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Chapter 3: Problem 18

Let \(A\) be a \(3 \times 3\) matrix and let \(\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) be vectors in \(\mathbb{R}^{3}\). Show that if the vectors \\[ \mathbf{y}_{1}=A \mathbf{x}_{1}, \quad \mathbf{y}_{2}=A \mathbf{x}_{2}, \quad \mathbf{y}_{3}=A \mathbf{x}_{3} \\] are linearly independent, then the matrix \(A\) must be nonsingular and the vectors \(\mathbf{x}_{1}, \mathbf{x}_{2},\) and \(\mathbf{x}_{3}\) must be linearly independent.

Short Answer

Expert verified
Since 饾懄鈧�, 饾懄鈧�, 饾懄鈧� are linearly independent, we can derive that 伪, 尾, 纬 must all be zero in the equation: \(伪\mathbf{y}_{1} + 尾\mathbf{y}_{2} + 纬\mathbf{y}_{3} = \mathbf{0}\). Substituting the given information about the relationship between 饾懄 and 饾懃 using matrix A, we deduce that matrix A must be nonsingular (i.e., it has an inverse), and the original vectors 饾懃鈧�, 饾懃鈧�, and 饾懃鈧� must be linearly independent.

Step by step solution

01

Write down the linear independence definition

Recall the definition of linear independence: A set of vectors is linearly independent if the only way to create a linear combination (using scalar coefficients) that equals the zero vector is by having all coefficients equal to zero. In other words, if 饾懄鈧�, 饾懄鈧�, 饾懄鈧� are linearly independent, then for any scalars 伪, 尾, 纬: \(伪\mathbf{y}_{1} + 尾\mathbf{y}_{2} + 纬\mathbf{y}_{3} = \mathbf{0}\) implies \(伪 = 尾 = 纬 = 0.\)
02

Use given information about 饾懄鈧�, 饾懄鈧�, 饾懄鈧� and substitute using A

We know that: \(\mathbf{y}_{1} = A\mathbf{x}_{1}\), \(\mathbf{y}_{2} = A\mathbf{x}_{2}\), \(\mathbf{y}_{3} = A\mathbf{x}_{3}\) Substitute these in the linear combination equation from Step 1: \( 伪A\mathbf{x}_{1} + 尾A\mathbf{x}_{2} + 纬A\mathbf{x}_{3} = \mathbf{0}\)
03

Factor out A from the linear combination

Since A is a matrix and 伪, 尾, 纬 are scalars, we can factor out A from the linear combination: \(A(伪\mathbf{x}_{1} + 尾\mathbf{x}_{2} + 纬\mathbf{x}_{3}) = \mathbf{0}\)
04

Derive the proof for nonsingular matrix A and linear independence of 饾懃鈧�, 饾懃鈧�, 饾懃鈧�

We want to show that A is nonsingular, which means it has an inverse. Suppose A is singular, which means it does not have an inverse. If this is the case, then there exists a nonzero vector 饾懅 such that: \(A\mathbf{z} = \mathbf{0}\) Now, let 饾懅 be the vector: \(饾懅 = 伪\mathbf{x}_{1} + 尾\mathbf{x}_{2} + 纬\mathbf{x}_{3}\) We have: \(A\mathbf{z} = A(伪\mathbf{x}_{1} + 尾\mathbf{x}_{2} + 纬\mathbf{x}_{3}) = \mathbf{0}\) This implies that 伪, 尾, 纬 cannot all be zero (since 饾懅 is nonzero), which contradicts our assumption that 饾懄鈧�, 饾懄鈧�, 饾懄鈧� are linearly independent, as we derived in Step 1. Therefore, A must be nonsingular, and there exists an inverse of A. Since A is nonsingular, we can multiply both sides of the equation by \(A^{-1}\) and get: \((A^{-1}A)(伪\mathbf{x}_{1} + 尾\mathbf{x}_{2} + 纬\mathbf{x}_{3}) = A^{-1}\mathbf{0}\) Since \(A^{-1}A = I\) and \(A^{-1}\mathbf{0} = \mathbf{0}\), we get: \(伪\mathbf{x}_{1} + 尾\mathbf{x}_{2} + 纬\mathbf{x}_{3} = \mathbf{0}\) This implies 伪, 尾, 纬 must all be zero, as 饾懄鈧�, 饾懄鈧�, 饾懄鈧� are linearly independent. So, 饾懃鈧�, 饾懃鈧�, and 饾懃鈧� are also linearly independent. In conclusion, if the vectors 饾懄鈧�, 饾懄鈧�, 饾懄鈧� (transformed using matrix A) are linearly independent, then matrix A must be nonsingular, and the original vectors 饾懃鈧�, 饾懃鈧�, 饾懃鈧� must be linearly independent.

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Key Concepts

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

Nonsingular Matrix
A nonsingular matrix, also known as an invertible or non-degenerate matrix, is one that possesses an inverse. In simpler terms, this means the matrix can be "undone" or reversed, allowing us to retrieve the original data used to create it. A crucial property of a nonsingular matrix is that its determinant is non-zero.

