Recent content by Bhatia

  1. B

    Proving Jordan Canonical Form for Similarity of Matrices with Same Polynomials

    I have to prove the following result: Let A,B be two n×n matrices over the field F and A,B have the same characteristic and minimal polynomials. If no eigenvalue has algebraic multiplicity greater than 3, then A and B are similar. I have to use the following result: If A,B are...
  2. B

    Proving Cyclic Decompositions: Let T be a Linear Operator on V

    Thanks for your reply. Given a linear operator T on a vector space V then a subspace W is T- admissible if i) W is invariant under T (ii) if f(T) β belongs to W, there exists a vector η in W such that f(T)β = f(T) η.
  3. B

    Proving Cyclic Decompositions: Let T be a Linear Operator on V

    Let T be a linear operator on the the finite dimensional space V, and let R be the range of T. (a) Prove that R has a complementary T-invariant subspace iff R is independent of the null space N of T. (b) If R and N are independent, prove that, N is the unique T-invariant subspace...
  4. B

    Connection between Monotone and One-to-one Functions

    Thanks for the insight. I get what you mean now.
  5. B

    Connection between Monotone and One-to-one Functions

    Thanks for your help with an example. I appreciate it. Thanks for yesterday's reply too. Regards, Bhatia
  6. B

    Connection between Monotone and One-to-one Functions

    Hello, Micromass Thanks for your reply. If f is injective and continuous then it is strictly monotone...that is clear My question is : If f is montone then is f continuous ?
  7. B

    Connection between Monotone and One-to-one Functions

    If f is bijective and monotone with both its range and domain being a closed and bounded interval...then what can we say about the continuity of f ?
  8. B

    Congruence: Solving Let p be a Prime Number

    Thank you so much... I agree with this...I realized my mistake x should be strictly less than p. Taking p to be odd (as suggested) p= 5 , then d =1, 2, 3, 4. Does 2 ^ (p-1)/d imply the following we get: for d=1, 2^ 4 = 16 works for d=1 that is , 16= 1 (mod 5). for d=2, 2 ^ 2= 4 works for...
  9. B

    Congruence: Solving Let p be a Prime Number

    Thanks for your reply :smile: Going by your idea Let p= 5, then since d / p-1 then I guess we have d= 1, 2, 4...then solving x^ d = 1 (mod 5) for d=1, x=1 for d=2, x= 4, 6,9,11,14, ... for d=4, x= 2, 3,4,6, 7 ... So I am thinking atleast d... But not sure...I did not study...
  10. B

    Congruence: Solving Let p be a Prime Number

    Let p be a prime number and d / p-1 . Then which of the following statements about the congruence? x ^d = 1( mod p) is / are correct : 1. It does not have a solution 2 atmost d incongurent solutions 3 exactly d incongruent solutions 4 aleast d incongruent solutions.
Back
Top