Question

Define \(T:{{\rm P}_2} \to {\mathbb{R}^3}\) by \(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\).

1. Find the image under\(T\)of\({\bf{p}}\left( t \right) = 5 + 3t\).
2. Show that \(T\) is a linear transformation.
3. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2}} \right\}\)for \({{\rm P}_2}\)and the standard basis for \({\mathbb{R}^3}\).

Short answer

1. The image of \({\bf{p}}\left( t \right) = 5 + 3t\) is \(\left( {\begin{aligned}2\\5\\8\end{aligned}} \right)\).

1. It is verified that \(T\) is a linear transformation as both the properties of the transformation are satisfied.

1. The matrix for \(T\) relative to \(\left\{ {1,t,{t^2}} \right\}\) for \({{\rm P}_2}\) and the standard basis for \({\mathbb{R}^3}\)is \(\left( {\begin{aligned}1&{}&{ - 1}&{}&1\\1&{}&0&{}&0\\1&{}&1&{}&1\end{aligned}} \right)\).

Step-by-step solution

Find the image under \(T\) of \({\bf{p}}\left( t \right) = 5 + 3t\) 

(a)

It is given that the image of \({\bf{p}}\left( t \right)\) is given by \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\).

Find \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\) by using \({\bf{p}}\left( t \right) = 5 + 3t\).

\(\begin{aligned}T\left( {{\bf{p}}\left( t \right)} \right) &= \left( {\begin{aligned}{5 + 3\left( { - 1} \right)}\\{5 + 3\left( 0 \right)}\\{5 + 3\left( 1 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{5 - 3}\\{5 + 0}\\{5 + 3}\end{aligned}} \right)\\ &= \left( {\begin{aligned}2\\5\\8\end{aligned}} \right)\end{aligned}\)

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

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

So, the image under \(T\)of \({\bf{p}}\left( t \right) = 5 + 3t\) is \(\left( {\begin{aligned}2\\5\\8\end{aligned}} \right)\).

Linear transformation 

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.

Check if \(T\) is a linear transformation

(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 \(T\left( {{\bf{p}}\left( t \right)} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\) and \(T\left( {{\bf{q}}\left( t \right)} \right) = \left( {\begin{aligned}{{\bf{q}}\left( { - 1} \right)}\\{{\bf{q}}\left( 0 \right)}\\{{\bf{q}}\left( 1 \right)}\end{aligned}} \right)\) respectively.

Check for the first property.

\(\begin{aligned}{c}T\left( {{\bf{p}}\left( t \right) + {\bf{q}}\left( t \right)} \right) &= \left( {\begin{aligned}{\left( {{\bf{p}} + {\bf{q}}} \right)\left( { - 1} \right)}\\{\left( {{\bf{p}} + {\bf{q}}} \right)\left( 0 \right)}\\{\left( {{\bf{p}} + {\bf{q}}} \right)\left( 1 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right) + {\bf{q}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right) + {\bf{q}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right) + {\bf{q}}\left( 1 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right) + \left( {\begin{aligned}{{\bf{q}}\left( { - 1} \right)}\\{{\bf{q}}\left( 0 \right)}\\{{\bf{q}}\left( 1 \right)}\end{aligned}} \right)\\ &= T\left( {{\bf{p}}\left( t \right)} \right) + T\left( {{\bf{q}}\left( t \right)} \right)\end{aligned}\)

The first property is satisfied; check for the second property.

Let \(c\) be any scaler.

\(\begin{aligned}T\left( {c \cdot {\bf{p}}\left( t \right)} \right) &= \left( {\begin{aligned}{\left( {c \cdot {\bf{p}}} \right)\left( { - 1} \right)}\\{\left( {c \cdot {\bf{p}}} \right)\left( 0 \right)}\\{\left( {c \cdot {\bf{p}}} \right)\left( 1 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{c \cdot \left( {{\bf{p}}\left( { - 1} \right)} \right)}\\{c \cdot \left( {{\bf{p}}\left( 0 \right)} \right)}\\{c \cdot \left( {{\bf{p}}\left( 1 \right)} \right)}\end{aligned}} \right)\\ &= c \cdot \left( {\begin{aligned}{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 0 \right)}\\{{\bf{p}}\left( 1 \right)}\end{aligned}} \right)\\ &= c \cdot T\left( {{\bf{p}}\left( t \right)} \right)\end{aligned}\)

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

The matrix for 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 \(W\).

Find the matrix for a linear transformation

(c)

Let \(B = \left\{ {1,t,{t^2}} \right\}\) and the standard basis for \({\mathbb{R}^3}\) be \(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\). Find \(T\left( {{{\bf{b}}_1}} \right)\), \(T\left( {{{\bf{b}}_2}} \right)\) and \(T\left( {{{\bf{b}}_3}} \right)\) for \(B = \left\{ {1,t,{t^2}} \right\}\).

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

\(\begin{aligned}{c}T\left( {{{\bf{b}}_2}} \right) &= T\left( t \right)\\ &= \left( {\begin{aligned}{ - 1}\\0\\1\end{aligned}} \right)\end{aligned}\)

\(\begin{aligned}{c}T\left( {{{\bf{b}}_3}} \right) &= T\left( {{t^2}} \right)\\ &= \left( {\begin{aligned}1\\0\\1\end{aligned}} \right)\end{aligned}\)

Find \({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_\varepsilon }\), \({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_\varepsilon }\) and \({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_\varepsilon }\).

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

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 \(n = 3\).

\(\begin{aligned}{\left( {T\left( {\bf{x}} \right)} \right)_\varepsilon } &= \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_\varepsilon }}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_\varepsilon }}&{{{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)}_\varepsilon }}\end{aligned}} \right)\\ &= \left( {\begin{aligned}1&{}&{ - 1}&{}&1\\1&{}&0&{}70\\1&{}&1&{}&1\end{aligned}} \right)\end{aligned}\)

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