Let\(T:{{\rm P}_2} \to {{\rm P}_3}\) be a linear transformation that maps a polynomial \({\bf{p}}\left( t \right)\) into the polynomial \(\left( {t + 5} \right){\bf{p}}\left( t \right)\).

1. Find the image of\({\bf{p}}\left( t \right) = 2 - t + {t^2}\).
2. Show that \(T\) is a linear transformation.
3. Find the matrix for \(T\) relative to the bases \(\left\{ {1,t,{t^2}} \right\}\) and \(\left\{ {1,t,{t^2},{t^3}} \right\}\).

(a)

It is given that the image of \({\bf{p}}\left( t \right)\) is given by \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {t + 5} \right){\bf{p}}\left( t \right)\).

Obtain the image \({\bf{p}}\left( t \right) = 2 - t + {t^2}\) by using \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {t + 5} \right){\bf{p}}\left( t \right)\).

\(\begin{aligned}T\left( {{\bf{p}}\left( t \right)} \right) = T\left( {2 - t + {t^2}} \right)\\ = \left( {t + 5} \right)\left( {2 - t + {t^2}} \right)\\ = 2t - {t^2} + {t^3} + 10 - 5t + 5{t^2}\\ = 10 - 3t + 4{t^2} + {t^3}\end{aligned}\)

So, the image of \({\bf{p}}\left( t \right) = 2 - t + {t^2}\) is \(10 - 3t + 4{t^2} + {t^3}\).

Let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be a transformation; this transformation is said to be a linear transformation if it satisfies the following two properties:

1. \(T\left( {u + v} \right) = T\left( u \right) + T\left( v \right)\)
2. \(T\left( {cu} \right) = cT\left( u \right)\)

Here,\(c\) is any scalar and \(u,v\) are vectors.

(b)

Let there be two polynomials, \({\bf{p}}\left( t \right)\) and \({\bf{q}}\left( t \right)\) in \({{\rm P}_2}\), then, their images will be \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {t + 5} \right){\bf{p}}\left( t \right)\) and \(T\left( {q\left( t \right)} \right) = \left( {t + 5} \right)q\left( t \right)\) , respectively.

Check for the first property.

\(\begin{aligned}T\left( {{\bf{p}}\left( t \right) + {\bf{q}}\left( t \right)} \right) &= \left( {t + 5} \right)\left( {{\bf{p}}\left( t \right) + {\bf{q}}\left( t \right)} \right)\\ &= \left( {t + 5} \right){\bf{p}}\left( t \right) + \left( {t + 5} \right){\bf{q}}\left( t \right)\\ &= T\left( {{\bf{p}}\left( t \right)} \right) + T\left( {{\bf{q}}\left( t \right)} \right)\end{aligned}\)

The first property is satisfied; check for the second property. Let \(c\) be any scaler.

\(\begin{aligned}T\left( {c \cdot {\bf{p}}\left( t \right)} \right) &= \left( {t + 5} \right)\left( {c \cdot {\bf{p}}\left( t \right)} \right)\\ &= c \cdot \left( {t + 5} \right){\bf{p}}\left( t \right)\\ &= c \cdot T\left( {{\bf{p}}\left( t \right)} \right)\end{aligned}\)

As both the properties of a linear transformation are satisfied, so \(T\) is a linear transformation.

A matrix associated with a linear transformation \(T\) for \(V\) and \(W\) is given by \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(V\) and \(W\) are \(n\) and \(m\)-dimensional subspaces respectively, and \(B\), and\(C\) are the bases for \(V\), and \(W\).

Let \(B = \left\{ {1,t,{t^2}} \right\}\), and\(C = \left\{ {1,t,{t^2},{t^3}} \right\}\).

Find \(T\left( {{{\bf{b}}_1}} \right)\), \(T\left( {{{\bf{b}}_2}} \right)\) and \(T\left( {{{\bf{b}}_3}} \right)\) for \(B = \left\{ {1,t,{t^2}} \right\}\) by using \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {t + 5} \right){\bf{p}}\left( t \right)\).

\(\begin{aligned}T\left( {{{\bf{b}}_1}} \right) &= T\left( 1 \right)\\ &= \left( {t + 5} \right)\left( 1 \right)\\ &= t + 5\end{aligned}\)

\(\begin{aligned}T\left( {{{\bf{b}}_2}} \right) &= T\left( t \right)\\ &= \left( {t + 5} \right)\left( t \right)\\ &= {t^2} + 5t\end{aligned}\)

\(\begin{aligned}T\left( {{{\bf{b}}_3}} \right) &= T\left( {{t^2}} \right)\\ &= \left( {t + 5} \right)\left( {{t^2}} \right)\\ &= {t^3} + 5{t^2}\end{aligned}\)

Write\({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_C}\), \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_C}\) and \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_C}\).

\({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_C} = \left( {\begin{aligned}5\\1\\0\\0\end{aligned}} \right)\), \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_C} = \left( {\begin{aligned}0\\5\\1\\0\end{aligned}} \right)\), \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_C} = \left( {\begin{aligned}0\\0\\5\\1\end{aligned}} \right)\)

Form a matrix \(T\) for the obtained vectors by using the formula \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(n = 3\).

\(\begin{aligned}{c}{\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)}_C}}\end{aligned}} \right)\\ = \left( {\begin{aligned}5&{}&0&{}&0\\1&{}&5&{}&0\\0&{}&1&{}&5\\0&{}&0&{}&1\end{aligned}} \right)\end{aligned}\)

So, the required matrix is \(\left( {\begin{aligned}5&{}&0&{}&0\\1&{}&5&{}&0\\0&{}&1&{}&5\\0&{}&0&{}&1\end{aligned}} \right)\).