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In Exercises 31–36, respond as comprehensively as possible, and justify your answer. If \(P\) is a \(5 \times 5\) matrix and Nul \(P\) is the zero subspace, what can you say about solutions of equations of the form \(P \mathbf{x}=\mathbf{b}\) for \(\mathbf{b}\) in \(\mathbb{R}^{5} ?\)

Short Answer

Expert verified
For any \(\mathbf{b}\) in \(\mathbb{R}^{5}\), the equation \(P \mathbf{x} = \mathbf{b}\) has a unique solution since \(P\) is invertible.

Step by step solution

01

Understand the Null Space

The null space of a matrix, Nul \(P\), is defined as the set of all vectors \(\mathbf{x}\) such that \(P \mathbf{x} = \mathbf{0}\). When it is given that Nul \(P\) is the zero subspace, it means the only solution to \(P \mathbf{x} = \mathbf{0}\) is the trivial solution, \(\mathbf{x} = \mathbf{0}\).
02

Recognize Implication of Zero Null Space on Rank

A zero null space implies that the matrix \(P\) is full rank. Since \(P\) is a \(5 \times 5\) matrix, it can have a maximum rank of 5. Thus, full rank here means rank 5, indicating \(P\) is invertible.
03

Investigate Solutions for Given Equation

An invertible matrix has the property that for any vector \(\mathbf{b}\) in \(\mathbb{R}^{5}\), the equation \(P \mathbf{x} = \mathbf{b}\) has exactly one solution. This is because the invertibility of \(P\) guarantees there is a unique inverse, \(P^{-1}\), such that \(\mathbf{x} = P^{-1}\mathbf{b}\).
04

Conclude on Solution Existence and Uniqueness

Since \(P\) is invertible, for every vector \(\mathbf{b}\) in \(\mathbb{R}^{5}\), the equation \(P \mathbf{x} = \mathbf{b}\) has a unique solution. Therefore, there exists exactly one \(\mathbf{x}\) such that \(P \mathbf{x} = \mathbf{b}\) for any given \(\mathbf{b}\).

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Key Concepts

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

Matrix Invertibility
Matrix invertibility is a key concept in linear algebra that reflects a matrix's ability to "reverse" its operations. When a matrix is invertible, it means there exists another matrix, called the inverse, which when multiplied by the original matrix results in the identity matrix. The identity matrix, denoted as \I\, has ones on the diagonal and zeros elsewhere. Mathematically, a matrix \P\ is invertible if there exists a matrix \P^{-1}\ such that \PP^{-1} = I\ and \P^{-1}P = I\.

To test if a matrix \( P \) is invertible, you can check its determinant. A non-zero determinant implies that a square matrix is invertible. For a \(5 \times 5\) matrix, having a non-zero determinant means it can perform unique transformations on vectors with no loss of dimensionality.
  • An invertible matrix allows for the system \( P \mathbf{x} = \mathbf{b} \) to have one and only one solution for any given vector \( \mathbf{b} \) in \( \mathbb{R}^5 \).
  • If a matrix is not invertible, some vectors \( \mathbf{b} \) will not have solutions, or multiple solutions may exist.
Null Space
The null space of a matrix \( P \) is the set of vectors \( \mathbf{x} \) that satisfy the equation \( P \mathbf{x} = \mathbf{0} \). In essence, it is where the transformation represented by \( P \) collapses inputs to the zero vector. The null space is a subspace, which means it contains the zero vector and is closed under vector addition and scalar multiplication.

When the null space is the zero subspace itself, i.e., only the zero vector satisfies \( P \mathbf{x} = \mathbf{0} \), it indicates significant properties about the matrix. Specifically:
  • The null space being trivial (only zero vector) implies that \( P \) has full column rank.
  • It also indicates the matrix \( P \) is injective (one-to-one), meaning different \( \mathbf{x} \)'s lead to different outputs \( P \mathbf{x} \).
This concept is crucial in confirming invertibility. If the null space is zero, \P\ can be inverted.
Rank of a Matrix
The rank of a matrix provides a powerful insight into its dimensional properties. Simply put, the rank is the dimension of the column space of the matrix. In other words, it indicates the maximum number of linearly independent column vectors in the matrix. For a \(5 \times 5\) matrix, the maximum rank is 5, which means all columns are linearly independent.

