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Math 327 Exam 4 - Practice Problems
- Let V = P 3 and let S = {t^3 − 1 , t^2 + 1, t + 1}. Determine whether or not each of the following are in the span of S: (a) 3t^3 + 4t^2 − 2 t − 1 We begin by translating this problem into matrix form as follows. Notice that each vector in the set S corresponds to a column in the matrix, and that we are looking at the form of an arbitrary element in span S:
1 0 0 a 0 1 0 b 0 0 1 c − 1 1 1 d
r^4 +^ r^1 →^ r^4
1 0 0 a 0 1 0 b 0 0 1 c 0 1 1 a + d
r^4 −^ r^2 →^ r^4
1 0 0 a 0 1 0 b 0 0 1 c 0 0 1 a + d − b
r^4 +^ r^1 →^ r^4
From this, we see that in order for an element to be in span S, it must satisfy the equation a + d − b = c or a + d = b + c. Notice that for 3t^3 + 4t^2 − 2 t − 1, a = 3, b = 4, c = −2, and d = −1. Then a + d = 2 and b + c = 2, hence 3 t^3 + 4t^2 − 2 t − 1 is in span S. (b) 5t^3 + 3t^2 − 2 t − 3 Using the same condition as above, notice that for 5t^3 + 3t^2 − 2 t − 3, a = 5, b = 3, c = −2, and d = −3. Then a + d = 2 and b + c = 1, hence 5t^3 + 3t^2 − 2 t − 3 is not in span S.
- Let V = M 22 and let S =
{[ 2 − 1
]
[ 1
]
[ 0
]}
. Determine whether or not each of the following are in the span of S: (a)
[ 7 − 1
]
We will once again use matrices to solve this problem. The general matrix associated with span S is as follows:
2 1 0 a − 1 0 1 b 0 − 1 1 c 1 2 0 d
r 2 r^1 ↔↔r 4 r 3
1 2 0 d 0 − 1 1 c − 1 0 1 b 2 1 0 a
r^ r^4 +2r^3 →r^4 3 +^ r 1 →^ r 3
1 2 0 d 0 − 1 1 c 0 2 1 b + d 0 1 2 a + 2b
r^ r^1 −r^3 →r^1 4 +^ r 2 →^ r 4
1 0 − 1 −b 0 − 1 1 c 0 2 1 b + d 0 0 3 a + 2b + c
r − 3 +2r r 2 →r 3 2 →^ r 2
1 0 − 1 −b 0 1 − 1 −c 0 0 3 b + 2c + d 0 0 3 a + 2b + c
From this, we see that in order to be in span S, we must have b + 2c + d = a + 2b + c. That is, a + b = c + d. Notice that for the matrix above, a = 7, b = −1, c = 4 and d = 5. So a + b = 6 and c + d = 9, so this matrix is not in span S.
(b)
[ 7
]
Using the same condition as above, notice that for this matrix, a = 7, b = 1, c = −1 and d = 7. So a + b = 8 and c + d = 6, so this matrix is not in span S.
- Which of the following sets span R 3?
(a) S =
{[ − 1 4 5 ]
[ 2 − 1 3 ]
[ 1 3 − 2 ]}
We begin by representing these vectors as a matrix. Notice that each vector is represented by a column in the matrix.
A =
Since det(A) = (−2) + (30) + (12) − (−5) − (−16) − (−9) = 70 6 = 0, then this set of vectors spans R 3. (b) S = {[^1 − 2 4 ]^ , [^3 1 − 2 ]^ , [^1 5 − 10 ]} We begin by representing these vectors as a matrix. Notice that each vector is represented by a column in the matrix.
A =
Since det(A) = (−10) + (60) + (4) − (4) − (60) − (−10) = 0, then this set of vectors does not span R 3. (c) S = {[^2 − 3 1 ]^ , [^1 − 1 3 ]^ , [^2 − 4 − 5 ]^ , [^3 − 4 3 ]} We begin by representing these vectors as a matrix. Notice that each vector is represented by a column in the matrix.
A =
Since this matrix is not square and it has more columns than rows, we put this matrix into row echelon form:
r 1 ↔ r 3
(^) r 3 r−^2 +3 2 rr^11 → →r^2 r 3
(^) r 32 r−^3 +3 8 rr 22 →→r^2 r 3
− 13 r 3 → r 3
Notice that the row echelon form has three leading 1’s, so this collection of vectors does span R 3.
