Question

Let T : R^2 - R^2 and S: R^2 - R^2 be linear operators defined by
T (x1;x2) = (x1+x2, x1-x2) and S(x1;x2) = (x1, x1+2x2)
respectively.
i) Find T ◦ S and S ◦ T.
ii) Let B = f(1, 0), (0, 1) be the standard basis of R^3. Verify that
[T ◦ S]B = [T]B ◦ [S]B.

Answer 1

Answer on Question #44643 – Math – Linear Algebra:

Let $T\colon R^2 \to R^2$ and $S\colon R^2 \to R^2$ be linear operators defined by:

$$\begin{array}{l} T(x_1, x_2) = (x_1 + x_2, x_1 - x_2); \\ S(x_1, x_2) = (x_1, x_1 + 2x_2); \end{array}$$

(a) Find $T\circ S$ and $S\circ T$.

(b) Let $B = \{(1,0)^T, (0,1)^T\}$ be the standard basis of $R^3$. Verify that

$$[T \circ S]B = [T]B \circ [S]B.$$

Solution.

(a)

$$\begin{array}{l} (T \circ S)(x_1, x_2) = T(S(x_1, x_2)) = \\ = T(x_1, x_1 + 2x_2) = (x_1 + (x_1 + 2x_2), x_1 - (x_1 + 2x_2)) = (2x_1 + 2x_2, -2x_2); \\ (S \circ T)(x_1, x_2) = S(T(x_1, x_2)) = \\ = S(x_1 + x_2, x_1 - x_2) = (x_1 + x_2, x_1 + x_2 + 2(x_1 - x_2)) = (x_1 + x_2, 3x_1 - x_2); \end{array}$$

(b)

Denote $e_1 = (0,1)^T, e_2 = (1,0)^T$.

$$\begin{array}{l} Te_1 = (1 + 0, 1 - 0) = (1, 1); \\ Te_2 = (0 + 1, 0 - 1) = (1, -1); \\ Se_1 = (1, 1 + 2 \cdot 0) = (1, 1); \\ Se_2 = (0, 0 + 2 \cdot 1) = (0, 2); \\ (T \circ S)e_1 = (2 \cdot 1 + 2 \cdot 0, -2 \cdot 0) = (2, 0); \\ (T \circ S)e_2 = (2 \cdot 0 + 2 \cdot 1, -2 \cdot 1) = (2, -2); \end{array}$$

Hence:

$$\begin{array}{l} (T \circ S)(x_1, x_2) = \left( \begin{array}{cc} 2 & 2 \\ 0 & -2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right); \\ T(x_1, x_2) = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right); \\ S(x_1, x_2) = \left( \begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right); \\ [T]B \circ [S]B = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array} \right) = \left( \begin{array}{cc} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 2 \\ 1 \cdot 1 + (-1) \cdot 1 & 1 \cdot 0 + (-1) \cdot 2 \end{array} \right) = \\ = \left( \begin{array}{cc} 2 & 2 \\ 0 & -2 \end{array} \right) = [T \circ S]B. \end{array}$$

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