11 1f A, B are two n x n symmetric matrices, then AB is an n x n symm.pdf

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11 1f A, B are two n x n symmetric matrices, then AB is an n x n symmetric matrix. 12 The union V U W of any two linear sub-space V, W of R^n is not a linear space. 13 If two square matrices A and B have the same characteristic polynomials, then trA = tr(B). 14 For any orthogonal matrix A, the matrix (A - A^-1) is skew symmetric. 15 For any 2 x 2 orthogonal matrix A, the linear transformation T : R^2 R^2 defined by vector x A vector x is a rotation. Solution 11) False AB is a symmentric matrix when A and B are symmetric iff A and B commute 12) True say V is a points on x-axis and w is a points of y-axis then their sum is not in the union hence it is not closed under addition hence not a subspace 13) False for 2- square matrix characterstic polynomial is given by t^2 + tr(A) + det(A) so, determinent can be different 14) True 15) True.

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Section 18.3-19.1. Today we will discuss finite-dimensional associative algebras and their representations. Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element a ∈ A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if all of its elements are. Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con- versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is A. Example. The algebra of all strictly lower- (or upper-) triangular n×n matrices is nilpotent. Proposition 3. If the algebra A is nilpotent, then An = 0 for some n ∈ Z+, that is the product of any n elements if the algebra A equals 0. Proof. Let B ⊂ A be the maximal subspace for which there exists n ∈ Z+ such that Bn = 0. Note that B is closed under multiplication, i.e. B ⊃ Bk for any k ∈ Z+. Assume B 6= A and choose an element a ∈ A\B. Since aBn = 0, there exists k ∈ Z+ such that aBk 6⊂ B. Replacing a with a non-zero element in aBk we obtain aB ⊂ B. Recall that there exists m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we have Cmn = 0 which contradicts the definition of the subspace B. � Unless A is commutative, the set of all nilpotent elements of A does not have to be an ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A, then so is I + J = {x + y |x ∈ I,y ∈ J} . Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal of A. Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A). The algebra A is semisimple if rad(A) = 0. If char(F) = 0, there exists an alternative description of semisimple algebras. Consider the regular representation of the algebra A ρ: A → L(A), ρ(a)(b) = ab, and define a “scalar product” on A via (a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)). One can see that (·, ·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc). Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥ is defined as I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} . Proposition 6. For any ideal I ⊂ A, its orthogonal complement I⊥ ⊂ A is also an ideal. Proof. For any a ∈ A, x ∈ I⊥, and y ∈ I we have (xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0. � 1 2 Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a ∈ A orthogonal to all of its powers is nilpotent. Proof. Let a ∈ A be such that (a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+. Let L ⊃ F be the splitting field of the characteristic polynomial f(x) of the operator ρ(a). Then, over L we have f(x) = tk0 s∏ i=1 (t−λi)ki, where λi are distinct for i = 1, . . . ,s, and tr ρ(a)n+1 = s∑ i=1 kiλ n+1 i = 0. Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous linear equations in variables k1, . . . ,ks. The determinant of this system equals λ21 . . .λ 2 nV (λ1, . . . ,λn), where V (λ1.

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RELATIONS A relation associates an element of one set with one or more elements of another set. If ''a'' is an element from set A which associates another element ''b'' from set B, then the elements can be written in an ordered pairs as (a,b)Thus we can define a relation as a set of ordered pairs.Some relations are denoted by letter R; in set notation a relation can be written asR = {(a, b): a is an element of the first set, b is an element of the second set} Example of a relation 1. 1. Mwajuma is a wife of Juma. 2. 2. Amina is a sister of Joyce. 3. 3. y = 2x + 3 4. Juma is tall, Anna is short. (Not a relation) NOTE If the relation R defines the set of all ordered pairs (x,y) such that . y = 2x + 3 this can be written symbolically as R = {(x, y): y=2x +3} PICTORIAL REPRESENTATION OF RELATIONS Relation can be represented pictorially; i) Arrow diagram. ii) Cartesian graph.

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2. The Mars Climate Orbiter (formerly the Mars Surveyor 98 Orbiter) was one of two spacecraft in the Mars Surveyor 98 program, the other being the Mars Polar Lander (formerly the Mars Surveyor 98 Lander). The two missions were to study the Martian weather, climate, and water and carbon dioxide badget in order to understand the reservoirs, behavior, and atmospheric role of volatiles and to warch for evidence of long-term and episodic climate changes. The Mars Climate Orbiter was destroyed when a navigation error caused the spacecraft to miss its intended 140-150 km altitude above Mars during orbit insertion, instead entering the Martian atmosphere at about 57 km. The spacecraft was destroyed by atmospherie stresses and friction at this low altitude. A review board found that some data was calculated on the ground in Imperial units (pound-seconds) and eported that way to the navigation team, who were expeeting the data in metric units (newton-seconds). Calculate the conversion between Newton-seconds and pound- seconds. (Hint: pound is the force of 1 slug being accelerated at 1 foot/see2, and 1 slug is 14.59 kg.) s. (Hint: 1 pound is the force of 1 slug being accelerated at 1 foot/see, and 1 slug Solution 1 newtons = 0.224808942443 pounds hence 1 newton -second = 0.224808942443 pounds -seconds.

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2. Identifying time lags Advocates of active policy face several obstacles when implementing discretionary fiscal or monetary policy, such as identifying the economy?s potential output level, the natural rate of unemployment, and the effects of a passive approach. Policymakers must also cope with a multitude of time lags that complicate the implementation of active policy. Identify the type of time lag Illustrated in the following scenario: Policymakers have implemented an expansionary monetary policy, but it takes several months for the change. In the federal funds rate to translate into a change in aggregate demand and output. Decision-making lag Implementation lag Recognition lag Effectiveness lag Solution Hi, The correct answer is effectiveness lag Explanation: After all of that, and the policy is finally put into place, it takes time for policy to hit the economy and take effect. For monetary policy if can be a year to a year and a half before the peak effect of the policy is felt. This is called effectiveness lag.

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