Practice Problems Subspaces, Bases & Dimension Math 201-105-RE
Practice Problems Subspaces, Bases & Dimension Math 201-105-RE
1. Let u1 = (3, −1, 2) and u2 = (3, 1, 5). 8. Let u1 = (2, 3, −1, ), u2 = (5, 4, −1), u3 = (5, −3, 2),
(a) Express the vector v = (9, 11, 27) as a linear combination u4 = (0, 6, −2), u5 = (0, −15, 5).
of u1 and u2 if possible. Determine whether each set is linearly independent or linearly
dependent. In each case, state whether the span of the set is a
(b) Find k such that the vector w = (−5, 4, k) is a linear com- point, line, plane, or R3 .
bination of u1 and u2 . (a) {u1 , u2 }
6 9 7 (b) {u1 , u2 , u3 }
2. Let a1 = 3, a2 = −3, and b = 6 .
(c) {u2 , u3 , u4 }
4 5 h
(d) {u2 , u3 , u4 , u5 }
(a) Find h so that b is in Span{a1 , a2 }.
(e) {u4 , u5 }
(b) For the h that you found in the previous part, express b as a 9. Let u1 = (0, −5, 5, ), u2 = (0, 3, −3), u3 = (1, 1, 1),
linear combination of a1 and a2 . u4 = (1, 0, 1), u5 = (2, 2, 0), and 0 = (0, 0, 0).
Determine whether each set is linearly independent or linearly
3. Let b1 = (h, 5, 7), b2 = (−1, 3, 7), and b3 = (1, 1, 2).
dependent. (LI or LD?) In each case, state whether the span of
Find h so that b3 ∈ Span{b1 , b2 }.
the set is a point, line, plane, or R3 .
4. (a) Express the plane x − 3y + 4z = 0 as a span of vectors. (a) {u1 }
6. Let u1 = (2, 0, 3, −1), u2 = (−4, 0, −6, 2), u3 = (5, 5, 0, 3), (f) {u4 , u5 }
u4 = (1, 3, −6, 5), 0 = (0, 0, 0, 0). (g) {u4 , u5 , 0}
Determine whether each set is linearly independent or linearly
(h) {0}
dependent.
10. Determine if the following sets are subspaces. For those that are,
(a) {u1 }
express the set as a span of vectors. For those that are not, pro-
(b) {u1 , u2 } vide a counter-example to show it is not closed under VA or SM.
(a) S = {(x, y, z) ∈ R3 | x = 4s − t, y = s + 3t, z = 6s}
(c) {u1 , u2 , u3 }
(b) S = {(x, y, z) ∈ R3 | 3x + 4y − z = 2}
(d) {u2 , u3 , u4 }
(c) S = {(x, y, z) ∈ R3 | z 2 = xy}
(e) {u3 , u4 } (d) S = {(x, y, z) ∈ R3 | x + 2y − 3z = 0}
(f) {u3 , u4 , 0} (e) S = {(x, y, z) ∈ R3 | y ≥ x}
7. Let u1 = (5, 2, −1, 6), u2 = (3, 1, 0, 2), (f) S = {(x, y) ∈ R2 | x = 4 + 2t, y = −6 − 3t}
u3 = (1, 1, −2, 6), u4 = (1, 1, −2, 1), u5 = (1, 0, 0, 0). (g) S = {(x, y, z) ∈ R3 | y + z ≥ −1}
Determine whether each set is linearly independent or linearly (h) S = {(x, y, z) ∈ R3 | z = 2x − 3y}
dependent.
x
(a) {u1 , u2 } (i) S = y ∈ R
3
xy + z = 0
z
(b) {u1 , u2 , u3 }
x
(c) {u1 , u2 , u3 , u4 }
(j) S = y ∈ R3 2x = y − z
z
(d) {u2 , u3 , u4 , u5 }
(e) {u3 , u4 , u5 } x
(k) S = y ∈ R3 x = 6t, y = 4t some t ∈ R
(f) {u5 }
z
" # " # " #
1 2 3 2 −6 5 3 −8 18
11. Let a1 = , a2 = , a3 = .
2 1 5 16. The matrix A = −3 9 −1 −5 −1 −36
0 0 4 8 −8 36
(a) Express a3 as linear combinations of a1 and a2 if possible.
1 −3 0 0 1 4
has reduced form R = 0 0 1 0 −2 −1 .
(b) Is {a1 } a basis for R2 ? Why or why not? 0 0 0 1 0 5
(a) Choose a basis for Col(A) from the columns of A.
(c) Is {a1 , a2 , a3 } a basis for R2 ? Why or why not? (b) Choose a basis for Nul(A).
5 4 1 0 1 13
(d) Is {a2 , a3 } a basis for R2 ? Why or why not?
4 5 −1 0 1 14
17. The matrix A =
−4 −4 0 0 0 −12
3 2 1 0 1 7
12. Let u1 = (4, 2, 5), u2 = (3, −1, −2), and u3 = (6, 2, 0)
1 0 1 0 0 1
(a) Is {u1 , u2 , u3 } a basis for R3 ? Justify. −1 0 0 2
0 1
reduces to R = .
