Fall 2000  Math 2090  Exam 3

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

(a) Can the vector b⃗ = (2, 3, 9, 4) be expressed as a linear combination of {v⃗_{1}, v⃗_{2}, v⃗_{3}}?

In other words, is b⃗ in the subspace span{v⃗_{1}, v⃗_{2}, v⃗_{3}}? The system c_{1}(1 − 1, 0, 2) + c_{2}(−1, 2, 1, −2) + c_{3}(0, 1, 3, 0) = (2, 3, 9, 4) is consistent. So b⃗ = (2, 3, 9, 4) = a(1 − 1, 0, 2) + b(−1, 2, 1, −2) + c(0, 1, 3, 0) is in span{v⃗_{1}, v⃗_{2}, v⃗_{3}}.

(b) Determine if the set of vectors {v⃗_{1}, v⃗_{2}, v⃗_{3}} is linearly independent in ℝ^{4}. The system c_{1}(1 − 1, 0, 2) + c_{2}(−1, 2, 1, −2) + c_{3}(0, 1, 3, 0) = (0, 0, 0, 0) has only the trivial solution. So this set of vectors is linearly independent.

(c) Explain why the set {v⃗_{1}, v⃗_{2}, v⃗_{3}} does not form a basis for ℝ^{4}.

A basis for ℝ^{4} must consist of 4 vectors since dim ℝ^{4} = 4.

2. Let P_{3} be the vector space of all polynomials (in the variable x) of degree less than 3 with real coefficients, P_{3} = {ax^{2} + bx + c : a, b, c real}. Consider the polynomials p_{1} = 1 + x, p_{2} = 1 + x^{2}, p_{3} = x + x^{2} in P_{3}. Determine if the set {p_{1}, p_{2}, p_{3}} forms a basis for P_{3}.

This set is a basis for P_{3}. Check that c_{1}(1 + x) + c_{2}(1 + x^{2}) + c_{3}(x + x^{2}) = ax^{2} + bx + c has a unique solution {c_{1}, c_{2}, c_{3}} for any a, b, c by comparing coefficients of 1 = x^{0}, x^{1}, x^{2}.

3. 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_{2} − x_{1}, x_{1})

This mapping is a linear transformation. Check that T (x⃗ + y⃗) = T (x⃗) + T (y⃗) and T (cx⃗) = cT (x⃗) for any x⃗, y⃗, and c.

(b) T (x_{1}, x_{2}) = (x_{2} − 1, x_{1})

This mapping is not a linear transformation. For instance, T (x⃗ + y⃗) = (x_{2} + y_{2} − 1, x_{1} + y_{1}) ≠ (x_{2} + y_{2} − 2, x_{1} + y_{1}) = T (x⃗) + T (y⃗).

(c) If T : ℝ^{2} → ℝ^{2} is a linear transformation and T (1, 0) = (2, 3) and T (0, 1) = (−1, 1), what is T (4, 2)?

T (4, 2) = T (4(1, 0) + 2(0, 1)) = 4T (1, 0) + 2T (0, 1) = 4(2, 3) + 2(−1, 1) = (6, 14)

4. Consider the matrix A = [\begin{matrix}1 & −3 & 2 \\ −3 & 9 & −6 \end{matrix} ] and the linear transformation T : ℝ^{3} → ℝ^{2} defined by T (x⃗) = Ax⃗.

(a) Find a basis for Rng(T), the range of the linear transformation T.

What is the dimension of Rng(T)?

Rng(T) = {b⃗ in ℝ^{2} : Ax⃗ = b⃗ is consistent} = {b⃗ in ℝ^{2} : b_{2} + 3b_{1} = 0}

= {(b_{1}, −3b_{1})} = span{(1, −3)} dim Rng(T) = 1

(b) Find the dimension of Ker(T), the kernel of the linear transformation T.

Ker(T) = {x⃗ in ℝ^{3} : Ax⃗ = 0} = {x⃗ in ℝ^{3} : x_{1} − 3x_{2} + 2x_{3} = 0}

= {(3x_{2} − 2x_{3}, x_{2}, x_{3})} = span{(3, 1, 0), (−2, 0, 1)} dim Ker(T) = 2

(c) Do the columns of the matrix A span ℝ^{2}? Explain.

They do not. Only vectors on the line span{(1, −3)} can be expressed as linear combinations of the columns of A.

5. Consider the matrix A = [\begin{matrix}3 & −2 & 1 \\ 0 & 2 & 0 \\ −2 & 4 & 0 \end{matrix} ].

(a) One eigenvalue of A is λ = 2. Find the eigenvectors corresponding to this eigenvalue. The eigenvectors corresponding to λ = 2 are the non-zero solutions of (A − 2I) v⃗ = 0. These vectors are of the form v⃗ = (2s − t, s, t), where s and t are not both zero.

(b) Find all of the eigenvalues of the matrix A.

The eigenvalues of A are the roots of the characteristic polynomial p(λ) = det(A − λI). Expand along row two to get p(λ) = (2 − λ)(λ^{2} − 3λ + 2) = (2 − λ)(λ − 2)(λ − 1). So the eigenvalues of A are λ_{1} = λ_{2} = 2 and λ_{3} = 1.

(c) Can the answer to part (b) be used to determine if the columns of the matrix A are linearly independent?

Explain.

It can. The columns of A are linearly dependent if and only if Ax⃗ = 0 has a non-trivial solution. In other words, A has dependent columns if and only if there is a non-zero vector x⃗ so that Ax⃗ = 0x⃗. So A has dependent columns if and only if 0 is an eigenvalue of A.