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Chapter 6: Problem 22

Let \(V\) be an \(n\) -dimensional vector space with basis \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\} .\) Let \(P\) be an invertible \(n \times n\) matrix and set \\[ \mathbf{u}_{i}=p_{1 i} \mathbf{v}_{1}+\cdots+p_{n i} \mathbf{v}_{n} \\] for \(i=1, \ldots, n .\) Prove that \(\mathcal{C}=\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right\\}\) is a basis for \(V\) and show that \(P=P_{B+C}\)

Short Answer

Expert verified
\(\mathcal{C}\) is a basis since \(P\) is invertible; thus \(P=P_{B+C}\).

Step by step solution

01

Definition of Basis

To show that \(\mathcal{C}\) is a basis of \(V\), we first note that \(\mathcal{B}\) is a basis of \(V\) meaning \(\mathcal{B}\) spans \(V\) and is linearly independent. We need to show that \(\mathcal{C}\) also satisfies these two properties.
02

Expressing \(\mathbf{u}_i\) using Matrix Product

We observe that each vector \(\mathbf{u}_i\) can be represented as a linear combination of \(\mathbf{v}_1, \ldots, \mathbf{v}_n\) with coefficients from the matrix \(P\). This is equivalent to multiplying the matrix \(P\) by the vector of basis vectors.
03

Showing Linear Independence

Since \(P\) is invertible, the columns of \(P\) are linearly independent. Hence, the linear combinations \(\mathbf{u}_i = p_{1i}\mathbf{v}_1 + \cdots + p_{ni}\mathbf{v}_n\) are linearly independent because the determinant of \(P\) is non-zero.
04

Showing \(\mathcal{C}\) Spans \(V\)

Since \(P\) is an invertible matrix, any vector in \(V\) can be expressed as a linear combination of \(\mathbf{u}_1, \ldots, \mathbf{u}_n\) by using the transformed set of basis vectors. This confirms that \(\mathcal{C}\) spans \(V\).
05

Conclusion on Basis \(\mathcal{C}\)

With linear independence and span demonstrated, the set \(\mathcal{C}\) is a basis for \(V\).
06

Proving \(P = P_{B+C}\)

The matrix \(P_{B+C}\) is defined as the change of basis matrix from \(\mathcal{B}\) to \(\mathcal{C}\). Since \[\mathbf{u}_i = p_{1i}\mathbf{v}_1 + \cdots + p_{ni}\mathbf{v}_n\], the matrix \(P\) itself serves as the change of basis matrix showing \(P = P_{B+C}\).

