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Q26E

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Linear Algebra and its Applications
Found in: Page 191
Linear Algebra and its Applications

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

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Short Answer

In statistical theory, a common requirement is that a matrix be of full rank. That is, the rank should be as large as possible. Explain why an m n matrix with more rows than columns has full rank if and only if its columns are linearly independent.

No, the new non-homogeneous system will not have any solution.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Describe the given statement

It is given that a homogeneous system has full rank. It implies that if the system is \(m \times n\), then the rank of the system is \(n\) as the rank is equal to the number of pivot positions.

Step 2: Use the rank theorem

Consider the homogeneous system \(Ax = 0\), where \(A\) is an \(m \times n\) matrix. As the system has full rank, \({\rm{rank }}A = n\) and there are \(n\) unknowns. By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= n - {\rm{rank}}\,A\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= n - n\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= 0\end{aligned}\)

Step 3: Draw a conclusion

As \({\rm{dim Nul}}\,A\) is 0 and the system is homogeneous, there must exist only a trivial solution. This is possible when matrix \(A\) has linearly independent columns.

Most popular questions for Math Textbooks

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Given vectors, \({u_{\bf{1}}}\),….,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),….,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution? Discuss.

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Suppose \({{\bf{p}}_{\bf{1}}}\), \({{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\) are specific polynomials that span a two-dimensional subspace H of \({P_{\bf{5}}}\). Describe how one can find a basis for H by examining the four polynomials and making almost no computations.

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)
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