/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Let \(A, B\) be \(n \times n\) m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 20

Let \(A, B\) be \(n \times n\) matrices over a field \(K\), all of whose components are cqual to 0 except possibly those of the first column. Assume \(A \neq 0\). Show that there exists an \(n \times n\) matrix \(C\) over \(K\) such that \(C A=B\). Hint: Consider first a special case where $$ A=\left(\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & 0 \end{array}\right) $$

Short Answer

Expert verified
The short answer is that given two matrices, $A$ and $B$, with only their first columns possibly having nonzero components, we can define a matrix $C$ such that $CA=B$ as follows: $$ C=\left(\begin{array}{cccc} k a_1 & k a_2 & \cdots & k a_n \\\ 0 & 0 & \cdots & 0 \\\ \vdots & \vdots & & \vdots \\\ 0 & 0 & \cdots & 0 \end{array}\right) $$ where $k=\frac{b_i}{a_i}$ for some nonzero component $a_i$ of $A$.

Step by step solution

01

Special Case Solution for A

For the special case, we have A as given in the hint: $$ A=\left(\begin{array}{cccc} 1 & 0 & \cdots & 0 \\\ 0 & 0 & \cdots & 0 \\\ \vdots & \vdots & & \vdots \\\ 0 & 0 & \cdots & 0 \end{array}\right) $$ Let B have a first column represented by the vector \(b = [b_1, b_2, \ldots, b_n]^T\). We want to find a matrix C such that CA = B. Since A only has a 1 in the first row of the first column, it is clear that the first row of C must be the first column of B for the matrix multiplication to result in matrix B. In other words, the first row of C should contain the components in reverse order: \([b_1, b_2, \ldots, b_n]\). For the other rows of C, they can simply be filled with zeroes, as their multiplication with A will have no effect on the first column of B. Thus, we have a specific matrix C that satisfies CA = B: $$ C=\left(\begin{array}{cccc} b_1 & b_2 & \cdots & b_n \\\ 0 & 0 & \cdots & 0 \\\ \vdots & \vdots & & \vdots \\\ 0 & 0 & \cdots & 0 \end{array}\right) $$
02

Generalization to Arbitrary A and B

Let A have its first column represented by the vector \(a = [a_1, a_2, \ldots, a_n]^T\) and assume that \(a \neq 0\). Let B have its first column represented by the vector \(b = [b_1, b_2, \ldots, b_n]^T\). Now we want to find a matrix C such that CA = B. The first row of C should be found by multiplying the first column of A by a scalar, k, to obtain the first column of B: \(k a = b\). To find \(k\), we can use any nonzero component of a. Assume that \(a_i \neq 0\), we find the scalar as \(k = \frac{b_i}{a_i}\). Now, define the first row of C as \([k a_1, k a_2, \ldots, k a_n ]\). For the other rows of C, they can be filled with zeroes as in the special case. Thus, we have a general matrix C that satisfies CA = B: $$ C=\left(\begin{array}{cccc} k a_1 & k a_2 & \cdots & k a_n \\\ 0 & 0 & \cdots & 0 \\\ \vdots & \vdots & & \vdots \\\ 0 & 0 & \cdots & 0 \end{array}\right) $$ This completes our proof of the existence of the matrix C such that CA = B for arbitrary matrices A and B with only their first columns possibly having nonzero components.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrices Over a Field
In linear algebra, the concept of a field is fundamental, as it provides a structured environment where scalar multiplication and addition are well defined. A field is a set accompanied by two operations, commonly addition and multiplication, satisfying certain axioms such as commutativity, associativity, and the existence of an identity element.

A matrix over a field is composed of elements that all belong to the same field. This matters because the properties of the field ensure that operations such as matrix addition, subtraction, and multiplication behave predictably. This is crucial when solving linear equations or transforming spaces. Consider field elements as building blocks that conform to specific rules, leading to reliable solutions in mathematical scenarios.
Nonzero Matrix
A nonzero matrix, as the name suggests, is a matrix that contains at least one entry that is not zero. It's important to distinguish between a zero and a nonzero matrix in linear algebra, as their properties and effects on operations like multiplication differ greatly.

Even one nonzero element in a matrix can contribute significantly to the outcome of a multiplication. As seen in the exercise, a matrix with a single nonzero element in the first column can still be used to transform another matrix. This property is a testament to the power each entry in a matrix holds and the influence it may have on matrix operations.
Scalar Multiplication
Scalar multiplication refers to multiplying a matrix by a single number, called a scalar. In this process, every entry in the matrix is multiplied by the same scalar. This is a central operation in linear algebra, used for resizing or rotating geometric figures or adjusting the weighting of certain variables in vector spaces.

