Search results for query: *

  1. P

    Indexed Family fo Subsets

    It is amazing the difference a space will make. :biggrin: Thank you so much!...I was able to do all parts of the problem and even presented them in class!
  2. P

    Indexed Family fo Subsets

    Homework Statement Let f: A \rightarrow B be given and let {X_{\alpha}} for \alpha \in I be an indexed family of subsets of A. Prove: a) f(U_{\alpha\inI} X_{\alpha}) = U_{\alpha\inI}f(X_{\alpha}) The Attempt at a Solution To prove these two things are equal I must show...
  3. P

    Linear Algebra - Direct Sums

    [SOLVED] Linear Algebra - Direct Sums Homework Statement Let W1, W2, K1, K2,..., Kp, M1, M2,..., Mq be subspaces of a vector space V such that W1 = K1 \oplusK2\oplus ... \oplusKp and W2 = M1 \oplusM2 \oplus...\oplusMq Prove that if W1 \capW2 = {0}, then W1 + W2 = W1 \oplusW2 = K1...
  4. P

    Linear algebra determinant of linear operator

    [SOLVED] linear algebra determinant of linear operator Homework Statement Let T be a linear operator on a finite-dimensional vector space V. Define the determinant of T as: det(T)=det([T]β) where β is any ordered basis for V. Prove that for any scalar λ and any ordered basis β for V...
  5. P

    Prove A is Diagonalizable

    yeah, i know... i just submitted the real one...can I delete this blank one?
  6. P

    Prove A is Diagonalizable (Actual Question)

    [SOLVED] Prove A is Diagonalizable (Actual Question) Homework Statement Suppose that A \in M^{nxn}(F) and has two distinct eigenvalues, \lambda_{1} and \lambda_{2}, and that dim(E(subscript \lambda_{1} ))= n-1. Prove that A is diagonalizable. The Attempt at a Solution So far, I...
  7. P

    Prove A is Diagonalizable

    [b]1.
  8. P

    Linear Algebra Dual Basis

    [SOLVED] Linear Algebra Dual Basis Let V= R3 and define f1, f2, f3 in V* as follows: f1 = x -2y f2 = x + y +z f3 = y -3z part (a): prove that {f1, f2, f3} is a basis for V* I did this by using the gauss jordan method and showing that {f1, f2, f3} is linearly independent. Now...
Back
Top