# 线性代数代写Linear Algebra|MATH1703 University of Plymouth Assignment

0

Assignment-daixieTM为您提供普利茅斯大学University of Plymouth MATH 1703 Linear Algebra线性代数代写代考辅导服务！

## Instructions:

Vectors and matrices are indeed fundamental concepts in mathematics and have wide-ranging applications in various fields, including statistics, physics, data science, and engineering. A vector is a quantity that has both magnitude and direction, while a matrix is a rectangular array of numbers or symbols arranged in rows and columns.

In linear algebra, the study of vectors and matrices is critical. Linear algebra deals with the algebraic properties of linear equations, linear mappings, and their representations in vector spaces and through matrices. It provides a powerful framework for modeling, analyzing, and solving problems that arise in many fields, including physics, engineering, economics, and computer science.

Vector spaces are mathematical structures that abstract the essential properties of vectors. They are defined as sets of vectors that satisfy certain axioms, including closure under addition and scalar multiplication. Linear transformations are mappings between vector spaces that preserve the structure of the vector space, and they are represented by matrices.

Analytic geometry is the branch of mathematics that deals with the study of geometry using algebraic techniques. The connection between vectors, matrices, and analytic geometry is fundamental, as vectors can be used to represent points and directions in space, and matrices can be used to transform and project them onto different coordinate systems.

In summary, the practical skills of handling vectors and matrices are essential in many applications, and their deep connections with linear spaces and analytic geometry make them a powerful tool in the study of mathematics and its applications.

A group of matrices includes $A B$ and $A^{-1}$ if it includes $A$ and $B$. “Products and inverses stay in the group.” Which of these sets are groups?
Lower triangular matrices $L$ with 1 ‘s on the diagonal, symmetric matrices $S$, positive matrices $M$, diagonal invertible matrices $D$, permutation matrices $P$, matrices with $Q^{\mathrm{T}}=Q^{-1}$. Invent two more matrix groups.

Yes, the lower triangular matrices $L$ with 1’s on the diagonal form a group. Clearly, the product of two is a third. Further, the Gauss-Jordan method shows that the inverse of one is another.

No, the symmetric matrices do not form a group. For example, here are two symmetric matrices $A$ and $B$ whose product $A B$ is not symmetric.
$$A=\left[\begin{array}{lll} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{array}\right], \quad B=\left[\begin{array}{lll} 1 & 2 & 3 \ 2 & 4 & 5 \ 3 & 5 & 6 \end{array}\right], \quad A B=\left[\begin{array}{lll} 2 & 4 & 5 \ 1 & 2 & 3 \ 3 & 5 & 6 \end{array}\right]$$
No, the positive matrices do not form a group. For example, $\left(\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right)$ is positive, but its inverse $\left(\begin{array}{rr}1 & -1 \ 0 & 1\end{array}\right)$ is not.
Yes, clearly, the diagonal invertible matrices form a group.
Yes, clearly, the permutation matrices matrices form a group.
Yes, the matrices with $Q^{\mathrm{T}}=Q^{-1}$ form a group. Indeed, if $A$ and $B$ are two such matrices, then so are $A B$ and $A^{-1}$, as
$$(A B)^{\mathrm{T}}=B^{\mathrm{T}} A^{\mathrm{T}}=B^{-1} A^{-1}=(A B)^{-1} \quad \text { and }\left(A^{-1}\right)^{\mathrm{T}}=\left(A^{\mathrm{T}}\right)^{-1}=A^{-1} .$$
There are many more matrix groups. For example, given two, the block matrices $\left(\begin{array}{cc}A & 0 \ 0 & B\end{array}\right)$ form a third as $A$ ranges over the first group and $B$ ranges over the second. Another example is the set of all products $c P$ where $c$ is a nonzero scalar and $P$ is a permutation matrix of given size.

