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Chapter 7: Q9SE (page 395)

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.)

Short Answer

Expert verified

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

Step by step solution

01

Gram matrix

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

02

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

Exercise 22 in section 6.5states 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.

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Most popular questions from this chapter

Question: Mark Each statement True or False. Justify each answer. In each part, A represents an \(n \times n\) matrix.

  1. If A is orthogonally diagonizable, then A is symmetric.
  2. If A is an orthogonal matrix, then A is symmetric.
  3. If A is an orthogonal matrix, then \(\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).
  4. The principal axes of a quadratic from \({{\bf{x}}^T}A{\bf{x}}\) can be the columns of any matrix P that diagonalizes A.
  5. If P is an \(n \times n\) matrix with orthogonal columns, then \({P^T} = {P^{ - {\bf{1}}}}\).
  6. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
  7. If \({{\bf{x}}^T}A{\bf{x}} > {\bf{0}}\) for some x, then the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is positive definite.
  8. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
  9. The largest value of a quadratic form \({{\bf{x}}^T}A{\bf{x}}\), for \(\left\| {\bf{x}} \right\| = {\bf{1}}\) is the largest entery on the diagonal A.
  10. The maximum value of a positive definite quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is the greatest eigenvalue of A.
  11. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable \({\bf{x}} = P{\bf{u}}\), for some orthogonal matrix P.
  12. An indefinite quadratic form is one whose eigenvalues are not definite.
  13. If P is an \(n \times n\) orthogonal matrix, then the change of variable \({\bf{x}} = P{\bf{u}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form whose matrix is \({P^{ - {\bf{1}}}}AP\).
  14. If U is \(m \times n\) with orthogonal columns, then \(U{U^T}{\bf{x}}\) is the orthogonal projection of x onto ColU.
  15. If B is \(m \times n\) and x is a unit vector in \({\mathbb{R}^n}\), then \(\left\| {B{\bf{x}}} \right\| \le {\sigma _{\bf{1}}}\), where \({\sigma _{\bf{1}}}\) is the first singular value of B.
  16. A singular value decomposition of an \(m \times n\) matrix B can be written as \(B = P\Sigma Q\), where P is an \(m \times n\) orthogonal matrix and \(\Sigma \) is an \(m \times n\) diagonal matrix.
  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

In Exercises 1 and 2,find the change of variable \({\rm{x}} = P{\rm{y}}\) that transforms the quadratic form \({{\rm{x}}^T}A{\rm{x}}\) into \({{\rm{y}}^T}D{\rm{y}}\) as shown.

2. \(3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{} = 7y_1^2 + 4y_2^2\).

Let B be an \(n \times n\) symmetric matrix such that \({B^{\bf{2}}} = B\). Any such matrix is called a projection matrix (or an orthogonal projection matrix.) Given any y in \({\mathbb{R}^n}\), let \({\bf{\hat y}} = B{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{\hat y}}\).

a) Show that z is orthogonal to \({\bf{\hat y}}\).

b) Let W be the column space of B. Show that y is the sum of a vector in W and a vector in \({W^ \bot }\). Why does this prove that By is the orthogonal projection of y onto the column space of B?

Suppose A is invertible and orthogonally diagonalizable. Explain why \({A^{ - {\bf{1}}}}\) is also orthogonally diagonalizable.

In Exercises 17鈥�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 \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)
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