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Key Concepts

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

Linear Independence
Linear independence is a fundamental concept in linear algebra that helps determine whether a set of vectors forms a basis for a vector space. Vectors are linearly independent if no vector in the set can be written as a linear combination of the others. This means they are not overlapping in direction.To better understand linear independence, consider the following points:
  • A set of vectors \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \) is linearly independent if the only solution to the equation \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_n\mathbf{v}_n = \mathbf{0} \) is when all coefficients \( c_1, c_2, \ldots, c_n \) are zero.
  • If a set of vectors is linearly independent, no vector in the set can be expressed as a linear combination of the others.
  • In the context of the invertible matrix \( P \), since the matrix is invertible, means that its columns (used as coefficients for vectors \( \mathbf{v}_i \)) are linearly independent.
Understanding linear independence is crucial for determining if a set of transformed vectors, such as \( \{ \mathbf{u}_1, \ldots, \mathbf{u}_n \} \), can serve as a basis for the vector space \( V \).
Vector Space
A vector space is one of the most essential structures in linear algebra. It is a collection of vectors, where you can perform vector addition and scalar multiplication while adhering to specific algebraic rules.Here are the important features of a vector space:
  • A vector space \( V \) over a field \( F \) must satisfy closure under addition and scalar multiplication. This means if you have vectors \( \mathbf{v} \) and \( \mathbf{w} \) in \( V \), then the sum \( \mathbf{v} + \mathbf{w} \) must also be in \( V \). Likewise, a scalar times a vector (\( c \mathbf{v} \)) is also in \( V \).
  • Vector spaces come with zero vector, a unique vector that, when added to any other vector in the space, yields the same vector.
  • Homogeneity and associativity rules apply, ensuring consistent operations within the space.
In the given exercise, \( V \) is described as an \( n \)-dimensional vector space, which simply means it is spanned by \( n \) linearly independent vectors. In other words, \( \mathcal{B} \) and \( \mathcal{C} \) both span the same vector space \( V \), illustrating the basis change idea.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, plays a crucial role in linear algebra. It is a square matrix that has an inverse, meaning there exists another matrix that can multiply with the original matrix to produce the identity matrix.Key features of an invertible matrix:
  • An invertible matrix \( P \) must be square, meaning it has the same number of rows and columns.
  • The determinant of \( P \) is non-zero. A non-zero determinant is essential because it implies that the matrix can be inverted.
  • If a matrix is invertible, the linear transformation it represents is bijective, meaning it is both injective (one-to-one) and surjective (onto).
  • Applying an invertible matrix to a set of vectors in a space transforms them while preserving linear independence.
In the exercise, the invertible matrix \( P \) guarantees that the new set of vectors \( \{ \mathbf{u}_1, \ldots, \mathbf{u}_n \} \) remains a basis for the vector space \( V \), thus both spanning the space and maintaining linear independence.
Linear Combinations
Linear combinations are expressions that take sums of scalar multiples of vectors to create new vectors. This concept is essential for understanding how vectors span a space and form bases.Let's explore some important aspects:
  • A linear combination of vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \) is formulated as \( a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \ldots + a_n\mathbf{v}_n \). The \( a_i \) are scalars (coefficients).
  • Every vector in a vector space can be expressed as a linear combination of the basis vectors. This property is crucial in the change of basis.
  • In the exercise, we see each \( \mathbf{u}_i \) defined as \( p_{1i}\mathbf{v}_1 + \ldots + p_{ni}\mathbf{v}_n \), depicting a change from the \( \mathcal{B} \)-basis to the \( \mathcal{C} \)-basis.
  • The matrix \( P \) serves as a transformation matrix, using its columns as coefficients to create new vectors (\( \mathbf{u}_i \)) through linear combination.
Understanding linear combinations is central to understanding how different sets of basis vectors are interrelated and how one can transition from one basis to another, as shown in this exercise.

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Most popular questions from this chapter

Let \(T: \mathscr{P}_{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation defined by \\[T\left(a+b x+c x^{2}\right)=\left[\begin{array}{l}a-b \\\b+c\end{array}\right]\\] (a) Which, if any, of the following polynomials are in \(\operatorname{ker}(T) ?\) (i) \(1+x\) (ii) \(x-x^{2}\) (iii) \(1+x-x^{2}\) (b) Which, if any, of the following vectors are in \(\operatorname{range}(T) ?\) (i) \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) (ii) \(\left[\begin{array}{l}1 \\ 0\end{array}\right]\) (iii) \(\left[\begin{array}{l}0 \\ 1\end{array}\right]\) (c) Describe ker( \(T\) ) and range(T).

Find the dimension of the vector space \(V\) and give a basis for \(V\). $$V=\left\\{p(x) \text { in } \mathscr{P}_{2}: x p^{\prime}(x)=p(x)\right\\}$$

Find either the nullity or the rank of \(T\) and then use the Rank Theorem to find the other. $$T: \mathscr{P}_{2} \rightarrow \mathbb{R} \text { defined by } T(p(x))=p^{\prime}(0)$$

Find a basis for \(\operatorname{span}(1,1+x, 2 x)\) in \(\mathscr{P}_{1}\).

Determine whether the linear transformation T is (a) one-to-one and (b) onto. $$\begin{array}{l}T: \mathscr{P}_{2} \rightarrow \mathbb{R}^{3} \text { defined by } \\\T\left(a+b x+c x^{2}\right)=\left[\begin{array}{c} 2 a-b \\\a+b-3 c \\\c-a\end{array}\right]\end{array}$$
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