Define a linear transformation by \(T\left( {\mathop{\rm p}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm p}\nolimits} \left( 0 \right)}\\{{\mathop{\rm p}\nolimits} \left( 0 \right)}\end{array}} \right)\). Find \(T:{{\mathop{\rm P}\nolimits} _2} \to {\mathbb{R}^2}\)polynomials \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _2}\) in \({{\mathop{\rm P}\nolimits} _2}\) that span the kernel of T, and describe the range of T.

Short Answer

Expert verified

The polynomials \({{\mathop{\rm p}\nolimits} _1}\left( t \right) = t\) and \({{\mathop{\rm p}\nolimits} _2}\left( t \right) = t\) span the kernel of T. The range of \(T\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}a\\a\end{array}} \right):a\,\,{\mathop{\rm real}\nolimits} } \right\}\).

Step by step solution

01

Determine the polynomials \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _2}\) that span the kernel of T

The kernel of \(T\) will contain any quadratic polynomial \({\mathop{\rm q}\nolimits} \) with \({\mathop{\rm q}\nolimits} \left( 0 \right) = 0\). The polynomial \({\mathop{\rm q}\nolimits} \) is \({\mathop{\rm q}\nolimits} = at + b{t^2}\).

Therefore, the polynomials \({{\mathop{\rm p}\nolimits} _1}\left( t \right) = t\) and \({{\mathop{\rm p}\nolimits} _2}\left( t \right) = t\) span the kernel of T.

02

Describe the range of T

When a vector is in the range of \(T\), it must be of the form \(\left( {\begin{array}{*{20}{c}}a\\a\end{array}} \right)\). When a vector is of the form, it is the image of the polynomial \({\mathop{\rm p}\nolimits} \left( t \right) = a\) in \({{\mathop{\rm P}\nolimits} _2}\).

Therefore, the range of \(T\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}a\\a\end{array}} \right):a\,\,{\mathop{\rm real}\nolimits} } \right\}\).

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Most popular questions from this chapter

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\) and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Show that the coordinate mapping is onto \({\mathbb{R}^n}\). That is, given any y in \({\mathbb{R}^n}\), with entries \({y_{\bf{1}}}\),….,\({y_n}\), produce u in V such that \({\left( {\bf{u}} \right)_B} = y\).

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

19. \(A = \left( {\begin{array}{*{20}{c}}{.9}&1&0\\0&{ - .9}&0\\0&0&{.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}0\\1\\1\end{array}} \right)\).

Let \(V\) and \(W\) be vector spaces, and let \(T:V \to W\) be a linear transformation. Given a subspace \(U\) of \(V\), let \(T\left( U \right)\) denote the set of all images of the form \(T\left( {\mathop{\rm x}\nolimits} \right)\), where x is in \(U\). Show that \(T\left( U \right)\) is a subspace of \(W\).

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