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Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Short Answer

Expert verified

The transformation \({T^{ - 1}}\) is one-to-one mapping.

Step by step solution

01

Show that T  maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\) 

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^m}\) be a linear transformation, and Abe the standard matrix for T. Then, according to theorem 12,

  1. Tmaps \({\mathbb{R}^n}\) onto \({\mathbb{R}^m}\) if and only if the columns of Aspan \({\mathbb{R}^m}\);
  2. T is one-to-one if and only if the columns of Aare linearly independent.

When T maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\), the columns of standard matrix Aspan \({\mathbb{R}^n}\), according to theorem 12.

A is invertible according to the invertible matrix theorem.

02

Show that \({T^{ - 1}}\) is one-to-one mapping

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a linear transformation and Abe the standard matrix for T. Then, according to theorem 9, Tis invertible if and only if Ais an invertible matrix. The linear transformation S, given by \[S\left( x \right) = {A^{ - 1}}{\mathop{\rm x}\nolimits} \], is a unique function satisfying the equations

  1. \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\), and
  2. \(T\left( {S\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \)for all x in \({\mathbb{R}^n}\).

Tis invertible and \({A^{ - 1}}\) is the standard matrix of \({T^{ - 1}}\), according to theorem 9.

By the invertible matrix theorem, the columns of \({A^{ - 1}}\) are linearly independent, and they span \({\mathbb{R}^n}\) since \({A^{ - 1}}\) is invertible. Thus, the transformation \({T^{ - 1}}\) is one-to-one mapping of \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\), according to theorem 12.

Thus, the transformation \({T^{ - 1}}\) is one-to-one mapping.

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

Suppose Ais \(n \times n\) and the equation \(A{\bf{x}} = {\bf{0}}\) has only the trivial solution. Explain why Ahas npivot columns and Ais row equivalent to \({I_n}\). By Theorem 7, this shows that Amust be invertible. (This exercise and Exercise 24 will be cited in Section 2.3.)

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4
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