Problems in Linear Algebra (1978), a book of solved problems
archive.org
archive.org
Some books also worth looking at in mathematics:
(1) Anti-Demidovich series
(2) Combinatorial Analysis, Ribnikov
(3) Problem series by Suprun
(4) Probability Theory, Zolotarievskaya
(5) Discrete Mathematics Problems, Evnin
(6) Theory of Surfaces, Finikov
... and several other problem book series for university mathematics.
Engineering (mechanics):
(1) Anti-Mescherski series
(2) Theoretical Mechanics, V. M. Starzhinski
(3) Strength of Materials, Stiopin
(4) Problems in Strength of Materials, Volmir
[0] https://urss.ru/cgi-bin/db.pl?lang=sp&blang=en&page=Bookstor... (apparently the headquarters are in Peru now)
I remember distinctly learning (as a teenager) Schaum's Outline Series of Vector Calculus teaching me quickly and in detail "grad div curl and all that" very effectively, so much so that I got basically 100% on all of my A-level maths and further maths papers because, well, if you can do a big book of problems like that, then exams really are just more of the same. At the time, I absolutely loved them -- big brown books, some of which are currently by my ankles.
I'm not sure it actually helped that much beyond early university though. Exams select for a very specific skill. What isn't taught is why something was discovered, or is useful, or how an idea came to be. Many more advanced ideas in mathematics are downright bizarre and it's the basic idea that you need to be able to come up with something similar to, not necessarily the detail.
On the other hand, there is an extremely rigorous book with all the proofs for differential geometry, by Kennington. It has 2400 pages. For tensors I would guess that I need more Algebra, specifically, modules. And to review smoothness and differentiability classes.
For those interested: topology.org
I went over that book fully. From cover to cover. It was absolutely fantastic, and as good as a "problems book" can be.
I loved it and would recommend it to anyone who is looking to get their feet wet with problems in Vector Calculus.
Although it is neither a rigorous treatment (look into Arfken, Weber, Harris for that), nor it is greatly healpful pedagogically (look into Mary L. Boas for that), nor is it easy (look into Riley, Hobson, Bence for that).
The greatest intro material into Linear Algebra still remains the one by Gilbert Strang.
I got perfect scores in both of the Mathematical Physics papers in college.
Many problems, with some variations did appear directly from Spiegel.
Best VC problems book ever.
Rybnikov
Zolotarevskaya
Stepin
Let me also add a recommendation for the calculus problems book by Günther (Gyunter) and Kuzmin (I don’t think it was ever translated, but how much translation do you really need in one?). It will not train you—every idea occurs once or maybe twice; it is not a book of exercises. (The joke goes that Demidovich is G&K with every problem repeated ten times.) But it is a book of problems, and it will teach you.
https://begriffs.com/posts/2016-07-24-best-linear-algebra-bo...
Non-solution: You are essentially being asked to show that given an invertible complex-symmetric matrix, you can find a complex-symmetric square root. You might be tempted to use the Taylor series of sqrt(z) centred at some z_0 to obtain K, but this won't work if M's eigenvalues can't be fit inside a half-plane not include the complex number 0. An example of this is when M is
⎡1 0 ⎤
⎢ ⎥
⎣0 -1⎦
Another non-solution: You might attempt to extend the spectral theorem to complex-symmetric matrices, but the only such extension I know of is complicated, and results in a block-diagonal (but not necessarily diagonal) canonical form.Solution: To actually extend the sqrt function from the complex numbers to matrices, use the Jordan canonical form of M. Then you only need to take the sqrt's of each Jordan block. Note that this sqrt won't exist if a Jordan block is non-zero and nilpotent, so we require M to be non-singular. This obtains K. Finally, observe that there exists a polynomial p such that p(M) = K (which can be obtained via Hermite interpolation), and therefore K^T = p(M)^T = p(M^T) = p(M) = K.
Since M is invertible and symmetric. It acts on R^N by independently scaling the coordinates of an orthogonal basis. Eg, M = O^T*A*O where O is orthogonal and A is a diagonal matrix. Let sqrt(A) be the diagonal matrix with entries equal to the square root of the corresponding entry in A. Then K = O^T*sqrt(A)*O is symmetric and K^2 = (O^T*sqrt(A)*O)*(O^T*sqrt(A)*O). Since O is orthogonal O^T*O = I, and K^2 = O^T*A*O = M.
This statement here:
It acts on R^N by independently scaling the coordinates of an orthogonal basis
is precisely the spectral theorem, which you can't use here.[edit]
Also, the corresponding claim for Hermitian (and real symmetric) matrices isn't true either. If an eigenvalue of a Hermitian matrix H is negative, then no square root of it can be Hermitian (or real symmetric).
Analog Stack Overflow?