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Chapter 2: Problem 43

Let \(\overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k}\) be vectors in \(\mathbb{R}^{n}\), and set \(V=\left[\vec{v}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k}\right]\). (a) Show that the set \(\overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k}\) is orthogonal if and only if \(\boldsymbol{V}^{\top} \boldsymbol{V}\) is diagonal. (b) Show that the set \(\overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k}\) is orthonormal if and only if \(V^{\top} V=I_{k}\).

Short Answer

Expert verified
(a) Orthogonal if \( V^{\top}V \) is diagonal. (b) Orthonormal if \( V^{\top}V = I_k \).

Step by step solution

01

Understanding Orthogonal Vectors

Vectors \( \overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k} \) are orthogonal if any pair of them is perpendicular, which means their dot product is zero, i.e., \( \overrightarrow{\mathbf{v}}_{i} \cdot \overrightarrow{\mathbf{v}}_{j} = 0 \) for all \( i eq j \).
02

Constructing the Matrix V

Construct matrix \( V \) by placing vectors \( \overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k} \) as columns: \( V = [\overrightarrow{\mathbf{v}}_{1}, \ldots, \overrightarrow{\mathbf{v}}_{k}] \). This yields \( V^{\top}V \) as a \( k \times k \) matrix where each entry is the dot product of corresponding vectors.
03

Showing Orthogonal Matrix Diagonal Structure

Analyze \( V^{\top}V \): since the vectors are orthogonal, off-diagonal entries \( (i eq j) \) become zero. Diagonal entries \( (i = j) \) become the squared magnitudes of vectors. Thus, \( V^{\top}V \) is diagonal if the set of vectors is orthogonal.
04

Validating Orthogonal Set Criterion

If \( V^{\top}V \) is diagonal, only the dot products of distinct vectors contribute off-diagonal entries, which are zero for orthogonality. Hence, the set is orthogonal if and only if \( V^{\top}V \) is diagonal.
05

Understanding Orthonormal Vectors

Orthonormal vectors are orthogonal vectors each of unit length, i.e., \( \|\overrightarrow{\mathbf{v}}_{i}\| = 1 \). This means the diagonal entries of \( V^{\top}V \) must all be 1.
06

Establishing Connection to Identity Matrix

In an orthonormal set, \( V^{\top}V \) is a diagonal matrix with 1s on its diagonal, i.e., \( V^{\top}V = I_k \) (identity matrix of size \( k \)). This indicates both orthogonality and unit-length properties are satisfied.
07

Confirming the Orthonormal Set Criterion

Verify: if \( V^{\top}V \) equals the identity matrix, then the set is both orthogonal and each vector has length 1, confirming the set is orthonormal as each diagonal entry of \( I_k \) is 1.

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Key Concepts

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

Dot Product
The concept of the dot product is foundational in vector algebra and plays a significant role in determining whether vectors are orthogonal. The dot product, also known as the scalar product, of two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) is calculated as follows: \[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = a_1b_1 + a_2b_2 + \cdots + a_nb_n \] where each \(a_i\) and \(b_i\) are the components of the vectors.
  • When the result of a dot product is zero, it indicates that the vectors are orthogonal, meaning they are perpendicular to each other.
  • The dot product also offers insight into the angle between vectors. A zero dot product specifically confirms a 90-degree angle.
Understanding this concept is crucial for further exploration into orthogonality and orthonormality.
Orthonormal Vectors
Orthonormal vectors take the idea of orthogonal vectors a step further. Not only are they perpendicular to one another, they also each have a unit magnitude, meaning their length is 1. When you place orthonormal vectors into a matrix, the relationships become more apparent.
  • These vector sets ensure that a matrix \( V^\top V \) becomes an identity matrix, \( I_k \), where \( k \) is the number of vectors.
  • To verify orthonormality, check that each vector has a length of 1 using the formula \( \| \overrightarrow{\mathbf{v}} \| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2} = 1 \).
  • Each dot product of two distinct vectors from the set equals zero, ensuring they are orthogonal as well.
This link between orthonormal vectors and the identity matrix provides a strong basis for understanding vector spaces and transformations.
Identity Matrix
An identity matrix is a special kind of diagonal matrix that serves as the neutral element in matrix multiplication. It holds significant importance in linear algebra, particularly for verifying orthonormal sets.
  • An identity matrix \( I_k \) is a square matrix with ones on its diagonal and zeroes elsewhere. This configuration acts like the number 1 in multiplication for matrices.
  • The equation \( V^\top V = I_k \) tells us two critical pieces of information about the set of vectors: they are orthogonal, and each one is of unit length (making them orthonormal).
  • A practical analogy: multiplying another matrix by the identity matrix results in the same matrix, retaining the original dimensions and properties.
Understanding identity matrices helps clarify why the property \( V^\top V = I_k \) is a key indicator of orthonormal vectors.
Matrix Transpose
The transpose of a matrix is a simple yet crucial operation in matrix algebra. Transposing a matrix involves flipping it over its diagonal, converting rows to columns and vice-versa. This concept directly applies to the matrices involved in determining orthogonal and orthonormal vectors.
  • For a given matrix \( V \), its transpose is denoted as \( V^\top \). If \( V \) is an \( m \times n \) matrix, \( V^\top \) will be an \( n \times m \) matrix.
  • Using the transpose in \( V^\top V \) allows the computation of the dot products of vectors within the original matrix, resulting in a square matrix where these products are stored.
  • The symmetry created by transposing ensures that any multiplication back with \( V \) (like \( V^\top V \)) retains essential properties of the vectors, such as orthogonality when yielding a diagonal matrix.
The step of transposing is indispensable, particularly in procedures where properties like orthogonality and orthonormality need to be verified.

