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Chapter 4: Q3E (page 191)

In Exercise 3, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

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

Short Answer

Expert verified

Vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\)

Step by step solution

01

Use the definition

The coordinates of x relative to basis\({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\)are the weights\({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\),such that \(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then,\({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

02

Find x

By the above definition, you get

\[\begin{array}{c}x = 3\left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right] + 0\left[ {\begin{array}{*{20}{c}}5\\2\\{ - 2}\end{array}} \right] + \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}}4\\{ - 7}\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3\\{ - 12}\\9\end{array}} \right] + 0 + \left[ {\begin{array}{*{20}{c}}{ - 4}\\7\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3 - 4}\\{ - 12 + 7}\\9\end{array}} \right]\\x = \left[ {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right].\end{array}\]

03

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\\9\end{array}} \right)\).

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Most popular questions from this chapter

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

\({\left( {{\bf{2}} - t} \right)^{\bf{3}}}\), \({\left( {{\bf{3}} - t} \right)^2}\), \({\bf{1}} + {\bf{6}}t - {\bf{5}}{t^{\bf{2}}} + {t^{\bf{3}}}\)

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

  1. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\). (Hint: Explain why every vector in the column space of \(AB\) is in the column space of \(A\).
  2. Show that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} B\). (Hint: Use part (a) to study rank\({\left( {AB} \right)^T}\).)

The null space of a \({\bf{5}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A.

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.)

Let \({\mathop{\rm u}\nolimits} = \left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]\). Find \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) such that \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&4\\2&{ - 6}&8\end{array}} \right] = {{\mathop{\rm uv}\nolimits} ^T}\) .
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