Fall 2001  Math 2090  Exam 3

1. Recall that ℝ^{3} denotes the vector space of all ordered 3-tuples of real numbers.

Consider the vectors v⃗_{1} = (1, 2, 1), v⃗_{2} = (2, 7, 3), v⃗_{3} = (−1, 0, 1) in ℝ^{3}.

(a) Can the vector b⃗ = (3, −4, 1) be expressed as a linear combination of v⃗_{1}, v⃗_{2}, and v⃗_{3}?

(b) Does the set of vectors {v⃗_{1}, v⃗_{2}, v⃗_{3}} form a basis for ℝ^{3}? Explain.

(a) Yes, solve Ax⃗ = b⃗, where A = [v⃗_{1} v⃗_{2} v⃗_{3}], to see that b⃗ = (12)v⃗_{1} + (−4)v⃗_{2} + (1)v⃗_{3}.

(b) Yes, {v⃗_{1}, v⃗_{2}, v⃗_{3}} is a linearly independent, spanning set, so is a basis.

Both of these properties follow from the fact that the matrix A = [v⃗_{1} v⃗_{2} v⃗_{3}] has rank three.

2. Let S be the set of vectors x⃗ = (x_{1}, x_{2}, x_{3}) in ℝ^{3} satisfying the condition x_{1} − 3x_{2} + x_{3} = 0.

(a) Show that S is a subspace of ℝ^{3}. (b) Find a basis for S, and determine the dimension of S.

(a) Note that S is non-empty. For instance, 0 = (0, 0, 0) is in S.

S is closed under vector addition:

If u⃗ = (u_{1}, u_{2}, u_{3}) and v⃗ = (v_{1}, v_{2}, v_{3}) are in S, then u_{1} − 3u_{2} + u_{3} = 0 and v_{1} − 3v_{2} + v_{3} = 0. It follows that u⃗ + v⃗ = (u_{1} + v_{1}, u_{2} + v_{2}, u_{3} + v_{3}) is also in S, since (u_{1} + v_{1}) − 3(u_{2} + v_{2}) + (u_{3} + v_{3}) = u_{1} − 3u_{2} + u_{3} + v_{1} − 3v_{2} + v_{3} = 0 + 0 = 0 S is closed under scalar multiplication: If u⃗ is in S as above, and k is a scalar, then ku⃗ = (ku_{1}, ku_{2}, ku_{3}) is in S, since ku_{1} − 3ku_{2} + ku_{3} = k(u_{1} − 3u_{2} + u_{3}) = k(0) = 0

(b) {(3, 1, 0), (−1, 0, 1)} is a basis for S. dim S = 2

3. Recall that M_{2}(ℝ) denotes the vector space of all 2 × 2 matrices with real entries.

Consider the matrices A_{1} = [ \begin{matrix}1 & 2 \\ 0 & 1 \end{matrix} ], A_{2} = [ \begin{matrix}2 & 4 \\ 0 & 1 \end{matrix} ], A_{3} = [ \begin{matrix}3 & 5 \\ 0 & 6 \end{matrix} ] in M_{2}(ℝ).

(a) Determine if the set {A_{1}, A_{2}, A_{3}} is linearly independent. (b) Explain why {A_{1}, A_{2}, A_{3}} does not span M_{2}(ℝ).

(a) This set is linearly independent. Check that the system of equations c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3} = 0 has only the trivial solution c_{1} = c_{2} = c_{3} = 0.

(b) For instance, the matrix [ \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ] can not be expressed as a linear combination of A_{1}, A_{2}, and A_{3}.

4. In parts (a) and (b), determine if the mapping T : ℝ^{2} → ℝ^{2} defines a linear transformation.

(a) T (x_{1}, x_{2}) = (x_{1} + 2x_{2}, 3x_{1}) (b) T (x_{1}, x_{2}) = (x_{1} + 2x_{2}, 3 + x_{1})

(c) Let T : ℝ^{2} → ℝ^{2} be a linear transformation satisfying T (1, 0) = (0, 3) and T (0, 1) = (1, −2). Find T (−1, 3), and determine the matrix of T.

(a) This mapping is a linear transformation. Check that T (x⃗ + y⃗) = T (x⃗) + T (y⃗) and T (kx⃗) = kT (x⃗).

(b) This is not a linear transformation. Neither of the two conditions above holds.

(c) T (−1, 3) = (3, −9). The matrix of T is [ \begin{matrix}0 & 1 \\ 3 & −2 \end{matrix} ].

5. Let T : ℝ^{3} → ℝ^{2} be the linear transformation defined by T (x⃗) = Ax⃗, where A = [ \begin{matrix}2 & −2 & 4 \\ 1 & −1 & 2 \end{matrix} ].

(a) Find a basis for Rng(T). (b) Find the dimension of Ker(T). (c) Do the columns of A span ℝ^{2}? Explain.

(a) {(2, 1)} is a basis for Rng(T).

(b) dim Ker(T) = 2.

(c) The columns of A do not span ℝ^{2}. Ax⃗ = b⃗ is consistent only when b⃗ = (b_{1}, b_{2}) satisfies b_{1} = 2b_{2}.