Questions with Solutions in Previous Midterm Exam | MATH 342, Exams of Linear Algebra

Download Questions with Solutions in Previous Midterm Exam | MATH 342 and more Exams Linear Algebra in PDF only on Docsity! Math 342 Solutions to Previous Midterm Exam 1. True/False If the statement is true, give a brief explanation; if it is false, provide a counterexample. T F A subset H of a vector space V is a subspace if the zero vector is in H. F A subspace must also be closed under addition and scalar multiplication. For example, H = {[ 0 0 ] , [ 1 0 ]} contains the zero vector. However, [ 1 0 ] ∈ H, but[ 1 0 ] + [ 1 0 ] = [ 2 0 ] /∈ H, so H is not a subspace. T F The column space of A is the range of the mapping ~x 7→ A~x. T The column space of A is the span of the columns of A. The range of ~x 7→ A~x is the set of all multiples A~x, where ~x is a vector of the appropriate size. But the span of the columns of A is set of all linear combinations of the columns of A. By the definition of matrix multiplication, each A~x is a linear combination of the columns of A, so the set of all such multiples is the set of all linear combinations of the columns of A. set of all linear combinations of the columns of A T F A single vector by itself is linearly dependent. F In the first place, vectors are not linearly dependent or independent, sets of vectors are. Even if we read the question to say ‘sets consisting of exactly one vector are dependent’, the statement is false. For example, {[ 1 1 ]} is a linearly independent set. T F If there exists a linearly dependent set {~v1, . . . , ~vp} in the vector space V , then dimV ≤ p. F The subset {~0} ⊆ R2 is linearly dependent (any set containing the zero vector is dependent), but dimR2 = 2 6≤ 1. T F The dimension of the row space and the column space of A are the same, even if A is not square. T This is one of the things the Rank Theorem says. Problems 2. If the null space of a 7 × 6 matrix A is 5-dimensional, what is the dimension of the column space of A? Solution: The Rank Theorem says rk(A) + dim Nul(A) = number of columns of A. Here we have rk(A) + 5 = 6, so rk(A) = dim Col(A) = 1. 3. If A is a 7× 5 matrix, what is the largest possible rank of A? Solution: Since rk(A) + dim Nul(A) = number of columns of A, and all the number involved are nonnegative integers, the largest the rank can be is the number of columns of A. The largest rank a 7× 5 matrix can have is 5. 4. Let H be the set of all points in R2 with rational coordinates. That is, H = {(r, s) | r, s ∈ Q}. Is H a subspace of R2? Explain. Solution: No , H is not a subspace of R2. For example, ~v = [ 1 0 ] ∈ H and π is a scalar, but π~v = [ π 0 ] /∈ H. Since it’s not closed under scalar multiplication, H is not a subspace. 5. Let F be a fixed 3 × 2 matrix and let H be the set of all 2 × 3 matrices A such that AF = 0 (the 2× 2 zero matrix). Show that H is a subspace of M2×3. Solution: We must check three things: a) H contains the zero vector. The zero of M2×3 is the 2 zero matrix—let’s call it O2×3. But O2×3F = O2×2, so O2×3 ∈ H b) H is closed under addition. Let A,B ∈ H. Then AF = O and BF = O. Hence, (A+B)F = AF +BF = O +O = O, so A+B ∈ H. c) H is closed under scalar multiplication. Let A ∈ H and let r be a scalar. Since A ∈ H, AF = O. But then (rA)F = r(AF ) = rO = O, so rA ∈ H. Therefore, H is indeed a subspace of M2×3. 6. Let T : M2×2 → M2×2 be T (A) = A − AT . This is a linear transformation. Describe the range and kernel of T . Solution: We can actually write a formula for this transformation. An arbitrary element of M2×2 is A = [ a b c d ] , where a, b, c, d are real numbers. Then T (A) = A−AT = [ a b c d ] − [ a c b d ] = [ 0 b− c c− b 0 ] . This equals zero precisely when b = c. Notice also that c− b = −(b− c). Hence, ker(T ) = {A | T (A) = O} = {[ r s s t ] : r, s, t ∈ R } range(T ) = {T (A) | A ∈M2×2} = {[ 0 z −z 0 ] : z ∈ R } We could also notice that ker(T ) is the set of symmetric matrices and range(T ) is the set of skew-symmetric matrices (in M2×2).