Determine whether T: R^2 → R^2 defined by T(x, y) = (x + y, 2x + 2y) is one-to-one.

Step 1: Write the standard matrix of T.

From:
T(x, y) = (x + y, 2x + 2y)

the standard matrix is:

A = [[1, 1], [2, 2]]

Step 2: Recall the condition for a linear transformation to be one-to-one.

A linear transformation is one-to-one if the equation:
Av = 0

has only the trivial solution v = 0.

Equivalently, every column of the standard matrix must contain a pivot.

Step 3: Row reduce the matrix.

Start with:
[[1, 1], [2, 2]]

Use:
R2 → R2 - 2R1

The matrix becomes:
[[1, 1], [0, 0]]

Step 4: Examine the pivots.

There is only one pivot, even though the matrix has two columns.

Therefore, there is a free variable.

Step 5: Show that a nonzero vector maps to zero.

From the equation:
x + y = 0

we get:
x = -y

Choose y = 1. Then x = -1.

So the nonzero vector (-1, 1) satisfies:
T(-1, 1) = (0, 0)

Step 6: State the conclusion.

Because a nonzero vector maps to the zero vector, the kernel contains more than just the zero vector.

Therefore, T is not one-to-one.