a formula for this inner product in R 1×n , i.e., when A T and B T are vectors in R n .
Solution: a. Let u, v ∈ R n×1 . (u, v) = Trace(u T v) = Trace(u · v) = u · v.
b. Let u, v ∈ R 1×n . (u, v) = Trace(u T v) = n
k=1 ukvk = u T · v T = uv T .
Problem 3. a. Consider an n × m matrix P and an m × n matrix Q. Show that
Trace(PQ) = Trace(QP).
b. Compare the following two inner product in R n×n ,
(A, B) = Trace(A T B), ((A, B)) = Trace(AB T ).
Solution: a. The jjth entry of PQ is m k=1 pjkqkj, so that Trace(PQ) = n m j=1 k=1 pjkqkj. Likewise
Trace(QP) = m n j=1 k=1 qjkqkj. Reversing the role of j and k and the order of summation we see that
Trace(PQ) = Trace(QP).
b. Using part a. and the fact that Trace(M) = Trace(MT ) for any square matrix M, we find that
(A, B) = Trace(A T B) = Trace(BA T ) = Trace((BA T ) T ) = Trace(AB T ) = ((A, B)).
1

2 MA 351 EXERCISES FOR SECTION 5.5 11/28/2011 YOLCU
Problem 4.
a. Consider the space P2 of polynomials of degree less than or equal to 2 with inner product
〈f, g〉 =
1
0
f(t)g(t) dt.
Find a basis of the space W of all functions in P2 that are orthogonal to f(t) = t. What is the dimension
of W?
b. Consider the space P2 of polynomials of degree less than or equal to 2 with inner product
(f, g) = 1
2
1
−1
f(t)g(t) dt.
Find an orthonormal basis of the space of all functions in P2 that are orthogonal to f(t) = t.
Solution: a. This one is similar to what is in b.
b. A function g(t) = a + bt + ct2 is orthogonal to f(t) = t if
0 = (f, g) = 1
2
1
−1
t(a + bt + ct 2 ) dt = 2
3 b.
meaning that b = 0. Thus, g(t) = a + ct2 belongs to span{1, t2 } = W. That is, 1 and t2 form a basis
of the space W of all functions in P2 orthogonal to f(t) = t. To find an orthonormal basis {g1(t), g2(t)}
we apply Gram-Schmidt. Now,
1 2 = (1, 1) = 1
1
1 dt = 1
2 −1
so we can take g1(t) = 1 as 1 = 1. Also,
g2(t) = t2 − (1, t2 )1
t2 − (1, t2 )1 = t2 − 1
3
t2 − 1
√
5
=
3 2 (3t2 − 1).
√
5
Therefore, {g1(t), g2(t)} = {1, 2 (3t2 − 1)} is an orthonormal basis for W. Moreover dim(W) = 2.
Problem 5. Which of the following is is an inner product in P2? Explain.
a. (f, g) = f(1)g(1) + f(2)g(2)
b. ((f, g)) = f(1)g(1) + f(2)g(2) + f(3)g(3)
Solution: a. (f, g) is not an inner pruduct in P2 because f(t) = (t − 1)(t − 2) implies (f, f) = 0 but
f = 0.
b. ((f, g)) is an inner product, and easy to check the four axioms. Note that ((f, f)) = f(1) 2 +
f(2) 2 + f(3) 2 = 0 when f(t) = 0 in P2, because any polynomial in P2 has at most 2 zeros. That is, if f
were nonzero polynomial, then at least one of f(1), f(2) and f(3) would be nonzero and ((f, f)) would
be strictly positive.

MA 351 EXERCISES FOR SECTION 5.5 11/28/2011 YOLCU 3
Problem 6. For which values of the constants b, c and d is the following an inner product in R2 ?
x1 y1
, = x1y1 + bx1y2 + cx2y1 + dx2y2
x2
y2
Solution: Using the four axioms, one can obtain b = c and d > b 2 . Work out the details!
Problem 7. a. Find an orthonormal basis of the P1 with the inner product
(f, g) =
1
0
f(t)g(t) dt.
b. Find the linear polynomial g(t) = at + b that best approximates the function f(t) = t 2 on the
interval [0, 1] in the least-squares sense. Draw a sketch.
Solution: a. We start with the standard basis {1, t} and use the Gram-Schmidt process to construct an
orthonormal basis {g1(t), g2(t)}. Note that
so that we can let g1(t) = 1. Then
g2(t) =
1 2 = (1, 1) =
t − (t, 1)1
t − (t, 1)1
1
0
1 dt = 1
= t − 1
2
t − 1
2 = √ 3(2t − 1).
Therefore, {g1(t), g2(t)} = {1, √ 3(2t − 1)} is an orthonormal basis of P1.
b. We are looking for
Note that
Therefore,
(t 2 , g1(t)) =
projP1 (t2 ) = (t 2 , g1(t))g1(t) + (t 2 , g2(t))g2(t).
1
0
t 2 dt = 1
3 , (t2 , g2(t)) =
projP1 (t2 ) = 1
3 +
√ 3
6
1
0
t 2√ 3(2t − 1) dt =
√ 3(2t − 1) = t − 1
6 .
√ 3
6 .

