Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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).

10. Show that if an \(n \times n\) matrix G is positive semidefinite and has rank r, then G is the Gram matrix of some \(r \times n\) matrix A. This is called a rank-revealing factorization of G. (Hint: Consider the spectral decomposition of G, and first write G as \(B{B^T}\) for an \(n \times r\) matrix B.)

Short Answer

Expert verified

It is proved that the G is the Gram matrix of some \(r \times n\) matrix A.

Step by step solution

01

Rank Theorem

The Rank Theoremstates that a \(m \times n\) matrix \(A\) has the same dimensions of the row space and column space. The number of pivot position in \(A\) is exactly equal to the rank of \(A\) and satisfy the equation \({\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = n\).

02

Show that G is the Gram matrix of some \(r \times n\) matrix A

When \({\mathop{\rm rank}\nolimits} G = r\) then \(\dim {\mathop{\rm Nul}\nolimits} G = n - r\), according to the Rank theorem. Therefore, the eigenvalue of \(G\) is 0 of multiplicity \(n - r\) , and the spectral decomposition of \(G\) is given by \(G = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + \ldots + {\lambda _r}{{\bf{u}}_r}{\bf{u}}_r^T\).

Moreover \({\lambda _1}, \ldots ,{\lambda _r}\) are positive since \(G\) is positive semidefinite. Then

\(G = \left( {\sqrt {{\lambda _1}} {{\bf{u}}_1}} \right)\left( {\sqrt {{\lambda _1}} {{\bf{u}}_1}^T} \right) + \ldots + \left( {\sqrt {{\lambda _r}} {{\bf{u}}_r}} \right)\left( {\sqrt {{\lambda _r}} {\bf{u}}_r^T} \right)\)


According to the column-row expansion of a matrix product, \(G = B{B^T}\), with \(B\) is the \(n \times r\) matrix \(B = \left( {\begin{array}{*{20}{c}}{\sqrt {{\lambda _1}} {{\bf{u}}_1}}& \ldots &{\sqrt {{\lambda _r}} {{\bf{u}}_r}}\end{array}} \right)\). Therefore, \(G = {A^T}A\) for \(A = {B^T}\).

Thus, it is proved that the G is the Gram matrix of some \(r \times n\) matrix A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

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

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

40. \(\left( {\begin{aligned}{{}}{\bf{8}}&{\bf{2}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{8}}&{\bf{2}}&{ - {\bf{6}}}&{\bf{9}}\\{\bf{2}}&{\bf{2}}&{\bf{8}}&{ - {\bf{6}}}&{\bf{9}}\\{ - {\bf{6}}}&{ - {\bf{6}}}&{ - {\bf{6}}}&{{\bf{24}}}&{\bf{9}}\\{\bf{9}}&{\bf{9}}&{\bf{9}}&{\bf{9}}&{ - {\bf{21}}}\end{aligned}} \right)\)

Question: Repeat Exercise 15 for the following SVD of a \({\bf{3 \times 4}}\) matrix \(A\):

\(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{.86}}}&{ - {\bf{.11}}}&{ - {\bf{.50}}}\\{{\bf{.31}}}&{{\bf{.68}}}&{ - {\bf{.67}}}\\{{\bf{.41}}}&{ - {\bf{.73}}}&{ - {\bf{5}}{\bf{.5}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\bf{12}}{\bf{.48}}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{{\bf{6}}{\bf{.34}}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right){\bf{ \times }}\left( {\begin{array}{*{20}{c}}{{\bf{.66}}}&{ - {\bf{.03}}}&{ - {\bf{.35}}}&{{\bf{.66}}}\\{ - {\bf{1}}{\bf{.3}}}&{ - {\bf{.90}}}&{ - {\bf{.39}}}&{ - {\bf{.13}}}\\{{\bf{.65}}}&{{\bf{.08}}}&{ - {\bf{.16}}}&{ - {\bf{.73}}}\\{ - {\bf{.34}}}&{{\bf{.42}}}&{ - {\bf{8}}{\bf{.4}}}&{ - {\bf{0}}{\bf{.8}}}\end{array}} \right)\)

Question:Find the principal components of the data for Exercise 1.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free