Scalar multiplication is also a key step in multiplying matrices where one matrix with a scalar non-zero entry can dictate the outcome of the product. In the context of our exercise, the scalar multiplication helps derive a matrix C that, when multiplied by A, produces B, highlighting its crucial role in matrix equations.
Proof in Linear Algebra
Proofs are the backbone of mathematics, serving as the means by which mathematicians establish the truth of a statement. In linear algebra, proofs often involve showing the existence of a particular matrix that fulfills a set of conditions, like the exercise given.

The step-by-step approach in the solution showcases how to build a proof: start with a specific case, deduce a formula, then generalize the result. The proof process in our exercise begins with a special case matrix and strategically scales up to a wider circumstance, demonstrating a technique widely used in linear algebra to establish the reliability of various operations and properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(V\) be a vector space over the field \(K\), and let \(A: V \rightarrow V\) be an endomorphism. Assume that \(A^{\prime}=I\) for some integer \(r \geq 1\). Let $$ T=I+A+\cdots+A^{\prime-1} $$ Let \(t_{0}\) be an element of \(V\). Show that the space generated by \(T_{\mathrm{r}}\) is an invariant subspace of \(A\), and that \(T_{v_{0}}\) is an eigenvector of \(A\). If \(T v_{0} \neq 0\), what is the eigenvalue?

Let \(V\) be a finite dimensional vector space over the field \(K\). Let \(R\) be the ring of \(K\) -linear maps of \(V\) into itself. Show that \(R\) has no two- sided ideals except \(\\{O\\}\) and \(R\) itself. [Hint: Let \(A \in R, A \neq O\). Let \(v_{1} \in V, v_{t} \neq 0\), and \(A v_{1} \neq 0\). Complete \(v_{1}\) to a basis \(\left\\{v_{1+\ldots, v_{n}}\right\\}\) of \(V\). Let \(\left\\{w_{1}, \ldots, w_{n}\right\\}\) be arbitrary elcments of \(V .\) For each \(i=1, \ldots, n\) there exists \(B_{i} \in R\) such that $$ B_{1} v_{i}=v_{1} \quad \text { and } \quad B_{i} v_{j}=0 \text { if } j \neq i_{,} $$ and there exists \(C_{i} \in R\) such that \(C_{1} A v_{1}=w_{i}\) (justify these two existence statements in detail). Let \(F=C_{1} A B_{1}+\cdots+C_{n} A B_{s^{+}}\) Show that \(F\left(v_{j}\right)=w_{i}\) for all \(i=1, \ldots, n .\) Conclude that the two-sided ideal generated by \(A\) is the whole ring \(R .]\)

A sequence of homomorphisms of abelian groups $$ A \stackrel{f}{\rightarrow} B \stackrel{a}{\rightarrow} C $$ is said to be exact if \(\operatorname{Im} f=\) Ker \(g\). Thus to say that \(0 \rightarrow A \stackrel{J}{\rightarrow} B\) is exact means that \(f\) is injective. Let \(R\) be a ring. If $$ 0 \rightarrow E \stackrel{I}{\rightarrow} E \stackrel{0}{\rightarrow} E^{v} $$ is an exact sequence of \(R\) -modules, show that for every \(R\) -module \(F\) $$ 0 \rightarrow \mathrm{Hom}_{R}\left(F, E^{\prime}\right) \rightarrow \mathrm{Hom}_{R}(F, E) \rightarrow \mathrm{Hom}_{R}\left(F, E^{\prime}\right) $$ is an exact sequence.

Let \(K\) be a field and \(R=K[X]\) the polynomial ring over \(K\). Let \(f(X)\) be a polynomial of degree \(d>0\) in \(K[X]\). Let \(J\) be the ideal generated by \(f(X)\). What is the dimension of \(R / J\) over \(K\) ? Exhibit a basis of \(R / J\) over \(K\). Show that \(R / J\) is an integral ring if and only if \(f\) is irreducible.

Let \(R\) be a ring. We define the center \(Z\) of \(R\) to be the subset of all elements \(=\in R\) such that \(z x=x=\) for all \(x \in R\) (a) Show that the center of a ring \(R\) is a subring (b) Let \(R=\mathrm{Mat}_{n \times=}(K)\) be the ring of \(n \times n\) matrices over the field \(K\). Show that the center is the set of scalar matrices \(c l\), with \(c \in K\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.