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Found in: Page 395

### 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

# 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.

See the step by step solution

## 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.