Consider T:R² → R² be the linear operator given by the formula T(x,y)=(x+ky,-y).

Show that T is one-one for every real value of k and that $T^{-1}=T.$

Question

Consider T:R² → R² be the linear operator given by the formula T(x,y)=(x+ky,-y).

Show that T is one-one for every real value of k and that [latex]T^{-1}=T.[/latex]

Accepted Answer

Here, the liner operator T(x,y)=(x+ky,-y) is given.

The standard matrix for this operator is [latex] T=\left[{\begin{array}{c c}{1}&{k}\ {0}&{-1}\end{array}} ight]\!\!.[/latex]

By Theorem 3.1, Ker(T)={0}, because [latex]|T|=−1≠0[/latex] for any value of k . T(X)=0 has a unique solution.

Therefore, T is the one-one transformation for any value of k .

[latex]T^{-1}={\frac{\mathrm{adj}T}{|T|}}={\frac{1}{-1}}{\left[\begin{array}{c c}{-1}&{-k}\ {0}&{1}\end{array} ight]} \ \qquad \ \ =\left[{\begin{array}{c c}{1}&{k}\ {0}&{-1}\end{array}} ight] \ \qquad =T.[/latex]

Hence, the proof.

Question 3.7: Consider T:R² → R² be the linear operator given by the formu......

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