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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

Determine which of the matrices in Exercises 1–6 are symmetric.

5. \(\left( {\begin{aligned}{{}{}}{ - 6}&2&0\\2&{ - 6}&2\\0&2&{ - 6}\end{aligned}} \right)\)

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.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

10.Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

10. \(\left( {\begin{aligned}{{}}{1/3}&{\,\,2/3}&{\,\,2/3}\\{2/3}&{\,\,1/3}&{ - 2/3}\\{2/3}&{ - 2/3}&{\,\,1/3}\end{aligned}} \right)\)

Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

Question: In Exercises 15 and 16, construct the pseudo-inverse of \(A\). Begin by using a matrix program to produce the SVD of \(A\), or, if that is not available, begin with an orthogonal diagonalization of \({A^T}A\). Use the pseudo-inverse to solve \(A{\rm{x}} = {\rm{b}}\), for \({\rm{b}} = \left( {6, - 1, - 4,6} \right)\) and let \(\mathop {\rm{x}}\limits^\^ \)be the solution. Make a calculation to verify that \(\mathop {\rm{x}}\limits^\^ \) is in Row \(A\). Find a nonzero vector \({\rm{u}}\) in Nul\(A\), and verify that \(\left\| {\mathop {\rm{x}}\limits^\^ } \right\| < \left\| {\mathop {\rm{x}}\limits^\^ + {\rm{u}}} \right\|\), which must be true by Exercise 13(c).

16. \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 1}&{ - 2}&0\\{ - 5}&0&3&5&0\\{\,\,\,2}&{\,\,0}&{ - 1}&{ - 2}&0\\{\,\,\,6}&{\,\,0}&{ - 3}&{ - 6}&0\end{array}} \right)\)
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