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Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]

Short Answer

Expert verified

It is proved that Sis a linear transformation.

Step by step solution

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01

Show that S preserves sums

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

It is given that u, and v are in \({\mathbb{R}^n}\).

If \[x = S\left( {\mathop{\rm u}\nolimits} \right)\] and \[y = S\left( {\mathop{\rm v}\nolimits} \right)\], then \[T\left( x \right) = T\left( {S\left( {\mathop{\rm u}\nolimits} \right)} \right) = {\mathop{\rm u}\nolimits} \] and \[T\left( y \right) = T\left( {S\left( {\mathop{\rm v}\nolimits} \right)} \right) = {\mathop{\rm v}\nolimits} \] from equation (2). Thus,

\[\begin{array}{c}S\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) = S\left( {T\left( x \right) + T\left( y \right)} \right)\\ = S\left( {T\left( {x + y} \right)} \right)\,\,\,\,{\rm{ (}}{\mathop{\rm Since}\nolimits} \,\,T\,\,{\mathop{\rm is}\nolimits} \,\,{\mathop{\rm linear}\nolimits} )\\ = x + y\,\,\,\,\,{\rm{ (}}By\,\,equation\,\,\left( 1 \right))\\ = S\left( {\mathop{\rm u}\nolimits} \right) + S\left( {\mathop{\rm v}\nolimits} \right).\end{array}\]

Therefore, Spreserves sums.

02

Show that S is a linear transformation

For any scalar r,

\[\begin{array}{c}S\left( {r{\mathop{\rm u}\nolimits} } \right) = S\left( {rT\left( {\mathop{\rm x}\nolimits} \right)} \right)\\ = S\left( {T\left( {r{\mathop{\rm x}\nolimits} } \right)} \right)\,\,\,\,{\rm{ (}}{\mathop{\rm Since}\nolimits} \,\,T\,\,is\,\,linear)\\ = r{\mathop{\rm x}\nolimits} \,\,\,{\rm{ (}}\,{\mathop{\rm By}\nolimits} \,\,equation\left( 1 \right))\\ = rS\left( {\mathop{\rm u}\nolimits} \right).\end{array}\]

Spreserves the scalar multiples, and hence,itis a linear transformation.

Thus, it is proved that Sis a linear transformation.

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Most popular questions from this chapter

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

5. \[\left[ {\begin{array}{*{20}{c}}A&B\\C&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&Y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\Z&{\bf{0}}\end{array}} \right]\]

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Explain why the columns of an \(n \times n\) matrix Aspan \({\mathbb{R}^{\bf{n}}}\) when

Ais invertible. (Hint:Review Theorem 4 in Section 1.4.)

In exercises 11 and 12, mark each statement True or False. Justify each answer.

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

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