MATH 2090  Answers to the Review Problems for Exam 3  Fall 2001

1. Consider the vectors v⃗_{1} = (1, 3, 4), v⃗_{2} = (0, 1, 1), v⃗_{3} = (−1, 0, −1), v⃗_{4} = (5, 3, −2) in ℝ^{3}.

(a) Does the set of vectors {v⃗_{1}, v⃗_{2}, v⃗_{3}, v⃗_{4}} span ℝ^{3}? (b) Is this set of vectors is linearly independent? This set of vectors does span ℝ^{3} No. For instance, v⃗_{1} − 3v⃗_{2} + v⃗_{3} = 0

(c) Does this set of vectors form a basis for ℝ^{3}? Explain.  No, it is not a linearly independent set.

2. Recall that the trace of a square matrix is the sum of the diagonal entries. Let S be the set of all 2 × 2 matrices with real entries and zero trace: S = \{[ \begin{matrix}a & b \\ c & d \end{matrix} ] : a, b, c, d real, and a + d = 0\} .

(a) Show that S is a subspace of 𝕄_{2}(ℝ), the vector space of all 2 × 2 matrices with real entries.

Show that S is closed under vector addition and scalar multiplication:

Let A = [ \begin{matrix}a & b \\ c & d \end{matrix} ] and B = [ \begin{matrix}x & y \\ z & w \end{matrix} ] be in S. Then a + d = 0, x + w = 0, and A + B = [ \begin{matrix}a + x & b + y \\ c + z & d + w \end{matrix} ].

Now A + B is in S if (a + x) + (d + w) = 0. But (a + x) + (d + w) = a + d + x + w = 0 + 0 = 0. So A + B is in S, and S is closed under vector addition. If A = [ \begin{matrix}a & b \\ c & d \end{matrix} ] is in S and k is a scalar, then a + d = 0 and kA = [ \begin{matrix}ka & kb \\ kc & kd \end{matrix} ]. Now kA is in S if ka + kd = 0. But ka + kd = k(a + d) = k · 0 = 0. So A + B is in S, and S is closed under scalar multiplication.

(b) Verify that \{[ \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ] , [ \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ] , [\begin{matrix}−1 & 1 \\ 0 & 1 \end{matrix} ]\} is a basis for S.

Write A_{1} = [ \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ], A_{2} = [ \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ], A_{3} = [\begin{matrix}−1 & 1 \\ 0 & 1 \end{matrix} ], 0 = [ \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ], and A = [ \begin{matrix}a & b \\ c & d \end{matrix} ].

Check that c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3} = 0 has only the trivial solution c_{1} = 0, c_{2} = 0, c_{3} = 0 to see that {A_{1}, A_{2}, A_{3}} is a linearly independent set. Check that c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3} = A is consistent whenever a + d = 0 to see that {A_{1}, A_{2}, A_{3}} spans S. Note that these are systems of four linear equations in three unknowns.

(c) Find the components of the matrix A = [ \begin{matrix}1 & 2 \\ 1 & −1 \end{matrix} ] with respect to the basis in part (b).

In other words, express the matrix A as a linear combination of the matrices given in part (b).

A = (2)A_{1} + (−1)A_{2} + (1)A_{3}.

3.(a) Consider the mapping T : ℝ^{3} → ℝ^{3} defined by T (x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2}, x_{3}, k), where k is a scalar. Find all values of k for which T is a linear transformation. Find the matrix of T for these values of k.

T is a linear transformation if k = 0. If k = 0, the matrix of T is [ \begin{matrix}1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ].

(b) If T : ℝ^{3} → ℝ is a linear transformation and T (1, 0, 0) = 1, T (1, 1, 0) = 3, T (1, 1, 1) = 5, find T (2, 4, 6).

Since (2, 4, 6) = −2(1, 0, 0) − 2(1, 1, 0) + 6(1, 1, 1), T (2, 4, 6) = −2T (1, 0, 0) − 2T (1, 1, 0) + 6T (1, 1, 1) = 22.

4. Consider the linear transformation T : ℝ^{3} → ℝ^{3} defined by T (x⃗) = Ax⃗, where A = [\begin{matrix}2 & 0 & −1 \\ 2 & 3 & 1 \\ −4 & 0 & 2 \end{matrix} ].

(a) Find a basis for Ker(T), the kernel of the linear transformation T. What is the dimension of Ker(T)? {(3, −4, 6)} is a basis for Ker(T). dim Ker(T) = 1.

(b) Find a basis for Rng(T), the range of the linear transformation T. What is the dimension of Rng(T)? {(2, 2, −4), (0, 3, 0)} is a basis for Rng(T). dim Rng(T) = 2.

(c) Do the rows of the matrix A form a basis for ℝ^{3}? Explain.  No. e.g., they are linearly dependent.