When a matrix is applied to a set of vectors, and these transformed vectors remain linearly independent, as in the case from our exercise, the matrix must be nonsingular. This is because only with a nonsingular matrix could we be certain that different vectors do not end up as the same or overlapping outcomes after the transformation.

The concept of a nonsingular matrix is critical in many areas of linear algebra, including solving systems of linear equations and in determining whether a matrix transformation preserves vector space dimensions.
Inverse of a Matrix
The inverse of a matrix is akin to a mirror image in that applying a matrix and then its inverse yields the original space unaffected. For a square matrix, the inverse is defined only if the matrix is nonsingular, meaning it has full rank and a non-zero determinant.

The inverse of a matrix, denoted by \(A^{-1}\) for a matrix \(A\), satisfies the condition \(A \cdot A^{-1} = I\), where \(I\) is the identity matrix. The identity matrix acts like the number 1 in matrix multiplication, leaving vectors unchanged.

In applications, finding a matrix's inverse allows us to solve linear equations of the form \(A\mathbf{x} = \,\mathbf{b}\) by rewriting it as \(\mathbf{x} = A^{-1}\mathbf{b}\). This reversibility is the essence behind matrix arithmetic operations and supports critical operations in transformations and systems analysis.
Linear Combination
A linear combination involves creating a new vector by adding together scaled versions of some initial vectors. The scaling factors are scalar values, and the process allows us to explore the "space" these vectors span.

For instance, if given vectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), we can form a linear combination by \(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3\), where \(c_1, c_2, \) and \(c_3\) are scalars.

In the exercise's context, proving a set of vectors like \(\mathbf{y}_1, \mathbf{y}_2, \) and \(\mathbf{y}_3\) as linearly independent via linear combinations meant showing the equation \(\alpha \mathbf{y}_1 + \beta \mathbf{y}_2 + \gamma \mathbf{y}_3 = \mathbf{0}\) holds exclusively if all scalars are zero. This property of linear combinations is fundamental in determining the span and dimensionality of vector spaces.
Matrix Transformation
Matrix transformations provide a framework for changing or translating vectors within a space. These transformations can encompass rotations, reflections, scaling, or even more complex linear transformations within linear algebra.

Using a transformation matrix \(A\), a vector \(\mathbf{x}\) can be transformed into \(\mathbf{y}\) using the equation \(\mathbf{y} = A\mathbf{x}\). This operation affects how the vector \(\mathbf{x}\) is represented within the new framework dictated by \(A\).

In linear independence scenarios, if a matrix transformation keeps resulting vectors \(\mathbf{y}_1, \mathbf{y}_2, \mathbf{y}_3\) linearly independent, it's indicative of a matrix's nonsingular nature, as was shown in the given exercise. Matrix transformations play critical roles in graphics, where they change object orientations and scales, and in data science for dimensionality reduction.

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Most popular questions from this chapter

Let \(\mathbf{v}_{1},\) and \(\mathbf{v}_{2}\) be two vectors in a vector space \(V\) Show that \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly dependent if and only if one of the vectors is a scalar multiple of the other.

Let \(A\) be a \(5 \times 3\) matrix of rank 3 and let \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right\\}\) be a basis for \(\mathbb{R}^{3}\) (a) Show that \(N(A)=\\{0\\}\) (b) Show that if \(\mathbf{y}_{1}=A \mathbf{x}_{1}, \mathbf{y}_{2}=A \mathbf{x}_{2},\) and \(\mathbf{y}_{3}=A \mathbf{x}_{3}\) then \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) are linearly independent. (c) Do the vectors \(\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\) from part \((\mathbf{b})\) form a basis for \(\mathbb{R}^{5}\) ? Explain.

Let \(A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{n \times r},\) and \(C=A B .\) Show that (a) if \(A\) and \(B\) both have linearly independent column vectors, then the column vectors of \(C\) will also be linearly independent. (b) if \(A\) and \(B\) both have linearly independent row vectors, then the row vectors of \(C\) will also be linearly independent. [Hint: Apply part (a) to \(\left.C^{T}\right]\).

An \(m \times n\) matrix \(A\) is said to have a right inverse if there exists an \(n \times m\) matrix \(C\) such that \(A C=I_{m}\) The matrix \(A\) is said to have a left inverse if there exists an \(n \times m\) matrix \(D\) such that \(D A=I_{n}\) (a) Show that if \(A\) has a right inverse, then the column vectors of \(A\) span \(\mathbb{R}^{m}\) (b) Is it possible for an \(m \times n\) matrix to have a right inverse if \(n

Let \(Z\) denote the set of all integers with addition defined in the usual way and define scalar multiplication, denoted \(\circ,\) by \\[ \alpha \circ k=[[\alpha]] \cdot k \quad \text { for all } \quad k \in Z \\] where \([[\alpha]]\) denotes the greatest integer less than or equal to \(\alpha .\) For example, \\[ 2.25 \circ 4=[[2.25]] \cdot 4=2 \cdot 4=8 \\] Show that \(Z\), together with these operations, is not a vector space. Which axioms fail to hold?
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