A I AI =
... Property 4: if A and B are similar, then det( A) det( B) Prove. Since A and B are similar, there is a nonsingular matrix S , such that A S 1BS det( A) det( S 1BS ) det( S 1 ) det( B) det( S ) det( B) since det(S 1 ) ...
... Property 4: if A and B are similar, then det( A) det( B) Prove. Since A and B are similar, there is a nonsingular matrix S , such that A S 1BS det( A) det( S 1BS ) det( S 1 ) det( B) det( S ) det( B) since det(S 1 ) ...
Linear Algebra Application~ Markov Chains
... rather that it is in a state of dynamic equilibrium such that the net movement into and out of a given state is zero. ...
... rather that it is in a state of dynamic equilibrium such that the net movement into and out of a given state is zero. ...
hw2
... 2. (1.2.11) Let V = {0} consist of a single vector 0 and define 0 + 0 = 0 and c0 = 0 for all scalars c in F . Prove that V is an F -vector space. 3. (1.2.12) A real-valued function f defined on the real line is called an even function if f (−t) = f (t) for each real number t. Prove that the set of e ...
... 2. (1.2.11) Let V = {0} consist of a single vector 0 and define 0 + 0 = 0 and c0 = 0 for all scalars c in F . Prove that V is an F -vector space. 3. (1.2.12) A real-valued function f defined on the real line is called an even function if f (−t) = f (t) for each real number t. Prove that the set of e ...
Solving simultaneous equations
... Mathematica has a built in Norm function (which takes lots of types of argument), but in this case gives the standard norm: Norm@v3D ...
... Mathematica has a built in Norm function (which takes lots of types of argument), but in this case gives the standard norm: Norm@v3D ...
ECON3120/4120 Mathematics 2, autumn 2005 Problem
... (For practical reasons some of the solutions may include problem parts that are not on the problem list for this seminar.) EMEA, 15.7.3 (= LA, 2.1.5) Using the definitions of vector addition and multiplication of a vector by a real number, we get 3(x, y, z) + 5(−1, 2, 3) = (4, 1, 3) ⇐⇒ (3x − 5, 3y + ...
... (For practical reasons some of the solutions may include problem parts that are not on the problem list for this seminar.) EMEA, 15.7.3 (= LA, 2.1.5) Using the definitions of vector addition and multiplication of a vector by a real number, we get 3(x, y, z) + 5(−1, 2, 3) = (4, 1, 3) ⇐⇒ (3x − 5, 3y + ...
PDF
... However, 1 = λa λb is not an eigenvalue of −I = AB; also, −2 = λa + λb is not an eigenvalue of O = A + B. (b) Show that λ ∈ R is an eigenvalue of AB iff λ ∈ R is an eigenvalue of BA. [Hint: Homework 8, Problem 1(a).] Solution. Suppose λ 6= 0. Then by Homework 8, Problem 1(a), λI − AB = λ[I − (λ−1 A) ...
... However, 1 = λa λb is not an eigenvalue of −I = AB; also, −2 = λa + λb is not an eigenvalue of O = A + B. (b) Show that λ ∈ R is an eigenvalue of AB iff λ ∈ R is an eigenvalue of BA. [Hint: Homework 8, Problem 1(a).] Solution. Suppose λ 6= 0. Then by Homework 8, Problem 1(a), λI − AB = λ[I − (λ−1 A) ...
LINEAR ALGEBRA (1) True or False? (No explanation required
... Explanations: matrices like ( 10 00 ) or ( 11 11 ) are nonzero but do not have an inverse. Matrices have an inverse if and only if they are nonsingular square matrices. If A and B are nonsingular, then so is AB, and its inverse clearly is B −1 A−1 since B −1 A−1 AB = B −1 IB = B −1 B = I. In general ...
... Explanations: matrices like ( 10 00 ) or ( 11 11 ) are nonzero but do not have an inverse. Matrices have an inverse if and only if they are nonsingular square matrices. If A and B are nonsingular, then so is AB, and its inverse clearly is B −1 A−1 since B −1 A−1 AB = B −1 IB = B −1 B = I. In general ...
1 Box Muller - NYU Courant
... It may seem odd that X and Y in (13) are independent given that they use the same R and Θ. Not only does our algebra shows that this is true, but we can test the independence computationally, and it will be confirmed. Part of this method was generating a point “at random” on the unit circle. We sugg ...
... It may seem odd that X and Y in (13) are independent given that they use the same R and Θ. Not only does our algebra shows that this is true, but we can test the independence computationally, and it will be confirmed. Part of this method was generating a point “at random” on the unit circle. We sugg ...