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Chapter 4: Q12Q (page 191)

In Exercises 11 and 12, use an inverse matrix to find for the given and \(B\).

12. \(B = \left\{ {\left( {\begin{array}{*{20}{c}}4\\5\end{array}} \right),\left( {\begin{array}{*{20}{c}}6\\7\end{array}} \right)} \right\},{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\)

Short Answer

Expert verified

The \(B\)-coordinate vector is \({\left( {\mathop{\rm x}\nolimits} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\).

Step by step solution

01

State the change of coordinate from B

Let \({P_B} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}& \cdots &{{{\mathop{\rm b}\nolimits} _n}}\end{array}} \right)\), then thevector equation\({\mathop{\rm x}\nolimits} = {c_1}{{\mathop{\rm b}\nolimits} _1} + {c_2}{{\mathop{\rm b}\nolimits} _2} + ... + {c_n}{{\mathop{\rm b}\nolimits} _n}\)becomes equivalent to \({\mathop{\rm x}\nolimits} = {P_B}{\left( {\mathop{\rm x}\nolimits} \right)_B}\). \({P_B}\) represents the change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^n}\).

Multiplication by \(P_B^{ - 1}\) on the L.H.S converts \({\mathop{\rm x}\nolimits} \) into its \(B\)-coordinate vector:

\(P_B^{ - 1}{\mathop{\rm x}\nolimits} = {\left( {\mathop{\rm x}\nolimits} \right)_B}\)

02

Use an inverse matrix to determine \({\left( {\mathop{\rm x}\nolimits}  \right)_B}\)

The change-of-coordinates matrix from \(B\) to the standard basis in \({\mathbb{R}^2}\) is shown below.

\(\begin{array}{c}{P_B} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm b}\nolimits} _1}}&{{{\mathop{\rm b}\nolimits} _2}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4&6\\5&7\end{array}} \right)\end{array}\)

Multiply the L.H.S of the equation \({\mathop{\rm x}\nolimits} = {P_B}{\left( {\mathop{\rm x}\nolimits} \right)_B}\) by \(P_B^{ - 1}\) to convert \({\mathop{\rm x}\nolimits} \) into its \(B\)-coordinate vector.

\(\begin{array}{c}{\left( {\mathop{\rm x}\nolimits} \right)_B} = P_B^{ - 1}{\mathop{\rm x}\nolimits} \\ = {\left( {\begin{array}{*{20}{c}}4&6\\5&7\end{array}} \right)^{ - 1}}\left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - \frac{7}{2}}&3\\{\frac{5}{2}}&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}2\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 7 + 0}\\{5 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\end{array}\)\({\left( {\mathop{\rm x}\nolimits} \right)_B}\)

Thus, the \(B\)-coordinate vector is \({\left( {\mathop{\rm x}\nolimits} \right)_B} = \left( {\begin{array}{*{20}{c}}{ - 7}\\5\end{array}} \right)\).

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Most popular questions from this chapter

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.)

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.)

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

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.)

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

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)
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