- Describe the subspace of R^3 spanned by the set S = {[^1 0 − 1 ]^ , [^2 2 2 ]^ , [^3 2 1 ]}
We begin by forming a matrix representing an arbitrary linear combination of these vectors, and then we put this matrix into reduced row echelon form:
1 2 3 a 0 2 2 b − 1 2 1 c
r 11 +r 3 →r 3 2 r^2 →^ r^2
1 2 3 a 0 1 1 b 2 0 4 4 a + c
(^) r 3 r−^1 − 42 rr^22 →→r^1 r 3
1 0 1 a − b 0 1 1 2 b 0 0 0 a + c − 2 b
From this, we see that spanS = {[^ a b c ]^ : a + c − 2 b = 0}
- For each of the following sets of vectors, determine whether or not the set is linearly independent. For those that are dependent, write one of the vectors as a linear combination of the others. (a) S = {[^2 − 3 ]^ , [^ − 1 3 ]^ , [^2 − 4 ]} Notice that the matrix representing the homogeneous system of equations is:
[ 2 − 1 2 0
]
Since this system has more unknowns than equations, it must have at least one free variable, meaning that it has non-trivial solutions. Hence this set is not linearly independent.
Case 1: If a 3 = 0, then this becomes a 1 v~ 1 + a 2 v~ 2 + (0) v~ 3 = a 1 v~ 1 + a 2 v~ 2 = ~0. Therefore, since S = { v~ 1 , ~v 2 } is a linearly independent set, then we must have a 1 = a 2 = 0. Then a 1 = a 2 = a 3 = 0 Case 2: If a 3 6 = 0, then a 1 v~ 1 + a 2 v~ 2 = −a 3 v~ 3 , so v~ 3 = − a a^13 v~ 1 − a a^23 v~ 2. But this contradicts the fact that v~ 3 is not in the span of S = { v~ 1 , ~v 2 }. So this case is not possible. Hence T = { v~ 1 , ~v 2 , ~v 3 } is linearly independent. 2.
- Let S 1 and S 2 be finite subsets of a vector space V. Suppose that S 1 ⊂ S 2 and that S 1 is linearly independent. Must S 2 be linearly independent? No.{[ Adding a vector to a linearly independent set could create a linear dependence. For example, suppose S 1 = 1 0 ]^ , [^0 1 ]}^ and S 2 = {[^1 0 ]^ , [^0 1 ]^ , [^1 1 ]}. Then S 1 is linearly independent (in fact, it is the standard basis for R 2 ), but S 2 is linearly dependent, since v~ 1 + v~ 2 − v~ 3 = ~0.
- Determine which of the following sets form a basis for R 3. For those that are not, state which part(s) of the definition of a basis are not satisfied. (a) S 1 = {[^4 − 1 5 ]^ , [^ − 2 3 1 ]} We know that dim(R 3 ) = 3, so this set is not a basis. Alternatively, we can see from the associated matrix:
4 − 2 a − 1 3 b 5 1 c
(^) that this set of vectors does not span R 3 and hence is not a basis for R 3. Notice that since these two vectors are not multiples of each other, this set is linearly independent. (b) S 2 = {[^4 3 5 ]^ , [^2 3 − 1 ]^ , [^2 0 6 ]}
We consider the associated matrix: M =
Notice that det(M ) = 72 + 0 − 6 − 30 − 36 − 0 = 0. Therefore, this set is linearly dependent, and hence is not a basis for R 3. In addition, since removing one vector that is a linear combination of the others will leave us with only two vectors, this set also does not span R 3. (c) S 3 = {[^ − 1 5 3 ]^ , [^3 − 2 1 ]^ , [^2 3 − 2 ]}
We consider the associated matrix: M =
Notice that det(M ) = −4 + 27 + 10 − (−12) − (−30) − (−3) = 78. Therefore, this set is a basis for R 3. (d) S 4 = {[^5 3 − 1 ]^ , [^2 − 1 4 ]^ , [^3 0 − 2 ]^ , [^4 1 − 3 ]}
We consider the associated matrix: M =
Using appropriate row operations (verify this), we find that the reduced row echelon form of this matrix is:
From this, we see that this set of vectors does span R^3 , but since there is a free variable, it is linearly dependent, so this set is not a basis for R 3.
