I'm teaching myself Georgi E. Shilov's Linear Algebra, and have just finished chapter 6 except the final two starred sections. So far, I haven't seen Singular Value Decomposition (SVD). I skimmed the rest of the book, and looked it up in the Index, but had no luck. I'd like to quit studying the book prematurely if it turns out not to cover the subject. You information is valuable. Thank you!
Does Shilov’s Linear Algebra cover SVD
linear algebrasvd
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I haven't bought this book, but I have access to a copy of it. On a cursory read, I think it isn't too advanced and its choices of topics are about right. It covers slightly more topics than one shall see in a typical modern introductory course that focuses only on the theory of linear algebra.
Whether the book is suitable for beginners depends very much on the readers. There are books (like Rudin's Principles of Mathematical Analysis or Spivak's Calculus on Manifolds) that are praised by many people but also found too hard by equally many. To me, Shilov's exposition looks clear, efficient and rigourous. It is harder (but not to a large extent) than most modern introductory texts on linear algebra, but much easier to read than Hoffman and Kunze's classic, Linear Algebra.
Shilov introduces and discusses determinants in chapter 1. This is usually one of the most difficult topics in an introductory course. If you can get past chapter 1 without difficulty, you should be able to understand the whole book (sans the chapters marked by asterisks).
However, given that the book was published in the 1970s, those readers who are accustomed to the new breed of more "conversational" texts may dislike its style. They may also find it lacking motivations or practical applications.
If you are currently interested only in the theory (but not applications) of linear algebra, I think Berberian's Linear Algebra is another inexpensive and viable choice for private study. Jim Hefferon's Linear Algebra also worths a look. It does cover some applications and it is a favourite text of Darij Grinberg, a well respected user of this site. The electronic copy is free and the paper copy is cheap.
For (1), you have to associate a unique, different element of $K$ to each element of $K'$, and vice versa. More importantly, this has to be the same map below that respects the sum and products. You're not just saying that $K$ and $K'$ have the same cardinality, but that the particular map the author is talking about is one-to-one.
For (2), yes.
For (3), yes, because this map $f$ has $f(-x) + f(x) = f(x + (-x)) = f(0) = 0$. (For the last part, note that $f(0) = f(0 + 0) = f(0) + f(0)$.)
I'm not sure what you mean in (4), but the idea is that $f(xy) = f(x)f(y)$.
The text in the quote is unclear and unnecessarily verbose; this is an instance where symbols make things much clearer. What the author means is that an isomorphism of fields $K$ and $K'$ is a function $f:K \to K'$ with the following properties:
- $f$ is bijective: For any $x'\in K'$, there exists a unique $x\in K$ with $f(x) = x'$.
- $f(x + y) = f(x) + f(y$) for any $x, y\in K$;
- $f(xy) = f(x) f(y)$ for any $x, y\in K$.
There are some other properties, like $f(1) = 1$ and $f(x^{-1}) = f(x)^{-1}$ for nonzero $x$, that follow immediately from these properties; but that's the definition itself.
Best Answer
No, it doesn't. SVD had yet to become a textbook topic in the English speaking world when Shilov's book was first published (1971). I think it entered the textbook scene by virtues of Gilbert Strang's Linear Algebra and Its Applications and Golub and Van Loan's Matrix Computations, which were published at later times.