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From: "Kenneth A. Ribet" <kri...@gmail.com>
Date: Sun, 12 Oct 2008 07:50:46 -0700
Local: Sun, Oct 12 2008 10:50 am
Subject: Re: A Few Questions on Chapter 6
> On page 99, it says that "we want to think of ||z||^2 as the inner For complex n-space, the usual "dot product" formula for the inner > product of z with itself," but his formula for it looks like the inner > product of z with the conjugate of z. I wanted to know what I am not > seeing. product is modified: you through in a complex conjugation. This has the result that distance can be expressed in terms of the inner product, as you say. In many linear algebra books, F can really be any old field, and one ends up with a strange definition -- namely, you do one thing in all cases *except* the case where F=C, when you do something special. In our book, there are only two F's, namely R and C. The definition is to one one thing in case of R and another, only slightly different, thing in the case of C. This seems nicer because C represents 50% of all fields. > On page 104, I am having trouble understanding the proof of the He's using the principle that the square of the length of a.v (where a > Cauchy- > Schwarz Inequality where Axler looks at what ||u||^2 is equal to based > on the orthogonal decomposition. How does he get to the second > equality (this equality is right before 6.8)? is a scalar and v is a vector) is a^2 times the square of the length of v. The scalar "a" here is a fraction whose denominator happens to be the square of the length of v, so there is some cancellation. Note that the denominator of a^2 is the fourth power of the length of v, so you're still left with the square of the length of v in the denominator after cancellation. This proof is a tad different from the very common proof that exploits > On page 105, in the proof of the Triangle Inequality, how do I know The inner product (u,v) is some number a+bi. Its absolute value is > that 2|<u,v>| is greater than or equal to 2Re<u,v>? the square root of a^2 + b^2. He's saying that a is less than or equal to this square root. If a is negative, this is obvious because the square root is always 0 or a positive number. If a is positive or zero, then you say that the square root of a^2+b^2 is at least as big as the square root of a^2, which is a. Geometrically, we're confronting the statement that the hypotenuse of the right triangle with sides |a| and |b| is bigger than either of the other sides. This is kind of the triangle inequality for numbers, and it's being parlayed into the triangle inequality for vectors. Good work if you can get it. Thanks for your questions! Ken R You must Sign in before you can post messages.
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