- (a) As above, let S 1 = {[^4 − 1 5 ]^ , [^ − 2 3 1 ]}. Find a set T such that S 1 ⊂ T and T is a basis for R 3.
As mentioned above, since dim(R^3 ) = 3, we must add a third vector that is linearly independent from the previous two in order to obtain a basis for R 3.
We will try adding the vector [^0 0 1 ].
We consider the associated matrix: M =
Notice that det(M ) = 12+0+0−(0)−(2)−(0) = 10. Therefore, the set T = {[^4 − 1 5 ]^ , [^ − 2 3 1 ]^ , [^0 0 1 ]} is a basis for R^3. (b) As above, let S 2 = {[^4 3 5 ]^ , [^2 3 − 1 ]^ , [^2 0 6 ]}. Either express the vector ~v = [^5 − 1 3 ]^ as a linear combination of elements in S 2 or show that this is not possible.
We consider the associated matrix: M =
Notice that the reduced row echelon form for this matrix is (verify this):
. Therefore, the vector ~v = [^5 −^1 3 ]^ cannot be written as a linear combination of elements in S 2. (c) As above, let S 3 = {[^ − 1 5 3 ]^ , [^3 − 2 1 ]^ , [^2 3 − 2 ]}. Either express the vector ~v = [^5 − 1 3 ] as a linear combination of elements in S 3 or show that this is not possible. As shown above, S 3 is a basis for R 3 , so any vector can be written as a linear combination of elements in S 3. (d) As above, let S 4 = {[^5 3 − 1 ]^ , [^2 − 1 4 ]^ , [^3 0 − 2 ]^ , [^4 1 − 3 ]}. Find a set T ⊂ S 4 so that T is a basis for R 3. In problem 10(d) above, we observed that the reduced row echelon form for the matrix associated with this set has the form:
From this, we see that taking the first three vectors in the original set (those corresponding to the leading 1’s in the reduced form) gives a basis for R 3. That is, the set T = {[^5 3 − 1 ]^ , [^2 − 1 4 ]^ , [^3 0 − 2 ]}^ is a basis for R 3.
- Let W be the subspace of P 3 spanned by the set S = {t^3 − t^2 + 2t − 1 , t^3 − t + 7, t^2 − 3 t + 8, t^2 + 2t − 1 }. Find a basis for W and find the dimension of W.
We begin by looking at the related matrix:
After performing the appropriate row operations to put this matrix into reduced row echelon form, we have:
Noting the placement and number of the leading 1’s, we see that dim W = 3, and that t = {t^3 − t^2 + 2t − 1 , t^3 − t + 7 , t^2 + 2t − 1 } is a basis for W.
Then, subtracting these, ~v − ~v = ~0 = (a 1 − b 1 ) v~ 1 + (a 2 − b 2 ) v~ 2 + · · · + (an − bn) v~n Since S is a basis for V , S is a linearly independent set. Therefore we must have ai − bi = 0 for each i = 1 · · · n. That is, ai = bi for all i = 1 · · · n. Hence the linear combination representing ~v is unique. 2.
- Prove Corollary 4.
Corollary 4.4: If a vector space V has dimension n, then any subset of m > n vectors must be linearly independent. Proof: Suppose that V has dimension n. Then V has a basis S consisting of exactly n vectors. Let m be a positive integer with m > n. Let T be a set of m distinct vectors drawn from V. Suppose that T is linearly independent. According to Theorem 4.10, any linearly independent set of vectors must satisfy m ≤ n. This contradicts out previous assumption that m > n. Hence T cannot be linearly independent. This T is linearly dependent. 2 Let ~v ∈ V. Since S is a basis, it spans V. Thus ~v = a 1 v~ 1 + a 2 v~ 2 + · · · + an v~n for some constants ai ∈ R. Suppose that ~v = b 1 v~ 1 + b 2 v~ 2 + · · · + bn v~n is another linear combination representing ~v. Then, subtracting these, ~v − ~v = ~0 = (a 1 − b 1 ) v~ 1 + (a 2 − b 2 ) v~ 2 + · · · + (an − bn) v~n Since S is a basis for V , S is a linearly independent set. Therefore we must have ai − bi = 0 for each i = 1 · · · n. That is, ai = bi for all i = 1 · · · n. Hence the linear combination representing ~v is unique.
