Question

Let T : R^2 -> R^2 and S: R^2 -> R^2 be linear operators defined by
T ( x(subscript1) , x(subscript2) ) = (x(subscript1) + x(subscript2) , x(subscript1) - x(subscript2)) and S( x(subscript1) , x(subscript2) ) = ( x(subscript1) , x(subscript1) + 2x(subscript2) )
respectively.
i) Find ToS and SoT.
ii) Let B ={ (1;0) , (0;1) } be the standard basis of R^3. Verify that
[ToS](subscriptB) = [T](subscriptB) o [S](subscriptB).

Answer 1

Answer on Question #46067 – Math - Linear Algebra

Problem.

Let $T: \mathbb{R}^{\wedge}2 \to \mathbb{R}^{\wedge}2$ and $S: \mathbb{R}^{\wedge}2 \to \mathbb{R}^{\wedge}2$ be linear operators defined by

$T(x(\text{subscript}1), x(\text{subscript}2)) = (x(\text{subscript}1) + x(\text{subscript}2), x(\text{subscript}1) - x(\text{subscript}2))$ and

$S(x(\text{subscript}1), x(\text{subscript}2)) = (x(\text{subscript}1), x(\text{subscript}1) + 2x(\text{subscript}2))$

respectively.

i) Find ToS and SoT.

ii) Let $B = \{(1;0), (0;1)\}$ be the standard basis of $\mathbb{R}^{\wedge}3$. Verify that

[ToS](subscriptB) = [T](subscriptB) o [S](subscriptB).

Solution:

$$T \left( \begin{array}{c} x _ {1} \\ x _ {2} \end{array} \right) = \left( \begin{array}{c} x _ {1} + x _ {2} \\ x _ {1} - x _ {2} \end{array} \right) \text{ and } S \left( \begin{array}{c} x _ {1} \\ x _ {2} \end{array} \right) = \left( \begin{array}{c} x _ {1} \\ x _ {1} + 2 x _ {2} \end{array} \right), \text{ so } T = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \text{ and } S = \left( \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right).$$

i) $TS = \left( \begin{array}{cc}1 & 1\\ 1 & -1 \end{array} \right)\left( \begin{array}{cc}1 & 1\\ 0 & 2 \end{array} \right) = \left( \begin{array}{cc}1 & 3\\ 1 & -1 \end{array} \right)$ and $ST = \left( \begin{array}{cc}1 & 1\\ 0 & 2 \end{array} \right)\left( \begin{array}{cc}1 & 1\\ 1 & -1 \end{array} \right) = \left( \begin{array}{cc}2 & 0\\ 2 & -2 \end{array} \right).$

ii) $S\left( \begin{array}{c}1\\ 0 \end{array} \right) = \left( \begin{array}{cc}1 & 1\\ 0 & 2 \end{array} \right)\left( \begin{array}{c}1\\ 0 \end{array} \right) = \left( \begin{array}{c}1\\ 0 \end{array} \right), T\left( \begin{array}{c}1\\ 0 \end{array} \right) = \left( \begin{array}{cc}1 & 1\\ 1 & -1 \end{array} \right)\left( \begin{array}{c}1\\ 0 \end{array} \right) = \left( \begin{array}{c}1\\ 1 \end{array} \right), \text{ so } T(S)\left( \begin{array}{c}1\\ 0 \end{array} \right) = \left( \begin{array}{c}1\\ 1 \end{array} \right).$

$$TS \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 3 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} 1 \\ 1 \end{array} \right) = T(S) \left( \begin{array}{c} 1 \\ 0 \end{array} \right).$$ $$S \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ 0 & 2 \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} 1 \\ 2 \end{array} \right), T \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} 1 \\ 2 \end{array} \right) = \left( \begin{array}{c} 3 \\ -1 \end{array} \right), \text{ so } T(S) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} 3 \\ -1 \end{array} \right)$$ $$TS \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 3 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} 3 \\ -1 \end{array} \right) = T(S) \left( \begin{array}{c} 0 \\ 1 \end{array} \right).$$

Hence $[TS]_B = [T]_B[S]_B$, as the left and the right operators transform basis to the same vectors.

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