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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: If A is \(m \times n\), then the matrix \(G = {A^T}A\) is called the Gram matrix of A. In this case, the entries of G are the inner products of the columns of A. (See Exercises 9 and 10).

9. Show that the Gram matrix of any matrix A is positive semidefinite, with the same rank as A. (See the Exercises in Section 6.5.)

It is proved that the Gram matrix of any matrix \(A\) is positive semidefinite with the same rank as A.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Gram matrix

When \(A\) is a \(m \times n\) matrix then the matrix \(G = {A^T}A\) is known as the Gram matrix of A.

Step 2: Show that the Gram matrix of any matrix A is positive semidefinite, with the same rank as A

Exercise 22 in section 6.5 states that \({\mathop{\rm rank}\nolimits} {A^T}A = {\mathop{\rm rank}\nolimits} A\).

When \(A\) is a \(m \times n\) matrix and \({\bf{x}}\) in \({\mathbb{R}^n}\) then;

\(\begin{array}{c}{{\bf{x}}^T}{A^T}A{\bf{x}} = {\left( {A{\bf{x}}} \right)^T}\left( {A{\bf{x}}} \right)\\ = {\left\| {A{\bf{x}}} \right\|^2} \ge 0\end{array}\)

Therefore, \(G = {A^T}A\) is a positive semidefinite. According to Exercise 22 in Section 6.5, \({\mathop{\rm rank}\nolimits} {A^T}A = {\mathop{\rm rank}\nolimits} A\).

Thus, it is proved that the Gram matrix of any matrix \(A\) is positive semidefinite with the same rank as A.

Most popular questions for Math Textbooks

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

21. Justify the statement in Example 2 that the second singular value of a matrix \(A\) is the maximum of \(\left\| {A{\bf{x}}} \right\|\) as \({\bf{x}}\) varies over all unit vectors orthogonal to \({{\bf{v}}_{\bf{1}}}\), with \({{\bf{v}}_{\bf{1}}}\) a right singular vector corresponding to the first singular value of \(A\). (Hint: Use Theorem 7 in Section 7.3.)

Suppose\(A = PR{P^{ - {\bf{1}}}}\), where P is orthogonal and R is upper triangular. Show that if A is symmetric, then R is symmetric and hence is actually a diagonal matrix.

Question 7: Prove that an \(n \times n\) A is positive definite if and only if A admits a Cholesky factorization, namely, \(A = {R^T}R\) for some invertible upper triangular matrix R whose diagonal entries are all positive. (Hint; Use a QR factorization and Exercise 26 in Section 7.2.)

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^{\bf{3}}}\).

a. \(3x_1^2 - 2x_2^2 + 5x_3^2 + 4{x_1}{x_2} - 6{x_1}{x_3}\)

b. \(4x_3^2 - 2{x_1}{x_2} + 4{x_2}{x_3}\)

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

5. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).
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