- Given the homogeneous linear system
x 1 + 2x 2 − x 3 + 4x 4 = 0 3 x 1 − 2 x 2 + 5x 3 − x 4 = 0 4 x 1 + x 2 − x 3 = 0 Find a basis for the solution space of this system. We begin by expressing this system as a matrix, which we then put into reduced row echelon form:
(^) r 3 r−^2 − 43 rr^11 → →r^2 r 3
(^) r 3 − + 7r^2 +rr 23 →→r^2 r 3
r 1 − 2 r 2 →r 1 − 321 r 3 → r 3
(^) r 2 r + 5^1 −^9 rr^33 → →r^1 r 2
From this, we have one free variable, x 4 = t. Then x 1 = 1332 t, x 2 = − 8932 t, and x 3 = − 3732 t
Hence, all solutions are of the form:
(^1332) t − 37 8932 t (^32) t t
. Thus the set^ S^ =
(^1332) − 37 8932 (^321)
^ is a basis for the solution space.
- Given the homogeneous linear system
2 x 1 − x 2 − 3 x 3 + x 4 = 0 3 x 1 + x 2 − 5 x 3 + 2x 4 = 0 x 1 − 3 x 2 − x 3 = 0 Find a basis for the solution space of this system. We begin by expressing this system as a matrix:
After carrying out the appropriate row operations (check the details), we see that the reduced row echelon form for this matrix is:
From this, we have two free variables, x 3 = s and x 4 = t. Then x 1 = 85 s − 35 t and x 2 = 15 s − 15 t
Hence, all solutions are of the form:
(^85) s − 35 t (^15) s − 15 t s t
. Thus the set^ S^ =
(^85) (^15) 1 0
^ is a basis for the solution space.
- Find a basis for the null space of the matrix A =
We consider the matrix for the related homogeneous system of equations:
After carrying out the appropriate row operations (check the details), we see that the reduced row echelon form for this matrix is:
From this, we have two free variables, x 4 = s and x 5 = t. Then x 1 = − 209 s − 2120 t, x 2 = 1320 s + 1720 t, and x 3 = − 12 s + 12 t
Hence, all solutions are of the form:
− 13 209 s − 2120 t −^20 1 s^ +^1720 t 2 s^ s+^12 t t
. Thus the set S =
−^20
(^201) 02 1
is a basis for the
solution space.
- For each of the following matrices, find all real numbers λ such that the homogeneous system (λIn − A)~x = ~0 has a nontrivial solution. (a) A =
[ 3
]
Let M =
[ (^) λ 0 0 λ
]
[ 3
]
[ (^) λ − 3 − 2 − 2 λ + 1
]
Then there is a nontrivial solution when det(M ) = (λ − 3)(λ + 1) − 4 = λ^2 − 2 λ − 7 = 0. Using the quadratic formula, we see that non-trivial solutions exist if and only if λ = 2 ±
2(1) =^2 ±
√ 32 2 = 1 ± 2 √2.
(b) A =
Let M =
λ 0 0 0 λ 0 0 0 λ
λ − 3 0 − 2 − 1 λ + 1 0 0 0 λ − 1
Then there is a nontrivial solution when det(M ) = (λ − 3)(λ + 1)(λ − 1) + 0 + 0 − 0 − 0 − 0 = 0. Therefore, we see that non-trivial solutions exist if and only if λ = 3, λ = −1, or λ = 1.
- Let S = {[^1 2 −^4 ]^ , [^3 −^1 2 ]^ , [^2 −^3 0 ]}. If ~v = [^1 2 3 ], find [~v]S.
We find the coordinate of ~v with respect to the ordered basis S by considering the following matrix:
(c) Find the transition matrix PS←T. To find PS←T , we begin with a partitioned matrix. The columns of the left half correspond to the vectors in S. The columns in the right half correspond to the vectors in T.
The reduced row echelon form of this matrix is (check this by carrying out row operations):
Hence PS←T =
(^792313) (^2913) − (^13) (^23 1 )
(d) Use the transition matrix PS←T to compute [~v]S.
Recall that [~v]S = PS←T [~v]T =
(^792313) (^2913) − (^13) (^23 1 )
(^119) (^79) (^73)