Suppose $\mathbf{S}$ and $\mathbf{T}$ are two subspaces of a vector space $\mathbf{V}$.
(a) Definition: The sum $\mathbf{S}+\mathbf{T}$ contains all sums $\mathbf{s}+\mathbf{t}$ of a vector $\mathbf{s}$ in $\mathbf{S}$ and a vector $\mathbf{t}$ in T. Show that $\mathbf{S}+\mathbf{T}$ satisfies the requirements (addition and scalar multiplication) for a vector space.
(b) If $\mathbf{S}$ and $\mathbf{T}$ are lines in $\mathbf{R}^m$, what is the difference between $\mathbf{S}+\mathbf{T}$ and $\mathbf{S} \cup \mathbf{T}$ ? That union contains all vectors from $\mathbf{S}$ and $\mathbf{T}$ or both. Explain this statement: The span of $\mathbf{S} \cup \mathbf{T}$ is $\mathbf{S}+\mathbf{T}$. (Section 3.5 returns to this word “span.”)

(a) Let $\mathbf{s}, \mathbf{s}^{\prime}$ be vectors in $\mathbf{S}$, let $\mathbf{t}, \mathbf{t}^{\prime}$ be vectors in $\mathbf{T}$, and let $c$ be a scalar. Then
$$(\mathbf{s}+\mathbf{t})+\left(\mathbf{s}^{\prime}+\mathbf{t}^{\prime}\right)=\left(\mathbf{s}+\mathbf{s}^{\prime}\right)+\left(\mathbf{t}+\mathbf{t}^{\prime}\right) \quad \text { and } \quad c(\mathbf{s}+\mathbf{t})=c \mathbf{s}+c \mathbf{t}$$
Thus $\mathbf{S}+\mathbf{T}$ is closed under addition and scalar multiplication; in other words, it satisfies the two requirements for a vector space.
(b) If $\mathbf{S}$ and $\mathbf{T}$ are distinct lines, then $\mathbf{S}+\mathbf{T}$ is a plane, whereas $\mathbf{S} \cup \mathbf{T}$ is not even closed under addition. The span of $\mathbf{S} \cup \mathbf{T}$ is the set of all combinations of vectors in this union. In particular, it contains all sums $\mathbf{s}+\mathbf{t}$ of a vector $\mathbf{s}$ in $\mathbf{S}$ and a vector $\mathbf{t}$ in $\mathbf{T}$, and these sums form $\mathbf{S}+\mathbf{T}$. On the other hand, $\mathbf{S}+\mathbf{T}$ contains both $\mathbf{S}$ and $\mathbf{T}$; so it contains $\mathbf{S} \cup \mathbf{T}$. Further, $\mathbf{S}+\mathbf{T}$ is a vector space. So it contains all combinations of vectors in itself; in particular, it contains the span of $\mathbf{S} \cup \mathbf{T}$. Thus the span of $\mathbf{S} \cup \mathbf{T}$ is $\mathbf{S}+\mathbf{T}$.

Section 3.1. Problem 32: Show that the matrices $A$ and $[A A B]$ (with extra columns) have the same column space. But find a square matrix with $\mathbf{C}\left(A^2\right)$ smaller than $\mathbf{C}(A)$. Important point:
An $n$ by $n$ matrix has $\mathbf{C}(A)=\mathbf{R}^n$ exactly when $A$ is an____ matrix.

Each column of $A B$ is a combination of the columns of $A$ (the combining coefficients are the entries in the corresponding column of $B$ ). So any combination of the columns of $[A A B]$ is a combination of the columns of $A$ alone. Thus $A$ and $[A A B]$ have the same column space.
Let $A=\left(\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right)$. Then $A^2=0$, so $\mathbf{C}\left(A^2\right)=\mathbf{Z}$. But $\mathbf{C}(A)$ is the line through $\left(\begin{array}{l}1 \ 0\end{array}\right)$.
An $n$ by $n$ matrix has $\mathbf{C}(A)=\mathbf{R}^n$ exactly when $A$ is an invertible matrix, because $A x=b$ is solvable for any given $b$ exactly when $A$ is invertible.