Question

Let\({{\rm{S}}_{\rm{0}}}\)be the vector space of all sequences of the form\(\left( {{y_0},{y_1},{y_2},...} \right)\)and define linear transformations\(T\)and\({\rm{D}}\)from\({{\rm{S}}_{\rm{0}}}\)into\({{\rm{S}}_{\rm{0}}}\)by

\(T\left( {{y_0},{y_1},{y_2},...} \right) = \left( {{y_1},{y_2},{y_3}....} \right)\)

\(D\left( {{y_0},{y_1},{y_2},...} \right) = \left( {0,{y_0},{y_1},{y_2}....} \right)\)

Show that\(TD = I\)(the identity transformation on\({{\rm{S}}_{\rm{0}}}\)) and yet\(DT \ne I\).

Short answer

The transformation \(TD\) results in the same sequence, whereas the transformation \(DT\) results in a different sequence.

\(TD = I\), whereas \(DT \ne I\).

Step-by-step solution

Compute the combined transformation \(TD\) on the sequence \(\left( {{y_0},{y_1},{y_2},...} \right)\)

Use the given linear transformations of T and D on the sequence to find the combined transformation.

\(\begin{aligned} \left( {TD} \right)\left( {{y_0},{y_1},{y_2},...} \right) &= T\left( {D\left( {{y_0},{y_1},{y_2}....} \right)} \right)\\ &= T\left( {0,\,{y_0},{y_1},{y_2},...} \right)\\ &= \left( {{y_0},{y_1},{y_2},...} \right)\end{aligned}\)

Compute the combined transformation \(DT\) on the sequence \(\left( {{y_0},{y_1},{y_2},...} \right)\)

Use the given linear transformations of T and D on the sequence again to find the combined transformation

\(\begin{aligned} \left( {DT} \right)\left( {{y_0},{y_1},{y_2},...} \right) &= D\left( {T\left( {{y_0},{y_1},{y_2}....} \right)} \right)\\ &= D\left( {{y_1},{y_2},{y_3},...} \right)\\ &= \left( {0,{y_0},{y_1},{y_2},...} \right)\end{aligned}\)

Draw a conclusion

Transformation \(TD\) results in the same sequence, whereas transformation \(DT\) results in a different sequence.

\(TD = I\), whereas\(DT \ne I\).