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Q22E

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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

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.

No, it is not possible to find any vector in \({\rm{Nul }}A\) which spans \({\rm{Nul }}A\).

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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 ten linear equations with twelve unknowns. The twelve unknowns are multiples of one fixed nonzero solution. This implies that the maximum rank of the matrix formed from the homogeneous system is 10 as it has 10 pivot places.

Step 2: Use the rank theorem

Consider a homogeneous system \(Ax = 0\), where \(A\) is \(10 \times 12\) matrix. The value of \(n\) of the unknown is 12 and \({\rm{rank}}\,A \le 10\). By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values to get:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{dim}}\,{\rm{Nul}}\,\,A &= n - {\rm{rank}}\,A\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 12 - 10\\{\rm{dim}}\,{\rm{Nul}}\,\,A &\ge 2\end{aligned}\)

Step 3: Draw a conclusion

As the value of \({\rm{dim}}\,{\rm{Nul }}A\) is 2 or greater than 2, the number of non-pivot columns is at least 12. Thus, it is not possible to find any vector in \({\rm{Nul }}A\) which spans \({\rm{Nul }}A\).

Most popular questions for Math Textbooks

Suppose \(A\) is \(m \times n\)and \(b\) is in \({\mathbb{R}^m}\). What has to be true about the two numbers rank \(\left[ {A\,\,\,{\rm{b}}} \right]\) and \({\rm{rank}}\,A\) in order for the equation \(Ax = b\) to be consistent?

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)

  1. Explain why \({{\mathop{\rm a}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _5}\) are in the column space of \(B\).
  2. Find a set of vectors that spans \({\mathop{\rm Nul}\nolimits} A\).
  3. Let \(T:{\mathbb{R}^5} \to {\mathbb{R}^4}\) be defined by \(T\left( x \right) = A{\mathop{\rm x}\nolimits} \). Explain why \(T\) is neither one-to-one nor onto.

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

Which of the subspaces \({\rm{Row }}A\) , \({\rm{Col }}A\), \({\rm{Nul }}A\), \({\rm{Row}}\,{A^T}\) , \({\rm{Col}}\,{A^T}\) , and \({\rm{Nul}}\,{A^T}\) are in \({\mathbb{R}^m}\) and which are in \({\mathbb{R}^n}\) ? How many distinct subspaces are in this list?.
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