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Chapter 4: Problem 8

Suppose \(U\) is an orthogonal \(n \times n\) matrix. Explain why \(\operatorname{rank}(U)=n\).

Short Answer

Expert verified
The rank of \( U \) is \( n \) because all \( n \) columns are linearly independent.

Step by step solution

01

Understand Orthogonal Matrices

An orthogonal matrix is a square matrix whose rows and columns are orthonormal vectors. For a matrix to be orthogonal, it must satisfy the condition: \[ U^T U = I \] where \( U^T \) is the transpose of \( U \) and \( I \) is the identity matrix.
02

Analyze the Implications of Orthogonality

Given \( U^T U = I \), this implies that the columns of \( U \) form an orthonormal set. Each column vector is orthogonal to the others, and each has a norm (length) of 1.
03

Determine the Rank of the Matrix

The rank of a matrix is the maximum number of linearly independent column vectors in the matrix. Since the columns of \( U \) form an orthonormal set, they are linearly independent. Therefore, all \( n \) columns in the \( n \times n \) orthogonal matrix \( U \) are linearly independent.
04

Conclude the Rank of \( U \)

Since there are \( n \) linearly independent columns, the rank of \( U \) is \( n \). That is, \[ \text{rank}(U) = n \]

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Key Concepts

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

Orthogonal Matrices
Orthogonal matrices are a special type of square matrix that play an important role in linear algebra. For a matrix to be orthogonal, its rows and columns must be orthonormal vectors. Orthonormal vectors are vectors that are both orthogonal (perpendicular to each other) and normalized (have a magnitude of 1). We represent this property formally as:

Linear Independence
Understanding linear independence is crucial when analyzing the rank of orthogonal matrices. Vectors are considered linearly independent if no vector can be written as a linear combination of the others. This means that there are no redundant vectors in the set.

In the context of an orthogonal matrix, the columns (or rows) are automatically linearly independent due to their orthonormality. Each column vector stands alone and cannot be expressed as a combination of the others. This implies a lack of redundancy, which directly influences the rank of the matrix.
Matrix Rank
The rank of a matrix is a measure of the number of linearly independent columns (or rows) it has. For an orthogonal matrix, the task of determining the rank becomes straightforward.

Given that an orthogonal matrix's columns are linearly independent, we can conclude that its rank is equivalent to the total number of columns (or rows). For an \( n \times n \) orthogonal matrix \( U \), all \( n \) columns are linearly independent. Thus, the rank of \( U \) is \( n \). Mathematically, this is expressed as:

\[ \text{rank}(U) = n \]

This concept underlines the relationship between orthogonality, linear independence, and matrix rank in a concise manner.

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Most popular questions from this chapter

Here are some vectors. $$ \left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ 2 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ -3 \\ -2 \end{array}\right],\left[\begin{array}{r} -1 \\ 1 \\ 2 \end{array}\right] $$ Now here is another vector: $$ \left[\begin{array}{r} 1 \\ 2 \\ -1 \end{array}\right] $$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

The wind blows from West to East at a speed of 50 miles per hour and an airplane which travels at 400 miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

The total force acting on an object is to be \(4 \vec{i}+2 \vec{j}-3 \vec{k}\) Newtons. A force of \(-3 \vec{i}-1 \vec{j}+8 \vec{k}\) Newtons is being applied. What other force should be applied to achieve the desired total force?

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. $$ \left[\begin{array}{r} 1 \\ 2 \\ 2 \\ -4 \end{array}\right],\left[\begin{array}{r} 3 \\ 4 \\ 1 \\ -4 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ 0 \\ 4 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ -2 \\ 5 \end{array}\right] $$

Here are some vectors. $$ \left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ 2 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ -3 \\ -2 \end{array}\right],\left[\begin{array}{r} -1 \\ 1 \\ 2 \end{array}\right] $$ Now here is another vector: $$ \left[\begin{array}{r} 2 \\ -3 \\ -4 \end{array}\right] $$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.
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