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Chapter 4: Q11E (page 191)

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{3}}}\)in the plane\(x + {\bf{2}}y + z = {\bf{0}}\). (Hint:Think of the equation as a “system” of homogeneous equations.)

Short Answer

Expert verified

A basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\).

Step by step solution

01

Define the null space of a matrix

The set of allhomogeneousequation solutions, \(A{\bf{x}} = 0\), is \(Nul A\), or it is called thenull space of A.

02

State the basis for the set of vectors

Consider the plane equation \(x + 2y + z = 0\).

In matrix form, it is\(A = \left( {\begin{array}{*{20}{c}}1&2&1\end{array}} \right)\).

The plane equation \(x + 2y + z = 0\)can also be written as\(x = - 2y - z\).

From the above equations, \(x\) corresponds to thepivot positions. So, \(x\) is the basic variable, and \(y\) and \(z\) are free variables.

Let \(y = a\), \(z = b\).

Substitute the values \(y = a\) and \(z = b\) in the equation \(x = - 2y - z\) to obtain the general solution

\(x = - 2a - b\).

Obtain the vector in the parametric form using \(x = - 2a - b\), \(y = a\), and \(z = b\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2a - b}\\a\\b\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2a}\\a\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - b}\\0\\b\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\end{array}\)

It can be written as

\(\left( {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right) = y\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + z\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\).

Thebasis of vectorsis shown below:

\( \Rightarrow \left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\)

Thus, \({\bf{x}} = y\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right) + z\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\), and a basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)} \right\}\).

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Most popular questions from this chapter

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

15. Let\(A\)be an\(m \times n\)matrix, and let\(B\)be a\(n \times p\)matrix such that\(AB = 0\). Show that\({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces\({\mathop{\rm Nul}\nolimits} A\),\({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and\({\mathop{\rm Col}\nolimits} B\)is contained in one of the other three subspaces.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).

If A is a \({\bf{4}} \times {\bf{3}}\) matrix, what is the largest possible dimension of the row space of A? If Ais a \({\bf{3}} \times {\bf{4}}\) matrix, what is the largest possible dimension of the row space of A? Explain.

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

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

Show the coordinate mapping is one to one. (Hint: Suppose \({\left( {\bf{u}} \right)_B} = {\left( {\bf{w}} \right)_B}\) for some u and w in V, and show that \({\bf{u}} = {\bf{w}}\)).
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