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Most popular questions from this chapter

Consider the following vectors: \(\left[\begin{array}{l}1 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{l}1 \\ 2 \\ 1\end{array}\right]\), and \(\left[\begin{array}{l}0 \\ 1 \\ \alpha\end{array}\right]\). (a) For what values of \(\alpha\) are these three vectors linearly dependent? (b) Show that for each such \(\alpha\) the three vectors lie in the same plane, and give an equation of the plane.

Bring the following matrices to echelon form, using row operations. (a) \(\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]\) (b) \(\left[\begin{array}{rrr}1 & -1 & 1 \\ -1 & 0 & 2 \\ -1 & 1 & 1\end{array}\right]\) (c) \(\left[\begin{array}{rrrr}1 & 2 & 3 & 5 \\ 2 & 3 & 0 & -1 \\ 0 & 1 & 2 & 3\end{array}\right]\) (d) \(\left[\begin{array}{rrrr}1 & 3 & -1 & 4 \\ 1 & 2 & 1 & 2 \\ 3 & 7 & 1 & 9\end{array}\right]\) (e) \(\left[\begin{array}{rrrr}1 & 1 & 1 & 1 \\ 2 & -3 & 3 & 3 \\ 1 & -4 & 2 & 2\end{array}\right]\)

Predict whether each of the following systems of equations will have a unique solution, no solution, or infinitely many solutions. Solve, using row operations. If your results do not confirm your predictions, can you suggest an explanation for the discrepancy? (a) \(\begin{aligned} 2 x+13 y-3 z &=-7 \\ x+y &=1 \\ x+7 z &=22 \end{aligned}\) (b) \(\begin{aligned} x-2 y-12 z &=12 \\ 2 x+2 y+2 z &=4 \\ 2 x+3 y+4 z &=3 \end{aligned}\) (c) \(\begin{aligned} x+y+z &=5 \\ x-y-z &=4 \\ 2 x+6 y+6 z &=12 \end{aligned}\) (d) $$ \begin{array}{rr} x+3 y+z= & 4 \\ -x-y+z= & -1 \\ 2 x+4 y= & 0 \end{array} $$ (e) \(\begin{aligned} x+2 y+z-4 w+v &=0 \\ x+2 y-z+2 w-v &=0 \\ 2 x+4 y+z-5 w+v &=0 \\ x+2 y+3 z-10 w+2 v &=0 \end{aligned}\)

Consider the mapping \(\mathbf{f}: \mathbb{R}^{2}-\left(\begin{array}{l}0 \\\ 0\end{array}\right) \rightarrow \mathbb{R}^{2}\) given by $$ \mathbf{f}\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{c} \left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right) \\ x y /\left(x^{2}+y^{2}\right) \end{array}\right) $$ Does \(\mathbf{f}\) have a local inverse at every point of \(\mathbb{R}^{2}\) ?

(a) For what value of \(a\) can you not write $$ \frac{x-1}{(x+1)\left(x^{2}+a x+5\right)}=\frac{A_{0}}{x+1}+\frac{B_{1} x+B_{0}}{x^{2}+a x+5} ? $$ (b) Why does this not contradict Proposition 2.5.15?
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