8 ms·

Problems in Linear Algebra (1978), a book of solved problems

160 points by happy-go-lucky 4y ago

isaacfrond 4y ago
Boy, you've got a lot of problems.
- iqkznnft 4y ago
  99, to be precise.
vippy 4y ago
I've got 1938 problems but proving that the subspace L of the unitary space R_n has an orthogonal basis consisting of the eigenvectors of the transformation φ ain't one.
WoahNoun 4y ago
Skimming through there are some interesting problems (the fibonacci sequence generated by determinants of nxn matrices with -1/1 on the lower subdiagonal/upper subdiagonal is pretty cool). But a lot of these problems just look like super tedious calculations.
- jedimastert 4y ago
  > a lot of these problems just look like super tedious calculations Analog Stack Overflow?
  - xdavidliu 4y ago
    stackoverflow is a site for programming questions. Where does one see tedious mathematical calculations on there?
    - burnished 4y ago
      The math related stackoverflows
    - WoahNoun 4y ago
      Mathoverflow is run by StackExchange. https://mathoverflow.net/ https://mathoverflow.net/
- mbustamanter 4y ago
  That's the general style of the book, but further along the way it has problems that require proving statements by referring to the properties or definitions of the actual structures involved (vector spaces).
ogogmad 4y ago
Exercise: Let M be a square complex matrix satisfying M = M^T, and furthermore assume that M is not singular. Prove that there exists a matrix K such that K^2 = M and K = K^T. 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.
- WoahNoun 4y ago
  There's a geometric solution as well! 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.
  - ogogmad 4y ago
    That doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. 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).
mbustamanter 4y ago
I have the version of this book in Spanish (Editorial MIR), the translated title is 2000 Problems in Linear Algebra. This publishing house (MIR) doesn't exist anymore but now goes under the name of URSS [0] and publishes modern versions of these same books, at least in Spanish and Russian. It has several books on problem solving, which was probably a pathway to demonstrating true "elite" status in eastern europe at the time (hence the known romanian, hungarian and russian way of teaching). It also contains several books on theory too, be it mathematics, physics or engineering. The important thing with textbooks on theory is, unlike the custom of modern textbooks to have few problems and often disconnected or that require to make a few "leaps" in the sense that the exposition does not clearly pave a way to solve harder problems, the amount of exercises is large and they frequently build on top of one another, thus making learning much more enjoyable in my opinion. 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=Bookstore https://urss.ru/cgi-bin/db.pl?lang=sp&blang=en&page=Bookstor... (apparently the headquarters are in Peru now)
- azalemeth 4y ago
  At a glance, probably Western equivalent of these are Schaum's Outline Series, which start by telling you something, giving you some worked examples on it, and then setting you lose on them. It's a great way to learn, quickly, particularly for a student. 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.
stiff 4y ago
Paul Halmos wrote a book I like a lot titled "Linear Algebra Problem Book" that teaches you linear algebra through a sequence of problems - its different from the one posted here in that the problems are often preceded by 2-3 pages of discussion/motivation and almost all of them are conceptual and interesting.
begriffs 4y ago
I put together a list of the best linear algebra books, classified by type. Proskuryakov's practice book is one of them. Here's the full list: https://begriffs.com/posts/2016-07-24-best-linear-algebra-books.html https://begriffs.com/posts/2016-07-24-best-linear-algebra-bo...

Boy, you've got a lot of problems.

iqkznnft 4y ago
99, to be precise.

I've got 1938 problems but proving that the subspace L of the unitary space R_n has an orthogonal basis consisting of the eigenvectors of the transformation φ ain't one.

WoahNoun 4y ago

Skimming through there are some interesting problems (the fibonacci sequence generated by determinants of nxn matrices with -1/1 on the lower subdiagonal/upper subdiagonal is pretty cool). But a lot of these problems just look like super tedious calculations.

jedimastert 4y ago
> a lot of these problems just look like super tedious calculations Analog Stack Overflow?
- xdavidliu 4y ago
  stackoverflow is a site for programming questions. Where does one see tedious mathematical calculations on there?
  - burnished 4y ago
    The math related stackoverflows
  - WoahNoun 4y ago
    Mathoverflow is run by StackExchange. https://mathoverflow.net/ https://mathoverflow.net/
mbustamanter 4y ago
That's the general style of the book, but further along the way it has problems that require proving statements by referring to the properties or definitions of the actual structures involved (vector spaces).

ogogmad 4y ago

Exercise: Let M be a square complex matrix satisfying M = M^T, and furthermore assume that M is not singular. Prove that there exists a matrix K such that K^2 = M and K = K^T. 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.

WoahNoun 4y ago
There's a geometric solution as well! 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.
- ogogmad 4y ago
  That doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. 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).

mbustamanter 4y ago

I have the version of this book in Spanish (Editorial MIR), the translated title is 2000 Problems in Linear Algebra. This publishing house (MIR) doesn't exist anymore but now goes under the name of URSS [0] and publishes modern versions of these same books, at least in Spanish and Russian. It has several books on problem solving, which was probably a pathway to demonstrating true "elite" status in eastern europe at the time (hence the known romanian, hungarian and russian way of teaching). It also contains several books on theory too, be it mathematics, physics or engineering. The important thing with textbooks on theory is, unlike the custom of modern textbooks to have few problems and often disconnected or that require to make a few "leaps" in the sense that the exposition does not clearly pave a way to solve harder problems, the amount of exercises is large and they frequently build on top of one another, thus making learning much more enjoyable in my opinion. 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=Bookstore https://urss.ru/cgi-bin/db.pl?lang=sp&blang=en&page=Bookstor... (apparently the headquarters are in Peru now)

azalemeth 4y ago
At a glance, probably Western equivalent of these are Schaum's Outline Series, which start by telling you something, giving you some worked examples on it, and then setting you lose on them. It's a great way to learn, quickly, particularly for a student. 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.

stiff 4y ago

Paul Halmos wrote a book I like a lot titled "Linear Algebra Problem Book" that teaches you linear algebra through a sequence of problems - its different from the one posted here in that the problems are often preceded by 2-3 pages of discussion/motivation and almost all of them are conceptual and interesting.

begriffs 4y ago

I put together a list of the best linear algebra books, classified by type. Proskuryakov's practice book is one of them. Here's the full list: https://begriffs.com/posts/2016-07-24-best-linear-algebra-books.html https://begriffs.com/posts/2016-07-24-best-linear-algebra-bo...