Worksheet 3

Due 10/20

1. During the October 13th lecture, I wrote down many statements equivalent to “S is a linearly independent set”. Do the same for “S is a spanning set”. The answer is in the notes but see what you can do from memory.

ANSWER: See notes

2. Let T : ℝ² → ℝ³ be a linear transformation. We know that there exists a matrix A such that T(x) = Ax.

   • Suppose we know that T(1, 0) = (2, 3, 4) and T(0, 1) = ( − 1, 2, 1). Can we determine A? If so, what is it? If not, why not?
   • Suppose instead we know that T(1, 0) = (2, 3, 4) and T(2, 0) = (4, 6, 8). Can we determine A? If so, what is it? If not, why not?
   • Suppose instead we know that T(1, 0) = (2, 3, 4) and T(1, 1) = ( − 1, 2, 1). Can we determine A? If so, what is it? If not, why not?
   • Suppose instead we know that T(x) = u and T(y) = v. Under what conditions on x and y, can we determine A?

ANSWER:

• Yes. The columns of the matrix are (2, 3, 4) and ( − 1, 2, 1).
• No. We don’t know what T(0, 1) is. There are infinitely many possibilities for A. Just set T(0, 1) to be whatever you like.
• Yes. We need to determine what T(0, 1) is. But (0, 1) = (1, 1) − (1, 0). So by linearity, T(0, 1) = T(1, 1) − T(1, 0) = (2, 3, 4) − ( − 1, 2, 1) = (3, 1, 3).
• When x, y are spanning (which is equivalent to linearly independent here!). More on this later.

3. Come up with a linear transform that is:

   • One-to-one and onto
   • One-to-one but not onto
   • Onto but not one-to-one
   • Not one-to-one nor onto

ANSWER:

• T(x, y) = (x, y).
• T(x) = (x, 0).
• T(x, y) = x.
• T(x, y) = (0, 0).

4. Is differentiation a linear transformation? The answer is yes. I just want you to think about why this is true.

5. Do a full exam from the exam archive here under test like conditions.