Question

By making suitable choices for the \(n\) -dimensional vectors a and b, show that if \(\tilde{\mathbf{a}} \mathbf{C b}=\tilde{\mathbf{a}} \mathbf{D} \mathbf{b}\) for any choices of a and \(\mathbf{b}\) (where \(\mathbf{C}\) and \(\mathbf{D}\) are \(n \times n\) matrices), then \(\mathbf{C}=\mathbf{D}\).

Short answer

By choosing standard basis vectors, the equality implies \(\mathbf{C} = \mathbf{D}\) by equating corresponding elements.

Step-by-step solution

Understand the Hypothesis

We are given that \(\tilde{\mathbf{a}} \mathbf{C} \mathbf{b} = \tilde{\mathbf{a}} \mathbf{D} \mathbf{b}\) for any vectors \(\mathbf{a}\) and \(\mathbf{b}\). Our goal is to show that this implies \(\mathbf{C} = \mathbf{D}\). Here, \(\tilde{\mathbf{a}}\) represents the transpose of vector \(\mathbf{a}\). The condition is valid for any choice of vectors \(\mathbf{a}\) and \(\mathbf{b}\).

Simplify with Zero Vectors

Let's choose \(\mathbf{a}\) and \(\mathbf{b}\) as zero vectors which gives: \(\tilde{\mathbf{0}} \mathbf{C} \mathbf{0} = \tilde{\mathbf{0}} \mathbf{D} \mathbf{0}\). This is trivially true because both sides equal zero. This choice doesn't help us conclude anything directly about \(\mathbf{C}\) and \(\mathbf{D}\).

Use Canonical Basis Vectors

Consider choosing \(\mathbf{a}\) and \(\mathbf{b}\) as canonical basis vectors, such as \(\mathbf{a} = \mathbf{e}_i\) and \(\mathbf{b} = \mathbf{e}_j\), where \(\mathbf{e}_i\) and \(\mathbf{e}_j\) are unit vectors with components \(1\) at the \(i\)-th and \(j\)-th positions, respectively, and zero elsewhere.

Using the Condition

Substituting \(\mathbf{a} = \mathbf{e}_i\) and \(\mathbf{b} = \mathbf{e}_j\) into the hypothesis, we have: \(\tilde{\mathbf{e}}_i \mathbf{C} \mathbf{e}_j = \tilde{\mathbf{e}}_i \mathbf{D} \mathbf{e}_j\). This simplifies to \(c_{ij} = d_{ij}\), where \(c_{ij}\) and \(d_{ij}\) are the \((i,j)\)-th entries of matrices \(\mathbf{C}\) and \(\mathbf{D}\), respectively.

Generalize for All Entries

Since the choice of \(\mathbf{a}\) and \(\mathbf{b}\) was arbitrary, \(c_{ij} = d_{ij}\) must hold for all \(i\) and \(j\). Thus, every corresponding element of \(\mathbf{C}\) and \(\mathbf{D}\) is equal. Therefore, \(\mathbf{C} = \mathbf{D}\).

Key concepts

Linear Algebra

Linear Algebra is a branch of mathematics that deals with vectors and matrices. It is the study of linear spaces, which are also known as vector spaces, and the linear transformations between them. This field has numerous applications, ranging from computer science to engineering and economics.

At its core, linear algebra helps us understand various transformations we can apply to vectors and matrices. Matrices, in particular, are a fundamental concept because they can represent and model complex data and systems. They're like grid-like structures composed of numbers arranged in rows and columns. The exercise presented involves proving matrix equality, which is a common task when dealing with linear systems.

• Vectors in linear algebra can represent anything from a point in space to the coefficients of an equation.
• Matrices are used to perform operations on these vectors, such as scaling, rotating, or transforming them.
• Matrix equality, as explored in the given problem, implies that two matrices are identical in every corresponding element.

In conclusion, the ability to equate matrices and prove such equalities is a powerful aspect of linear algebra, helping solve problems where complex systems and data sets are involved.

Matrix Manipulation

Matrix manipulation is all about performing operations on matrices. It includes actions like matrix addition, multiplication, and transposition. The problem in the exercise focuses on how manipulations can demonstrate that two matrices are equal.

One of the most common operations is matrix multiplication. This involves taking two matrices and producing a new matrix. It follows a specific set of rules whereby you take the dot product of rows and columns from the input matrices. This operation is key in the solution process, which uses canonical basis vectors to prove the equality.

• The transpose of a matrix is obtained by flipping it over its diagonal, which is crucial in the step-by-step solution where vector \( \tilde{\mathbf{a}} \) represents a transposed vector.
• Using the properties of matrix multiplication with basis vectors, the proof simplifies, showing matrix entries correspond directly.
• The manipulation shown in this problem is not arbitrary but rather leverages the inherent properties of matrices to simplify complex expressions into recognizable formats.

Through matrix manipulation, it becomes clear how every element of matrices \( \mathbf{C} \) and \( \mathbf{D} \) compare, highlighting their equality under the given conditions.

Basis Vectors

Basis vectors are the building blocks of vector spaces. They provide a minimal and sufficient set of vectors through which we can express every vector within that space. In the context of the exercise, basis vectors are used to prove matrix equality effectively.

Basis vectors are often represented as unit vectors along the axes of a space. They are special because they form a "basis" for creating other vectors in their space. In most cases, basis vectors are assumed to be orthogonal, meaning they are at right angles to each other.

• Canonical basis vectors like \( \mathbf{e}_i \) are defined such that only one element is 1 and all others are 0. This simplicity helps in probing matrices.
• By examining matrices through the lens of these vectors, individual entries such as \( c_{ij} \) and \( d_{ij} \) can be isolated, facilitating easy comparisons.
• Using basis vectors in this exercise demonstrated that all corresponding elements in matrices \( \mathbf{C} \) and \( \mathbf{D} \) were indeed equal.

By employing basis vectors, the exercise provided profound insights into how matrices function and interact, making them an indispensable tool in matrix equality and linear algebra.