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Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**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).**

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**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*.

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

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

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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*.

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