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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

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

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

The inverse of \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\C&I&{\bf{0}}\\A&B&I\end{array}} \right]\) is \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\Z&I&{\bf{0}}\\X&Y&I\end{array}} \right]\). Find X, Y, and Z.

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.

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

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 \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

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