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Chapter 4: Q23E (page 191)

A homogeneous system of twelve linear equations in eight unknowns has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. Can the set of all solutions be described with fewer than twelve homogeneous linear equations? If so, how many? Discuss.

Short Answer

Expert verified

Yes, the set of all solutions can be described with six homogeneous linear equations.

Step by step solution

01

Describe the given statement

It is given that a homogeneous system has twelve linear equations with eight unknowns. The system has two fixed solutions that are not multiples of each other, and all other solutions are linear combinations of these two solutions. This implies that the number of non-pivot columns is 2, that is, the value of \({\rm{dim}}\,{\rm{Nul }}A\) is 2.

02

Use the rank theorem

Consider the homogeneous system \(Ax = 0\), where \(A\) is \(12 \times 8\) matrix. The value of \(n\) of the unknown is 8, and the linear equations (rows) are 12. By the rank theorem, \({\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A = n\).

Put the values as shown below:

\(\begin{aligned} {\rm{rank}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,A &= n\\{\rm{rank}}\,A &= n - {\rm{dim}}\,{\rm{Nul}}\,\,A\\{\rm{rank}}\,A &= 8 - 2\\{\rm{rank}}\,A &= 6\end{aligned}\)

03

Draw a conclusion

As the value of \({\rm{rank }}A\) is 6, the value of \({\rm{dimcol}}\,A\) is also 6. It means the echelon form of matrix \(A\) has six non-zero rows. Thus, the set of all solutions can be described with six homogeneous linear equations.

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Most popular questions from this chapter

The first four Hermite polynomials are \(1,2t, - 2 + 4{t^2},\) and \( - 12t + 8{t^3}\). These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Question: In Exercises 13 and 14, assume that A is row equivalent to B. Find bases for NulA and Col A.

13. \(A = \left[ {\begin{array}{*{20}{c}}{ - 2}&4&{ - 2}&{ - 4}\\2&{ - 6}&{ - 3}&1\\{ - 3}&8&2&{ - 3}\end{array}} \right]\), \(B = \left[ {\begin{array}{*{20}{c}}1&0&6&5\\0&2&5&3\\0&0&0&0\end{array}} \right]\)

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 \(T\).

Show that the space \(C\left( \mathbb{R} \right)\) of all continuous functions defined on the real line is infinite-dimensional.

In Exercises 7-10, let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for \({\mathbb{R}^2}\). In each exercise, find the change-of-coordinates matrix from \(B\) to \(C\) and the change-of-coordinates matrix from \(C\) to \(B\).

10. \({{\mathop{\rm b}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}7\\{ - 2}\end{array}} \right),{{\mathop{\rm b}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right),{{\mathop{\rm c}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}4\\1\end{array}} \right),{{\mathop{\rm c}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right)\).
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