0 0 0 0 1 0
0 0 0 0 0 0
(b) Is it possible to express u3 as a linear combination of u1
(a) Choose a basis for Col(A) from the columns of A.
and u2 ? Justify without solving.
(b) Choose a basis for Nul(A).
6 −9 2 −12 1 8
(c) Is it possible to express the vector (9, 5, 2) as a linear com- −6 9 5 54 2 −1
bination of u1 , u2 , and u3 ? Justify without solving. 18. The matrix A =
8 −12 1 −26 0 9
0 0 3 18 3 3
2 1 0 −1
1 −3/2 0 −4 0 1
13. Let A = 3 1 1 and v = 2 .
0 0 1 6 0 1
reduces to R =
1 0 1 3 0
0 0 0 1 0
(a) Is v in Nul(A)? Justify your answer. 0 0 0 0 0 0
0 d 0 2 1 6
(b) Find a basis for Nul(A).
1 0 4 −1 0 0
has reduced form 0 1 0 2 0 5 .
(c) Is v1 in Nul(A)? Justify your answer. 0 0 0 0 1 1
Find a, b, c, d, e, and f .
(d) Is v1 in Col(A)? Justify your answer.
2 a 5 3 b c
20. The matrix d 9 e −5 −1 −36
(e) Is v2 in Nul(A)? Justify your answer.
0 0 4 f −8 36
1 −3 0 0 1 4
(f) Is v2 in Col(A)? Justify your answer. has reduced form 0
0 1 0 −2 −1 .
0 0 0 1 0 5
15. Let a1 = (2, 3, −1, 1), a2 = (−2, −3, 1, −1), Find a, b, c, d, e, and f .
a3 = (2, 3, 1, 5), a4 = (2, 3, 2, 7), a5 = (4, 6, 3, 12).
Find a basis for S = Span{a1 , a2 , a3 , a4 , a5 }. 21. Suppose A is n × m, Dim( Col(A)) = 2, Dim( Nul(A)) = 3
and Dim( Nul(AT )) = 4. Find n and m.
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22. Suppose A is an 5 × 8 matrix. What is the general (parametric) solution to Ax = b?
(a) What is the minimum nullity of A? 26.
Suppose
the
general
solution
Ax = b is given by
to
(b) Can the system Ax = 0 have a unique solution? x 1 1 0
y =2 + s 0 + t 1.
(c) What is minimum nullity of AT ?
z 3 −2 5
(d) Can the system AT x = 0 have a unique solution?
23. Suppose A is a 6 × 4 matrix, and that the nullity of AT is 3. (a) Find a non-zero solution to the homogeneous equation
(a) Find the nullity of A. Ax = 0
(b) Does the system Ax = 0 have a unique solution? (b) Find the general solution to Ax = 2b.
(c) Are the columns of A linearly independent? (Hint: A(2x)=2(Ax).)
24. Suppose A is 5 × 3 and the general −2 5 6 x
solution
to the equation
27. Let u = 1 , v = 0, w = −4, and x = y .
x 1 5 −2
Ax = b is given by y = 0 +s1+t 0 .
0 1 3 z
z −3 0 1
Suppose A is 7 × 3, and that Nul(A) = Span{u, v}.
(a) Find the general solution to Ax = 0.
(a) Find the nullity of A.
(b) Find the rank of A.
(b) Find the rank of A.
−3
5
25. Suppose Nul(A) = Span 4 , 0 for a matrix A,
(c) Give the general solution of Ax = 0 in parametric form.
2 1
(d) Give the general solution of Ax = Aw in parametric form.
6
and that u = −1 is one particular solution to Ax = b.
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A NSWERS:
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2
5 3 21. n = 6, m = 5.
16. (a) Basis for Col(A): −3 , −1 , −5
22. (a) Max Rank of A = 5, so Min Nullity of A = 8 − 5 = 3.
0 4 8
(b) No. Solution must have at least 3 parameters.
3 −1 −4
(c) Min Nullity of AT is 0.
1 0 0
(d) Yes. Solution is unique when Nullity of A is 0.
0 2 1
(b) Basis for Nul(A): , ,
0 0 −5 23. (a) 1.
0 1 0
(b) No. There will be one parameter in the solution.
0 0 1
(c) No. There is a non-trivial solution to Ax = 0.
5 4 1
4 5 1
x 5 −2
17. (a) Basis for Col(A) : , ,
24. (a) y =s1+t 0
−4 −4 0
z 0 1
3 2 1
(b) Rank of A = 1.
−1 0 −1
x = 6 − 3s + 5t
−2
1 0
y = −1 + 4s
1 0 0
25.
(b) Basis for Nul(A) : , , .
z = 2 + 2s + t
0 1 0
0 0 0
26. (a) (x, y, z) = (1, 0, −2), for example when s = 1 and t = 0.
0 0 1
x 2 1 0
18. (a) Basis for Col(A) : {a1 , a3 , a5 }. (b) y =4 + s 0 + t 1.
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