Question

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)

Short answer

The value of\({\left( {T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)} \right)_B}\)is \(\left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\), and \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right) = 24{{\bf{b}}_1} - 20{{\bf{b}}_2} + 11{{\bf{b}}_3}\).

Step-by-step solution

The vector form

Let \({\left( {\bf{x}} \right)_B} = 3{{\bf{b}}_1} - 4{{\bf{b}}_2}\), which can also be written as \({\left( {\bf{x}} \right)_B} = 3{{\bf{b}}_1} - 4{{\bf{b}}_2} + 0{{\bf{b}}_3}\).

Write \({\left( {\bf{x}} \right)_B}\) in the vector form.

\({\left( {\bf{x}} \right)_B} = \left( {\begin{aligned}3\\{ - 4}\\0\end{aligned}} \right)\)

Linear transformation forms \(V\) into \(V\) 

The \(B\)-matrix for \(T\) or a matrix for \(T\) relative to \(B\) for \(T:V \to V\) is given by,

\({\left( {T\left( {\bf{x}} \right)} \right)_B} = {\left( T \right)_B}{\left( {\bf{x}} \right)_B}\)

Find linear transformation form \(V\) into \(V\)

The given matrix is \({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\).

Write\({\left( {T\left( {\bf{x}} \right)} \right)_B} = {\left( T \right)_B}{\left( {\bf{x}} \right)_B}\).

\(\begin{aligned}{\left( {T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)} \right)_B} &= {\left( T \right)_B}{\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)_B}\\ &= \left( {\begin{aligned}0&{ - 6}&1\\0&5&{ - 1}\\1&{ - 2}&7\end{aligned}} \right)\left( {\begin{aligned}3\\{ - 4}\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{0\left( 3 \right) + \left( { - 6} \right)\left( { - 4} \right) + 1\left( 0 \right)}\\{0\left( 3 \right) + \left( 5 \right)\left( { - 4} \right) + \left( { - 1} \right)\left( 0 \right)}\\{1\left( 3 \right) + \left( { - 2} \right)\left( { - 4} \right) + 7\left( 0 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\end{aligned}\)

So, the required matrix is \(\left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\).

Thus, \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right) = 24{{\bf{b}}_1} - 20{{\bf{b}}_2} + 11{{\bf{b}}_3}\).