数学代写|MATH2922 Modern Algebra

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这是悉尼大学现代代数课程的代写成功案例。

MATH2922课程简介

Linear and abstract algebra is one of the cornerstones of mathematics and it is at the heart of many applications of mathematics and statistics in the sciences and engineering. This unit is an advanced version of MATH2022, with more emphasis on the underlying concepts and on mathematical rigour. This unit investigates and explores properties of vector spaces, matrices and linear transformations, developing general principles relating to the solution sets of homogeneous and inhomogeneous linear equations, including differential equations. Linear independence is introduced as a way of understanding and solving linear systems of arbitrary dimension. Linear operators on real spaces are investigated, paying particular attention to the geometrical significance of eigenvalues and eigenvectors, extending ideas from first year linear algebra. To better understand symmetry, matrix and permutation groups are introduced and used to motivate the study of abstract groups theory. The unit culminates in studying inner spaces, quadratic forms and normal forms of matrices together with their applications to problems both in mathematics and in the sciences and engineering.

Prerequisites 

At the completion of this unit, you should be able to:

• LO1. appreciate the basic concepts and problems of linear algebra and be able to apply linear algebra to solve problems in mathematics, science and engineering
• LO2. understand the definitions of fields and vector spaces and be able to perform calculations in real and complex vector spaces, both algebraically and geometrically
• LO3. determine if a system of equations is consistent and find its general solution
• LO4. compute the rank of a matrix and understand how the rank of a matrix relates to the solution set of a corresponding system of linear equations
• LO5. compute the eigenvalues, eigenvectors, minimal polynomials and normal forms for linear transformations
• LO6. use the definition and properties of linear transformations and matrices of linear transformations and change of basis, including kernel, range and isomorphism
• LO7. compute inner products and determine orthogonality on vector spaces, including Gram-Schmidt orthogonalisation
• LO8. identify self-adjoint transformations and apply the spectral theorem and orthogonal decomposition of inner product spaces, and the Jordan canonical form, to solving systems of ordinary differential equations
• LO9. calculate the exponential of a matrix and use it to solve a linear system of ordinary differential equations with constant coefficients
• LO10. identify special properties of a matrix, such as symmetric of Hermitian, positive definite, etc., and use this information to facilitate the calculation of matrix characteristics
• LO11. demonstrate accurate and efficient use of advanced algebraic techniques and the capacity for mathematical reasoning through analysing, proving and explaining concepts from advanced algebra
• LO12. apply problem-solving using advanced algebraic techniques applied to diverse situations in physics, engineering and other mathematical contexts

MATH2922 Modern Algebra HELP（EXAM HELP， ONLINE TUTOR）

Step 1: $999=9 \cdot 101+90$. Step 2: $101=1 \cdot 90+11$. Step 3: $90=8 \cdot 11+2$. Step 4: $11=5 \cdot 2+1$. We now back-substitute: $1=11-5 \cdot 2=11-5 \cdot(90-8 \cdot 11)=41 \cdot 11-5 \cdot 90=$ $41 \cdot(101-1 \cdot 90)-5 \cdot 90=41 \cdot 101-46 \cdot 90=41 \cdot 101-46 \cdot(999-9 \cdot 101)=455 \cdot 101-46 \cdot 999$. Taking both sides $\bmod 999$ gives us $[455] \odot[101]=[1]$.

We use the generalized Euclidean algorithm (or trial and error) to discover the reciprocal of 24 modulo 35 , which is 19 . Multiplying, we get $19 \cdot 24 x \equiv 19 \cdot 10$, or $x \equiv 190 \equiv 15$ $(\bmod 35)$. Hence we get the single equivalence class $[15]$, modulo 35 .

We use our wonderful theorem with $a=5$, and conclude that this linear modular equation is equivalent to $5 x \equiv 2(\bmod 7)$. We now use the generalized Euclidean algorithm (or trial and error) to discover the reciprocal of 5 modulo 7 , which is 3 . Multiplying, we get $3 \cdot 5 x \equiv 3 \cdot 2$, or $x \equiv 6(\bmod 7)$. Hence there is a single solution modulo 7 , but the problem is about mod 35 . There are five equivalence classes modulo 35 solving the equation, namely $[6],[13],[20],[27],[34]$.

We will prove that there are no solutions, by contradiction. A solution would have $35 \mid(25 x-11)$, which would give some $k \in \mathbb{Z}$ with $35 k=25 x-11$. Rearranging, we get $5(-7 k+5 x)=11$. This would give us $5 \mid 11$, which we know is impossible.

Suppose that $a \in R$ is a unit and a zero divisor. Then $a \neq 0$, and there are nonzero $b, c \in R$ with $1=a b$ and $0=a c$. We now have $c=c \cdot 1=c(a b)=(c a) b=0 b=0$. This is a contradiction, as $c$ is nonzero.

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