Question

Show that T(⟨x,y⟩)=⟨x+y,x) is a linear transformation, find its matrix, and draw the basic box.

Answer 1

To prove that T(\langle x,y \rangle)=\langle x+y, x \rangle is a linear transformation, we verify its additivity and homogeneity properties, both of which are satisfied. The matrix corresponding to the linear transformation T is [ [1 1] [1 0] ], and we can draw the basic box by applying T to the corners of the unit square.

Step-by-step explanation:

To show that T(\langle x,y \rangle)=\langle x+y, x \rangle is a linear transformation, we must verify two main properties: additivity and homogeneity. A linear transformation T(u + v) should equal T(u) + T(v), and T(cv) should be equal to cT(v) for any vectors u, v, and scalar c.

Let's verify additivity:

• T(\langle x_1, y_1 \rangle + \langle x_2, y_2 \rangle) = T(\langle x_1 + x_2, y_1 + y_2 \rangle)
• = \langle (x_1 + x_2) + (y_1 + y_2), (x_1 + x_2) \rangle
• = \langle x_1 + y_1, x_1 \rangle + \langle x_2 + y_2, x_2 \rangle
• = T(\langle x_1, y_1 \rangle) + T(\langle x_2, y_2 \rangle)

Now let's verify homogeneity:

• T(c\langle x, y \rangle) = T(\langle cx, cy \rangle)
• = \langle cx + cy, cx \rangle
• = c\langle x + y, x \rangle
• = cT(\langle x, y \rangle)

Since both properties hold, T is a linear transformation. The matrix for this transformation can be derived by applying T to the basis vectors:

1. T(\langle 1, 0 \rangle) = \langle 1, 1 \rangle
2. T(\langle 0, 1 \rangle) = \langle 1, 0 \rangle

Therefore, the matrix of T with respect to the standard basis is:

[ [1 1] [1 0] ]

To draw the basic box, we can visualize the transformation applied to the unit square with corners at (0,0), (1,0), (0,1), and (1,1). The transformed square would have its corners at:

• T(0,0) = (0,0)
• T(1,0) = (1,1)
• T(0,1) = (1,0)
• T(1,1) = (2,1)