Which ones of the following linear transformations are NOT invertible? (Choose one or more)

Question

BlackTom AI · Accepted Answer

Invertibility of a linear map A: R^2 -> R^2 requires A to be one-to-one (and onto), equivalently the two column vectors v1 = A(e1) and v2 = A(e2) must be linearly independent (rank 2). Geometrically the unit square [0,1]×[0,1] is mapped to the parallelogram with vertices 0, v1, v2, v1+v2. If the area (absolute determinant det([v1 v2])) is nonzero the map is invertible; if det = 0 the image has lower dimension (a line or point) and A is not invertible.

Step-by-step reasoning and calculations:

1) Option (a): vertices given are 0, (2,0), (1,1), (1,-1). Choose v1=(1,1) and v2=(1,-1). Then v1+v2 = (2,0), matching the four vertices as 0, v1, v2, v1+v2. Compute determinant:
det([v1 v2]) = det([[1,1],[1,-1]]) = (1)(-1) - (1)(1) = -1 - 1 = -2.
Since det ≠ 0, rank = 2, so A is invertible. Option (a) is invertible.

2) Option (b): vertices 0, (2,0), (0,5), (2,5) correspond to v1=(2,0), v2=(0,5) (then v1+v2=(2,5)). Compute determinant:
det([[2,0],[0,5]]) = 2·5 - 0·0 = 10 ≠ 0.
Again invertible (rank 2). Option (b) is invertible.

3) Option (c): image is the line segment from 0 to (1,3). That means every image vector is collinear with (1,3); v1 = a(1,3), v2 = b(1,3) for some scalars a,b. Columns are proportional so det([v1 v2]) = 0, rank = 1. The map collapses 2D to a 1D line, so not invertible. Option (c) is NOT invertible.

4) Option (d): image is the single point 0, so v1 = v2 = 0. det([0 0]) = 0, rank = 0. The map is the zero map, not invertible. Option (d) is NOT invertible.

Why other options are incorrect: (a) and (b) give nonzero-area parallelograms (determinants -2 and 10), so they preserve two-dimensionality and are invertible. Options (c) and (d) collapse dimension, so non-invertible.

Conclusion: The non-invertible transformations are c and d, which matches the provided answer.

Which ones of the following linear transformations are NOT invertible? (Choose one or more)多项选择题

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