Having full rank is synonymous with being invertible for square matrices. If \( P \) is a \(5 \times 5\) matrix and has a rank of 5, it not only ensures invertibility, but also uniqueness of solutions to any linear system \( P \mathbf{x} = \mathbf{b} \).
  • Rank determines consistency of systems. Full rank implies that the equation \( P \mathbf{x} = \mathbf{b} \) is always consistent for any \( \mathbf{b} \).
  • When rank is less than the number of columns, the system may not have solutions, or may have infinitely many solutions depending on \( \mathbf{b} \).
A full-rank condition simplifies analysis and provides precise solutions in applications across physics, engineering, and computer graphics.

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Most popular questions from this chapter

Exercises \(9-12\) display a matrix \(A\) and an echelon form of \(A\) . Find bases for \(\operatorname{Col} A\) and \(\mathrm{Nul} A,\) and then state the dimensions of these subspaces. $$ A=\left[\begin{array}{rrrr}{1} & {-3} & {2} & {-4} \\ {-3} & {9} & {-1} & {5} \\\ {2} & {-6} & {4} & {-3} \\ {-4} & {12} & {2} & {7}\end{array}\right] \sim\left[\begin{array}{rrrr}{1} & {-3} & {2} & {-4} \\ {0} & {0} & {5} & {-7} \\ {0} & {0} & {0} & {5} \\ {0} & {0} & {0} & {0}\end{array}\right] $$

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations. [M] Use at least three pairs of random \(4 \times 4\) matrices \(A\) and \(B\) to test the equalities \((A+B)^{T}=A^{T}+B^{T}\) and \((A B)^{T}=\) \(A^{T} B^{T}\) . (See Exercise \(37 . )\) Report your conclusions. [Note: Most matrix programs use \(A^{\prime}\) for \(A^{T} . ]\)

IM Suppose an experiment leads to the following system of equations: \(4.5 x_{1}+3.1 x_{2}=19.249\) 1.6 \(x_{1}+1.1 x_{2}=6.843\) a. Solve system \((3),\) and then solve system \((4),\) below, in which the data on the right have been rounded to two decimal places. In each case, find the exact solution. \(4.5 x_{1}+3.1 x_{2}=19.25\) 1.6 \(x_{1}+1.1 x_{2}=6.84\) b. The entries in \((4)\) differ from those in \((3)\) by less than .05\(\% .\) Find the percentage error when using the solution of \((4)\) as an approximation for the solution of \((3) .\) c. Use your matrix program to produce the condition number of the coefficient matrix in \((3) .\)

In Exercises 17 and \(18,\) mark each statement True or False. Justify each answer. Here \(A\) is an \(m \times n\) matrix. a. If \(\mathcal{B}\) is a basis for a subspace \(H,\) then each vector in \(H\) can be written in only one way as a linear combination of the vectors in \(\mathcal{B}\) . bectors in \(\mathcal{B}\) . b. If \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is a basis for a subspace \(H\) of \(\mathbb{R}^{n},\) then the correspondence \(\mathbf{x} \mapsto[\mathbf{x}]_{\mathcal{B}}\) makes \(H\) look and act the same as \(\mathbb{R}^{p}\) . c. The dimension of Nul \(A\) is the number of variables in the equation \(A \mathbf{x}=\mathbf{0} .\) d. The dimension of the column space of \(A\) is rank \(A .\) e. If \(H\) is a \(p\) -dimensional subspace of \(\mathbb{R}^{n},\) then a linearly independent set of \(p\) vectors in \(H\) is a basis for \(H .\)

In Exercises 33 and \(34, T\) is a linear transformation from \(\mathbb{R}^{2}\) into \(\mathbb{R}^{2} .\) Show that \(T\) is invertible and find a formula for \(T^{-1} .\) $$ T\left(x_{1}, x_{2}\right)=\left(-5 x_{1}+9 x_{2}, 4 x_{1}-7 x_{2}\right) $$

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