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Math 3013
Problem Set 5
Problems from §2.1 (pgs. 134-136 of text): 1,3,11,12,13,16,
Problems from §2.2 (pgs. 140-141 of text): 1,3,5,7,
Problems from §2.3 (pgs. 152-154 of text): 1,2,3,4,5,7,13,15,19,
- (Problem 2.1.1 in text). Give a geometric criterion for a set of two distinct nonzero vectors in R^2 to be dependent.
- If two vectors v 1 and v 2 are linearly dependent, then there must exist a solution of c 1 v 1 + c 2 v 2 = 0 with at least one of the coefficients c 1 , c 2 not zero. Suppose (without loss of generality) that c 2 6 = 0. Then c 1 can not equal zero either (otherwise we’d have c 2 v 2 = 0 with neither c 2 or v 2 zero). Then we can multiply both sides of this equation by 1/c 2 to obtain c 1 c 2
v 1 + v 2 = 0 ⇒ v 2 = −
c 1 c 2
v 1
So v 2 must be a non-zero scalar multiple of v 1. But then, this implies that v 2 is either parallel (or anti-parallel) to v 1.
- (Problem 2.1.3 in text). Give a geometric criterion for a set of two distinct nonzero vectors in R^3 to be dependent.
- By exactly the same reasoning we used in Problem 1, we can conclude that if two distinct non-zero vectors in R^3 are dependent then they must be parallel (or anti-parallel).
- (Problem 2.1.11 in text). Find a basis for the subspace spanned by the vectors [1, 2 , 1 , 1], [2, 1 , 0 , −1], [− 1 , 4 , 3 , 8], [0, 3 , 2 , 5] ∈ R^4.
- First we form a 4 × 4 matrix A whose columns correspond to the above set of vectors.
A =
Now we row-reduce A to row-echelon form.
R 2 → R 2 − 2 R 1 R 3 → R 3 − R 1 R 4 → R 4 − R 1 −−−−−−−−−−−−−→
R 2 → − 13 R 2
R 3 → R 3 − 23 R 2
R 4 → R 4 − R 2
R 3 ↔ R 4
The pivots of the final matrix (a row-echelon form of A) are in the first three columns. Hence, the first three columns {[1, 2 , 1 , 1], [2, 1 , 0 , −1], [− 1 , 4 , 3 , 8]} of A will form a basis for the column space ColSp(A) = span ([1, 2 , 1 , 1], [2, 1 , 0 , −1], [− 1 , 4 , 3 , 8], [0, 3 , 2 , 5]) 1
- (Problem 2.1.12 in text). Find a basis for the column space of the matrix
A =
- We’ll apply the same technique used in Problem 3.
The pivots in the row-echelon form of A are in the first two columns. Therefore, the first two columns of A {[2, 5 , 1 , 6] , [3, 2 , 7 , −2]} will form a basis for the column space of A.
- (Problem 2.1.13 in text). Find a basis for the row space of the matrix
A =
The row space of A is the span of the row vectors {[1, 3 , 5 , 7] , [2. 0 , 4 , 2] , [3, 2 , 8 , 7]} of A To find a basis for the span of these vectors we arrange them as the columns of a new matrix A′
A′^ =
which happens to be the transpose of our original matrix A. We now row-reduce A′.
=^ H
The pivots of H are contained in the first two columns, therefore the first two columns of A′^ form a basis for the column space of A′, which is indentical to row space of our original matrix A. Thus,
{[1, 3 , 5 , 7] , [2, 0 , 4 , 2]}
is a basis for the row space of A.
- (Problems 2.1.16 and 2.1.23 in text). Determine whether the following sets of vectors are dependent or independent.
(a) {[1, 3] , [− 2 , −6]} in R^2.
- Let v 1 = [1, 3] and v 2 = [− 2 , −6] and Note that 2 v 1 + v 2 = 2[1, 3] + [− 2 , −6] = [0, 0] so c 1 = 2 and c 2 = 1 is a non-trivial solution to c 1 v 1 + c 2 v 2 = 0. Thus, v 1 and v 2 are linearly dependent.
