Module 5 Practice: Linear Transformations
10 problems covering definitions, standard matrices, kernel, range, and the Rank-Nullity Theorem.
Problem 1
Verify whether T(x, y) = (2x + y, x - 3y) is a linear transformation.
Solution
Check T(u + v) = T(u) + T(v) and T(cu) = cT(u). Let u = (x1, y1), v = (x2, y2).
T(u + v) = (2(x1+x2) + (y1+y2), (x1+x2) - 3(y1+y2)) = (2x1+y1, x1-3y1) + (2x2+y2, x2-3y2) = T(u) + T(v).
T(cu) = (2cx1 + cy1, cx1 - 3cy1) = c(2x1+y1, x1-3y1) = cT(u). Yes, T is linear.
Problem 2
Is T(x, y) = (x + 1, 2y) a linear transformation?
Solution
T(0, 0) = (0 + 1, 0) = (1, 0) which is not the zero vector. A linear transformation must map zero to zero. Therefore T is not linear.
Problem 3
Find the standard matrix for the linear transformation T(x, y) = (x + 2y, 3x - y, x).
Solution
Compute T(e1) and T(e2): T(1,0) = (1, 3, 1), T(0,1) = (2, -1, 0).
Standard matrix A = [1 2; 3 -1; 1 0]. This is the 3x2 matrix whose columns are T(e1) and T(e2).
Problem 4
Find the standard matrix for a 90-degree counterclockwise rotation in R^2.
Solution
T(e1) = T(1,0) = (0,1) and T(e2) = T(0,1) = (-1,0).
Standard matrix A = [0 -1; 1 0].
Check: A*(1,0)^T = (0,1)^T and A*(0,1)^T = (-1,0)^T. Correct.
Problem 5
Let A = [1 2 3; 0 1 1]. Find the kernel of the transformation T(x) = Ax.
Solution
Solve Ax = 0. Row reduce [1 2 3; 0 1 1] → [1 0 1; 0 1 1]. Free variable: x3 = t.
x1 = -t, x2 = -t, x3 = t. Kernel = span{(-1, -1, 1)}. Nullity = 1.
Problem 6
For the matrix in Problem 5, find the range of T.
Solution
The range is the column space of A. The pivot columns (columns 1 and 2) form a basis for the range.
Range = span{(1, 0), (2, 1)} = R^2. Rank = 2.
Problem 7
Verify the Rank-Nullity Theorem for A = [1 2 3; 0 1 1].
Solution
From Problems 5 and 6: rank = 2, nullity = 1. Number of columns n = 3.
rank + nullity = 2 + 1 = 3 = n. The theorem is verified.
Problem 8
T : R^4 → R^3 has rank 2. What is the nullity? Is T onto? Is T one-to-one?
Solution
By Rank-Nullity: nullity = 4 - 2 = 2.
T is not onto because rank = 2 < 3 (the codomain dimension).
T is not one-to-one because nullity = 2 > 0 (the kernel is nontrivial).
Problem 9
Find the standard matrix for the projection onto the x-axis in R^2, then find its kernel and range.
Solution
T(x,y) = (x, 0). Standard matrix: A = [1 0; 0 0].
Kernel: solve Ax = 0 gives x = 0 free, y = t. Kernel = span{(0,1)} = the y-axis.
Range: column space = span{(1,0)} = the x-axis. Rank = 1, nullity = 1, and 1+1 = 2 columns.
Problem 10
Let T : R^3 → R^3 have standard matrix A = [1 0 -1; 2 1 0; 0 1 2]. Find the kernel of T.
Solution
Row reduce [1 0 -1; 2 1 0; 0 1 2]. R2 = R2 - 2R1: [1 0 -1; 0 1 2; 0 1 2]. R3 = R3 - R2: [1 0 -1; 0 1 2; 0 0 0].
Free variable x3 = t. x2 = -2t, x1 = t. Kernel = span{(1, -2, 1)}. Nullity = 1, rank = 2.