One has . Taking the square root of both sides gives the result.
Equality holds precisely when one is a nonnegative multiple of the other. This is a consequence of the analogous assertion of the next problem.
The first assertion is the triangle inequality. I claim that equality holds precisely when one vector is a non-positive multiple of the other.
for some real
, then substituting shows that the inequality
is equivalent to and clearly equality holds if a is non-positive. Similarly, one has equality if for some real .
Conversely, if equality holds, then , and so . By Theorem 1-1 (2), it follows that and are linearly dependent. If for some real , then substituting back into the equality shows that must be non-positive or must be 0. The case where is treated similarly.
If , then the inequality to be proved is just which is just the triangle inequality. On the other hand, if , then the result follows from the first case by swapping the roles of and .
The inequality follows from Theorem 1-1(3):
Geometrically, if , , and are the vertices of a triangle, then the inequality says that the length of a side is no larger than the sum of the lengths of the other two sides.
Theorem 1-1(2) implies the inequality of Riemann sums:
Taking the limit as the mesh approaches 0, one gets the desired inequality.
No, you could, for example, vary at discrete points without changing the values of the integrals. If and are continuous, then the assertion is true.
In fact, suppose that for each , there is an with . Then the inequality holds true in an open neighborhood of since and are continuous. So since the integrand is always non-negative and is positive on some subinterval of . Expanding out gives for all . Since the quadratic has no solutions, it must be that its discriminant is negative.
Let , , and for all in for . Then part (a) gives the inequality of Theorem 1-1 (2). Note, however, that the equality condition does not follow from (a).
If is inner product preserving, then one has by Theorem 1-1 (4):
Similarly, if is norm preserving, then the polarization identity together with the linearity of T give: .
Let be norm preserving. Then implies , i.e. the kernel of is trivial. So T is 1-1. Since is a 1-1 linear map of a finite dimensional vector space into itself, it follows that is also onto. In particular, has an inverse. Further, given , there is a with , and so , since is norm preserving. Thus is norm preserving, and hence also inner product preserving.
Assume is norm preserving. By Problem 1-7, is inner product preserving. So .
The assertion is false. For example, if , , , , and , then . Now, , but showing that T is not angle preserving.
To correct the situation, add the condition that the be pairwise orthogonal, i.e. for all . Using bilinearity, this means that: because all the cross terms are zero.
Suppose all the are equal in absolute value. Then one has because all the are equal and cancel out. So, this condition suffices to make be angle preserving.
Now suppose that for some and and that . Then
since . So, this condition suffices to make not be angle preserving.
The angle preserving are precisely those which can be expressed in the form where U is angle preserving of the kind in part (b), V is norm preserving, and the operation is functional composition.
Clearly, any of this form is angle preserving as the composition of two angle preserving linear transformations is angle preserving. For the converse, suppose that is angle preserving. Let be an orthogonal basis of . Define to be the linear transformation such that for each . Since the are pairwise orthogonal and is angle preserving, the are also pairwise orthogonal. In particular, because the cross terms all cancel out. This proves that is norm preserving. Now define to be the linear transformation . Then clearly and is angle preserving because it is the composition of two angle preserving maps. Further, maps each to a scalar multiple of itself; so is a map of the type in part (b). This completes the characterization.
The transformation is 1-1 by Cramer's Rule because the determinant of its matrix is 1. Further, is norm preserving since
by the Pythagorean Theorem. By Problem 8(a), it follows that is angle preserving.
If , then one has . Further, since is norm preserving, . By the definition of angle, it follows that .
Let be the maximum of the absolute values of the entries in the matrix of and . One has . (Correction: Bound should be square root of n times mN.)
One needs to verify the trivial results that (a) is a linear tranformation and (b) . These follow from bilinearity; the proofs are omitted. Together these imply that is a linear transformation.
Since for , has no non-zero vectors in its kernel and so is 1-1. Since the dual space has dimension n, it follows that is also onto. This proves the last assertion.
By bilinearity of the inner product, one has for perpendicular and :