(b) {[1, − 4 , 3] , [3, − 11 , 2] , [1, − 3 , −4]} in R^3.
so the vectors
{[ 3
2 ,^ −^
13 8 ,^1 ,^0
]
[
]}
form a basis for the null space of A. Finally, rank (A) = dim (ColSp (A)) = dim (RowSp (A)) = 2
(b)
A =
- We proceed as in Part (a). The matrix A is row equivalent to the following matrix in row-echelon form
A′^ =
Only third column of A′^ lacks a pivot; so the first, second and fourth columns of A will provide a basis for the column space of A. The matrix A′^ has three non-zero rows and these will provide a basis for the row space of A. Finally, the solution of Ax = 0 will be x 1 − x 3 = 0 x 2 + x 3 = 0 x 4 = 0 0 = 0
=⇒ x =
x 3 −x 3 x 3 0
=^ x^3
so
basis for ColSp (A) =
basis for RowSp (A) = {[1, 0 , − 1 , 0] , [0, 1 , 1 , 0] , [0, 0 , 0 , 1]}
basis for N ullSp (A) =
(c)
A =
- This matrix row-reduces to
H =
The pivots of H lie in the first three columns. Therefore, the first three columns of A will be a basis for the column space of A. Each row of H is non-zero; therefore, each row of H will be a basis vector for the row space of A. The solution set of Ax = 0 coincides with the solution set of Hx = 0 , or 2 x 1 + x 2 + 2x 4 = 0 x 2 + 2x 3 + x 4 = 0 − 3 x 3 − x 4 = 0
x 1 = − 56 x 4 x 2 = − 13 x 4 x 3 = − 13 x 4
So every solution is a vector of the form
x =
− 56 x 4 − 13 x 4 − 13 x 4 x 4
=^ x^4
∈^ span
Thus,
{[
]}
is a basis for the null space of A. The rank of A is equal to the number of basis vectors for the column space of A, which is 3.
- (Problem 2.2.7 in text). Determine whether the following matrix is invertible, by finding its rank.
A =
- This matrix is row equivalent to
H =
The rank of A is equal to the dimension of its row space, which is equal to the number of non-zero rows in a row-echelon form of A. Hence, rank(A) = 3. This implies that the dimension of the null space of A is 1 since
of columns of A =rank (A) − dim (null space of A)
Hence, the matrix is not invertible, since it rank does does equal its number of columns (See Theorem 2.6 in text.)
- (Problem 2.2.11 in text). Determine whether the following statements are true or false.
(a) The number of independent row vectors in a matrix is the same as the number of independent column vectors.
- True. This number is the rank of a matrix.
(b) If H is a row-echelon form of a matrix A, then the nonzero column vectors of H form a basis for the column space of A.
- False. It is the columns of A that corresponding to column vectors of H that contain pivots that form a basis for the column space of A.
(c) If H is a row-echelon form of a matrix A, then the nonzero row vectors of H from a basis for the row space of A.
(d) If an n × n matrix A is invertible then rank (A) = n.
- True. (See Theorem 2.6 in text.)
(e) For every matrix A we have rank(A) > 0.
- False. The rank of a zero matrix will be zero.
(j) For all positive integers m and n, with m ≥ n, the nullity of an m × n matrix might be any number from 0 to n.
- True. In Part (h) we found n − min (m, n) ≤ dim (null space) ≤ n − 0 = n But if m ≥ n, then n − min (m, n) = 0, and so we have 0 ≤ dim (null space) ≤ n
- (Problems 2.3.1, 2.3.2, 2.3.3, and 2.3.4 in text). Determine which of the following mappings are linear transformations.
(a) T : R^3 → R^2 : T ([x 1 , x 2 , x 3 ]) = [x 1 + x 2 , x 1 − 3 x 2 ]
- This mapping is linear since if v = [x 1 , x 2 , x 3 ] T (λv) = T (λ [x 1 , x 2 , x 3 ]) = T ([λx 1 , λx 2 , λx 3 ]) = [λx 1 + λx 2 , λx 1 − 3 λx 2 ] = λ [x 1 + x 2 , x 1 − 3 x 2 ] = λT ([x 1 , x 2 , x 3 ]) = λT (v) (T preserves scalar multiplication) and if v = [x 1 , x 2 , x 3 ] and v′= [x′ 1 , x′ 2 , x′ 3 ] T (v + v′) = T ([x 1 + x′ 1 , x 2 + x′ 2 , x 3 + x′ 3 ]) = [x 1 + x′ 1 + x 2 + x′ 2 , x 1 + x′ 1 − 3(x 2 + x′ 2 )] = [x 1 + x 2 , x 1 − 3 x 2 ] + [x′ 1 + x′ 2 , x′ 1 − 3 x′ 2 ] = T (v) + T (v′) (T preserves vector addition)
(b) T : R^3 → R^4 : T ([x 1 , x 2 , x 3 ]) = [0, 0 , 0 , 0]
- This mapping is linear since if v = [x 1 , x 2 , x 3 ] T (λv) = T ([λx 1 , λx 2 , λx 3 ]) = [0, 0 , 0 , 0] = λ [0, 0 , 0 , 0] = λT ([x 1 , x 2 , x 3 ]) = λT (v) (T preserves scalar multiplication) and if v = [x 1 , x 2 , x 3 ] and v′= [x′ 1 , x′ 2 , x′ 3 ] T (v + v′) = T ([x 1 + x′ 1 , x 2 + x′ 2 , x 3 + x′ 3 ]) = [0, 0 , 0 , 0] = [0, 0 , 0 , 0] + [0, 0 , 0 , 0] = T (v) + T (v′) (T preserves vector addition)
(c) T : R^3 → R^4 : T ([x 1 , x 2 , x 3 ]) = [1, 1 , 1 , 1]
- This mapping is not linear since if v = [x 1 , x 2 , x 3 ]
T (v) = [1, 1 , 1 , 1] T (2v) = [1, 1 , 1 , 1] 6 = 2 [1, 1 , 1 , 1] = 2T (v)
So the mapping does not preserve scalar multiplication.