4 MA 351 EXERCISES FOR SECTION 5.5 11/28/2011 YOLCU
Problem 8. Consider the inner product
(u, v) = u T
1 2
v, u, v ∈ R
2 8
2 .
a. Find all vectors in R2
1
that are perpendicular to with respect to the inner product defined
0
above.
b. Find an orthonormal basis of R2 with respect to this inner product.
c. Is the following inequality true? Explain why.
(u1v1 + 2u1v2 + 2u2v1 + 8u2v2) 2
≤ (u1 + 2u2) 2 + 4u 2
2 (v1 + 2v2) 2 + 4v 2
2 .
x1 1
−2
Solution: a. , = x1 + 2x2 = 0 when x1 = 2x2. This is the line spanned by vector .
x2 0
1
1 −2
b. From part a. we noticed that and are orthogonal, we merely have to normalize them.
0 1
Note that
1
0
2
= 1 0 1 2
2 8
1
= 1,
0
−2 2
=
1
−2 1
1 2 −2
= 4.
2 8 1
1 −1
Thus, , is an orthonormal basis for R
0 1/2
2 with respect to the inner product defined above.
c. From (tu + v, tu + v) = tu + v2 ≥ 0 for all t, we see that t2 (u, u) + 2t(u, v) + (v, v) ≥ 0
holds true for all t. Recall that the discriminant b2 − 4ac ≤ 0 when at2 + bt + c ≥ 0. Therefore, we
get |(u, v)| 2 ≤ (u, u)(v, v). It remains to verify that the inequality claimed in part c is as the same as
|(u, v)| 2 ≤ (u, u)(v, v). Therefore, the answer is YES.
Problem 9. In the space P1 of the polynomials of degree ≤ 1, we define the inner product
(f, g) = 1
(f(0)g(0) + f(1)g(1)) .
2
Find an orthonormal basis for this inner product space.
Solution: We start with the standard basis {1, t} of P1 and use the Gram-Schmidt process to construct
and orthonormal basis {g1(t), g2(t)}. Note that
so that we can let g1(t) = 1. Then
1 2 = (1, 1) = 1
(1 · 1 + 1 · 1) = 1,
2
g2(t) =
t − (t, 1)1
t − (t, 1)1
= t − 1
2
t − 1
2
= 2t − 1.
Summary: The functions g1(t) = 1 and g2(t) = 2t − 1 form an orthonormal basis for P1.

MA 351 EXERCISES FOR SECTION 5.5 11/28/2011 YOLCU 5
Problem 10. Consider the linear space of all polynomials with inner product
(f, g) =
1
0
f(t)g(t) dt.
For three polynomials f, g, and h we are given the following inner products:
(f, f) = 4, (f, g) = 0, (f, h) = 8, (g, g) = 1, (g, h) = 3, (h, h) = 50.
a. Compute (f, g + h).
b. Find g + h.
c. Find projW(h), where W = span{f, g}, interms of f and g.
d. Find an orthonormal basis of span{f, g, h} in terms of f, g and h.
Solution: a. (f, g + h) = (f, g) + (f, h) = 8.
b. g + h2 = (g + h, g + h) = (g, g) + 2(g, h) + (h, h) = 57 imples g + h = √ 57.
c. Note that (f, g) = 0, g = 1 and f = 2. Therefore, { 1
2f, g} is an orthonormal basis for W.
Thus,
projW(h) = h, f
f
+ (h, g)g = 2f + 3g.
2 2
d. We already know that {g1 = 1
2f, g2 = g} is an orthonormal basis for W. To find an orthonormal
basis of span{f, g, h} in terms of f, g and h we need to obtain the third polynomial:
g3 =
h − projW(h) h − 2f − 3g
=
h − projW(h) 5
Thus, an orthonormal basis of span{f, g, h} is
{g1, g2, g3} =
1
2
f, g, −2
5
= − 2
5
3 1
f − g +
5 5 h
.
3 1
f − g +
5 5 h.