/*! 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 19 Prove that if \(\left\\{A_{1}, A... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 19

Prove that if \(\left\\{A_{1}, A_{2}, \ldots, A_{k}\right\\}\) is a linearly independent subset of \(\mathrm{M}_{n \times n}(F)\), then $\left\\{A_{1}^{t}, A_{2}^{t}, \ldots, A_{k}^{t}\right\\}$ is also linearly independent.

Short Answer

Expert verified
Assume a linear combination of the transposed matrices equals the zero matrix, i.e., \(\sum_{i=1}^{k} c_i A_i^t = 0\). Transpose both sides, yielding \((\sum_{i=1}^{k} c_i A_i^t)^t = 0^t \implies (A_1c_1 + A_2c_2 + \cdots + A_kc_k)^t = 0^t\). Since \(\{A_1, A_2, \ldots, A_k\}\) is linearly independent, \(c_1 = c_2 = \cdots = c_k = 0\). Thus, \(\{A_1^t, A_2^t, \ldots, A_k^t\}\) is also linearly independent.

Step by step solution

01

Definition of linear independence

A set of vectors (or matrices) is linearly independent if a linear combination of the vectors (matrices) equals the zero vector (matrix) only when all the coefficients in the combination are zero. In our case, we want to show that if \[ \sum_{i=1}^{k} c_i A_i^t = 0, \] then \(c_1 = c_2 = \cdots = c_k = 0\).
02

Finding a relation between linear combinations of \(A_i\) and \(A_i^t\)

To find a relationship between linear combinations of the matrices \(A_i\) and their transposes \(A_i^t\), we first write the linear combination of the transposed matrices as \[ \sum_{i=1}^{k} c_i A_i^t = 0 \implies (A_1^t)c_1 + (A_2^t)c_2 + \cdots + (A_k^t)c_k = 0. \] Now, let's transpose both sides of this equation to see what we get: \[ ((A_1^t)c_1 + (A_2^t)c_2 + \cdots + (A_k^t)c_k)^t = 0^t. \] Recall that the transpose of the sum of matrices is the sum of their transposes and the transpose of a scalar multiple of a matrix is the scalar multiple of the transpose of the matrix. This gives us \[ (A_1c_1 + A_2c_2 + \cdots + A_kc_k)^t = 0^t. \]
03

Applying the linear independence of the set \(\{A_1, A_2, \ldots, A_k\}\)

We know that \(\{A_1, A_2, \ldots, A_k\}\) is a linearly independent subset of \(M_{n\times n}(F)\). So, the linear combination of these matrices is equal to the zero matrix only when all the coefficients are zero: \[ A_1c_1 + A_2c_2 + \cdots + A_kc_k = 0^t \implies c_1 = c_2 = \cdots = c_k = 0. \]
04

Concluding linear independence of \(\{A_1^t, A_2^t, \ldots, A_k^t\}\)

We have shown that if a linear combination of the transpose matrices is equal to the zero matrix, then all of the coefficients in that linear combination must be zero. This means that the set \(\{A_1^t, A_2^t, \ldots, A_k^t\}\) is also linearly independent. In conclusion, if \(\{A_1, A_2, \ldots, A_k\}\) is a linearly independent subset of \(M_{n\times n}(F)\), then \(\{A_1^t, A_2^t, \ldots, A_k^t\}\) is also linearly independent.

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!

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

Prove that the norm in \(\mathbf{C}^{n}\) satisfies the following laws: \(\left[\mathrm{N}_{1}\right]\) For any vector \(u,\|u\| \geq 0 ;\) and \(\|u\|=0\) if and only if \(u=0\) \(\left[\mathrm{N}_{2}\right]\) For any vector \(u\) and complex number \(z,\|z u\|=| z\|u\|\) \(\left[\mathrm{N}_{3}\right]\) For any vectors \(u\) and \(v,\|u+v\| \leq\|u\|+\|v\|\)

Given \(u=[2,1,3], v=[4,-2,2], w=[1,1,5],\) find: (a) \(u \times v\) (b) \(u \times w\) (c) \( v \times w\)

Prove that the diagonals of a parallelogram bisect each other.

Let \(V\) denote the set of ordered pairs of real numbers. If $\left(a_{1}, a_{2}\right)\( and \)\left(b_{1}, b_{2}\right)\( are elements of \)\mathrm{V}$ and \(c \in R\), define $$ \left(a_{1}, a_{2}\right)+\left(b_{1}, b_{2}\right)=\left(a_{1}+b_{1}, a_{2} b_{2}\right) \quad \text { and } \quad c\left(a_{1}, a_{2}\right)=\left(c a_{1}, a_{2}\right) . $$ Is \(\mathrm{V}\) a vector space over \(R\) with these operations? Justify your answer.

Let \(S\) be a nonempty set and \(F\) a field. Let \(\mathcal{C}(S, F)\) denote the set of all functions \(f \in \mathcal{F}(S, F)\) such that \(f(s)=0\) for all but a finite number of elements of \(S\). Prove that \(\mathcal{C}(S, F)\) is a subspace of \(\mathcal{F}(S, F)\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.