Download Worksheet, March 14th

Worksheet, March 14th
James McIvor
1. Let V = P2 (R), the space of real polynomials with degree at most two. Define an inner product on V
by
Z 1
p(x)q(x)dx.
hp(x), q(x)i =
0
Verify that this satisfies each of the axioms for an inner product. Then use the Gram-Schmidt process
on the standard basis (1, x, x2 ) to obtain an orthonormal basis for V .
Solution:
(Positivity) For any real number x and any polynomial p, p(x)p(x) = |p(x)|2 , which is a nonnegative
R1
R1
real number. Thus hp(x), p(x)i = 0 p(x)p(x)dx = 0 |p(x)|2 dx is also a nonnegative real number.
R1
(Definiteness) As above, hp(x), p(x)i = 0 |p(x)|2 dx, which is zero if and only if |p(x)| = 0, if and only
if p is the zero polynomial (why?1 ).
(Additivity, Homogeneity in the first slot) These follow from the distributivity of polynomial multiplication and linearity of the integral. Details omitted.
(Conjugate Symmetry) We compute
Z
hq(x), p(x)i =
1
Z
q(x)p(x)dx =
0
1
Z
0
1
q(x)p(x)dx = hp(x), q(x)i
q(x)p(x)dx =
0
Now we apply Gram-Schmidt to the basis (1, x, x2 ). To match the notation of the book, we rename
our basis: let v1 = 1, v2 = x, and v3 = x2 .
• e1 is obtained from v1 by dividing by the length of v1 . But kv1 k2 =
already a unit vector, so actually e1 = v1 .
• e2 is obtained from v2 by first subtracting off
Z
hv2 , e1 ie1 =
1
0
R1
0
1 · 1dx = 1, i.e., v1 is
1
1 · xdx e1 = e1
2
to get v2 − 12 e1 = x − 21 . This thing is orthogonal to e1 , but may not be a unit vector, so we divide
by the length, giving
x − 12
v2 − hv2 , e1 ie1
12
6
e2 =
=
=
x−
1
kv2 − hv2 , e1 ie1 k
5
5
kx − 2 k
• I’m too lazy to type the third one, I’ll leave it to you.
2. Consider the map R2 × R2 → R given by
x
w
h
,
i= x
y
z
1 Really
y
1
b
a
1
w
z
.
I’m asking: why is it that if a polynomial is zero on [0, 1] ⊂ R, it must actually be zero on all of R?
For which values of a, b ∈ R does this define an inner product on R2 ?
Solution: First of all, the symmetry property forces a = b. Next we consider what a must be in order
to satisfy positivity. We require that
1 a
x + ay
x
x y
= x y
= x2 + 2axy + y 2 ≥ 0
a 1
y
ax + y
for all x, y ∈ R. What a can we choose so that this inequality is true for all x, y? It’s certainly true
when x = y = 0, so we may assume one of them is nonzero, say y. Without loss of generality2 we may
take y = 1. Then we have to choose a so that x2 + 2ax + 1 ≥ 0, and in fact definiteness forces this
inequality to be strict. So for which a is x2 + 2ax + 1 > 0 for all x? By the quadratic formula, we must
have (2a)2 − 4 < 0, i.e., |a| < 1.
3. Similar to the previous question, but now consider the map C2 × C2 → C given by
1 a
w
x
w
.
h
,
i= x y
z
y
z
b 1
For which a, b ∈ C does this define an inner product on C2 ? [Hint: You may find direct computation
significantly trickier here. Consider looking at (nonzero) eigenvectors of the matrix.]
1 a
x
Solution: Following the hint, we consider the eigenvalues of the matrix A =
. Suppose
b 1
y
is a nonzero eigenvector, with eigenvalue λ. Then the positivity and definiteness conditions require
that
1 a
x
x
x y
= x y λ
= λ(|x|2 + |y|2 ) > 0.
y
y
b 1
Note that the inequality must be strict because the vector is nonzero (definiteness). Thus, for this
function to be an inner product, the eigenvalues of the matrix A must be positive real numbers. Also,
the conjugate-symmetry condition for an inner product forces a = b, so we have to compute the
eigenvalues of the matrix
1 a
.
a 1
We’ll cheat and use determinants, even though they haven’t yet been defined in this class. The
characteristic equation is
(1 − λ)2 − aa = λ2 − 2λ + (1 − |a|2 ) = 0
By the quadratic formula, the solutions are 1 ± |a|, which are both positive if and only if |a| < 1.
4. Let V be a complex inner product space, and T an operator on V satisfying hT v, vi = hv, T vi. Prove
that all the eigenvalues of T are real.
Solution: Suppose λ ∈ C is an eigenvalue of T . Then there is a nonzero vector v ∈ V such that
T v = λv. Therefore we have
hλv, vi = hv, λvi.
Using homogeneity in the first slot and conjugate homogeneity in the second, we get
λhv, vi = λhv, vi
Since v is a nonzero vector, hv, vi is a nonzero complex number (by definiteness), so we may cancel
hv, vi to obtain λ = λ, which means λ is real.
2 Equivalently,
we may divide the inequality by y 2 and then relabel x/y as x.
2
5. (Arveson Final Exam, ’96) Let A be a 3 × 3 matrix satisfying A = −AT (such a matrix is called
skew-symmetric). Show that for all x ∈ R3 and c ∈ R,
kx + cAxk ≥ kxk,
where k · k is the Euclidean norm on R3 .
Solution: First we claim that for any 3 × 3 matrix A and any vector x ∈ R3 ,
hx, Axi = hAT x, xi.
You should check this for yourself.3 Then since our A is skew-symmetric, we have
hx, Axi = hAT x, xi = −hAx, xi = −hx, Axi,
where we used the skew-symmetry of A in the second equality, and the symmetric property of the
inner product in the third. This means the number hx, Axi is equal to its negative, so it must be zero.
Therefore for any x, x and Ax are orthogonal. Thus for any c, x and cAx are orthogonal, also. Then
by the Pythagorean theorem,
kx + cAxk2 = kxk2 + kcAxk2 ≥ kxk2 ,
and taking square roots proves the result.
3 The
easiest way to see this is to notice that for the Euclidean inner product, hx, yi = xT y.
3