Answer to Question #86543 in Linear Algebra for RAKESH DEY

Question

Let T:R^3 be defined by
T(x1,x2,x3)=( x1+x3,x3,x2-x3).
Is T invertible? If yes,find a rule for T^-1 like the one which defines T. If T is not invertible, check whether T satisfies Cayley-Hamiltion theorem.

Answer 1

Solution:

Let $e_1 = (1, 0, 0)$, $e_2 = (0, 1, 0)$, $e_3 = (0, 0, 1)$ be the natural basis of $\R^3$. We have $T \left( e_1 \right) = e_1$, $T \left( e_2 \right) = e_3$, and $T \left( e_3 \right) = e_1 + e_2 - e_3$. The matrix of this transformation with respect to the introduced basis is then

$$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{pmatrix} \, .$$

This matrix is non-degenerate (its determinant is equal to -1), therefore, T is invertible. To find the inverse, we observe that $T^{-1} \left( e_1 \right) = e_1$, $T^{-1} \left( e_3 \right) = e_2$, and $T^{-1} \left( e_1 + e_2 - e_3 \right) = e_3$. From the last equation, we have $T^{-1} \left( e_2 \right) = e_3 - T^{-1} \left( e_1 \right) + T^{-1} \left( e_3 \right) = - e_1 + e_2 + e_3$. Then, since $\left( x_1 , x_2 , x_3 \right) = x_1 e_1 + x_2 e_2 + x_3 e_3$, we obtain

$$T^{-1} \left( x_1 , x_2 , x_3 \right) = T^{-1} \left( x_1 e_1 + x_2 e_2 + x_3 e_3 \right) \\ {} = x_1 T^{-1} \left( e_1 \right) + x_2 T^{-1} \left( e_2 \right) + x_3 T^{-1} \left( e_3 \right) \\ {} = x_1 e_1 + x_2 \left( - e_1 + e_2 + e_3 \right) + x_3 e_2 \\ {} = \left( x_1 - x_2 , x_2 + x_3 , x_2 \right) \, .$$

Answer: T is invertible, with the rule $T^{-1} \left( x_1 , x_2 , x_3 \right) = \left( x_1 - x_2 , x_2 + x_3 , x_2 \right)$.