Let \(a\) and \(b\) be nonzero numbers. Show that the mapping \(T\) defined by \(T\left\{ {{y_k}} \right\} = \left\{ {{w_k}} \right\}\), where \({w_k} = {y_{k + 2}} + a{y_{k + 1}} + b{y_k}\) is a linear transformation from S into S.

If a vector is linear, then it must follow the distributive property, that is, \(T\left( {\left\{ {{y_k}} \right\} + \left\{ {{z_k}} \right\}} \right) = T\left\{ {{y_k}} \right\} + T\left\{ {{z_k}} \right\}\) and associative property of vectors, that is, \(T\left( {r\left\{ {{y_k}} \right\}} \right) = rT\left\{ {{y_k}} \right\}\). This is true for vectors \(\left\{ {{y_k}} \right\}\) and \(\left\{ {{z_k}} \right\}\) in S and a scalar quantity \(r\).

As \({w_k} = {y_{k + 2}} + a{y_{k + 1}} + b{y_k}\),

\(\begin{aligned} T\left\{ {{y_k}} \right\} &= \left\{ {{w_k}} \right\}\\ &= {y_{k + 2}} + a{y_{k + 1}} + b{y_k}.\end{aligned}\)

To check the distributive property, replace \(\left\{ {{y_k}} \right\}\)by \(\left\{ {{y_k}} \right\} + \left\{ {{z_k}} \right\}\)and simplify as shown below:

\(\begin{aligned} T\left( {\left\{ {{y_k}} \right\} + \left\{ {{z_k}} \right\}} \right) &= T\left( {{y_k} + {z_k}} \right)\\ &= {y_{k + 2}} + {z_{k + 2}} + a\left( {{y_{k + 1}} + {z_{k + 2}}} \right) + b\left( {{y_k} + {z_k}} \right)\\ &= \left( {{y_{k + 2}} + a{y_{k + 1}} + b{y_k}} \right) + \left( {{z_{k + 2}} + a{z_{k + 1}} + b{z_k}} \right)\\ &= T\left\{ {{y_k}} \right\} + T\left\{ {{z_k}} \right\}\end{aligned}\)

To check the associative property, replace \(\left\{ {{y_k}} \right\}\)by \(\left\{ {r{y_k}} \right\}\)and simplify as shown below:

\(\begin{aligned} T\left( {r\left\{ {{y_k}} \right\}} \right) &= T\left( {r{y_k}} \right)\\ &= r{y_{k + 2}} + a\left( {r{y_{k + 1}}} \right) + b\left( {r{y_k}} \right)\\ &= r\left( {{y_{k + 2}} + a{y_{k + 1}} + b{y_k}} \right)\\ &= rT\left\{ {{y_k}} \right\}\end{aligned}\)

As vector \(T\) satisfies both properties, the given mapping is a linear transformation from S into S.