(d) T : R^2 → R^3 : T ([x 1 , x 2 ]) = [x 1 − x 2 , x 2 + 1, 3 x 1 − 2 x 2 ]
- This mapping is not linear since, e.g., if v = [1, 1]
T (v) = [0, 2 , 1] T (2v) = T ([2, 2]) = [0, 3 , 2] 6 = [0, 4 , 2] = 2T (v)
So the mapping does not preserve scalar multiplication.
- (Problems 2.3.5 and 2.3.7 in text). For each of the following, assume T is a linear transformation, from the data given, compute the specified value.
(a) Given T ([1, 0]) = [3, −1], and T ([0, 1]) = [− 2 , 5], find T ([4, −6]).
- Because linear transformations preserve scalar multiplication and vector addition, they also preserve linear combinations: T (c 1 v 1 + c 2 v 2 ) = c 1 T (v 1 ) + c 2 T (v 2 ) Now take e 1 = [1, 0] and e 2 = [0, 1]. Then
T ([4, −6]) = T (4e 1 − 6 e 2 ) = 4 T (e 1 ) − 6 T (e 2 ) = 4 [3, −1] − 6 [− 2 , 5] = [12 + 12, − 4 − 30] = [24, −34]
(b) Given T ([1, 0 , 0]) = [3, 1 , 2], T ([0, 1 , 0]) = [2, − 1 , 4], and T ([0, 0 , 1]) = [6, 0 , 1], find T ([2, − 5 , 1]).
- As in Part (a), we set e 1 = [1, 0 , 0], e 2 = [0, 1 , 0], and e 3 = [0, 0 , 1] and then compute
T ([2, − 5 , 1]) = T (2e 1 − 5 e 2 + e 3 ) = 2 T (e 1 ) − 5 T (e 2 ) + T (e 3 ) = 2 [3, 1 , 2] − 5 [2, − 1 , 4] + [6, 0 , 1] = [6 − 10 + 6, 2 + 5 + 0, 4 − 20 + 1] = [2, 7 , −15]
- (Problems 2.3.13 and 2.3.15 in text). Find the standard matrix representations of the following linear transformations.
(a) T ([x 1 , x 2 ]) = [x 1 + x 2 , x 1 − 3 x 2 ]
- False. In order to be a linear transformation a function f : Rn^ → Rm^ must preserve scalar multi- plication and vector addtion.
(c) Composition of linear transformations corresponds to multiplication of their standard matrix represen- tations.
(d) Function composition is associative.
(e) An invertible linear transformation mapping Rn^ to itself has a unique inverse.
- True. (This follows from the corresponding theorem about invertible matrices.)
(f) The same matrix may be the standard matrix representation for several different linear transformations.
- False. (Unless one allows more general vector spaces - but idea won’t be broached until Chapter 3.)
(g) A linear transformation having an m × n matrix as its standard matrix representation maps Rn^ into Rm.
(h) If T and T ′^ are different linear transformations mapping Rn^ into Rm, then we may have T (ei) = T ′^ (ei) for all standard basis vectors ei of Rn.
- False. Linear transformations are determined uniquely by their standard matrix representations.
(i) If T and T ′^ are different linear transformations mapping Rn^ into Rm, then we may have T (ei) = T ′^ (ei) for some standard basis vectors ei of Rn.
- True. (So long as they are not all the same.)
(j) If B = {b 1 , b 2 ,... , bn} is a basis for Rn^ and T and T ′^ are linear transformations from Rn^ into Rm, then T (x) = T ′^ (x) for all x ∈ Rn^ if and only if T (bi) = T ′^ (bi) for i = 1, 2 ,... , n.
