Problems about Vector spaces
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Problems about Vector spaces
Level 1 problems
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Show that the set |
Since (0,0,0) is not in V, V is not a subspace of RxRxR.
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Show that the set |
(1,1,1) and (1,1,-1) are in V but the sum (1,2,0) is not!
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Show that the set |
For each r,s in R, and for each (x,y,z) and (x',y',z') in V, we have
x + 2y + z = 0 ; x' + 2y' + z' = 0 and
r(x,y,z)+s(x',y',z') = (rx+sx',ry+sy',rz+sz')
Now, it is sufficient to prove that (rx+sx',ry+sy',rz+sz') is in V.
(rx+sx',ry+sy',rz+sz') is in V
<=> rx + sx'+2ry + 2 rsy' + rz + sz' = 0
<=> r(x + 2y + z) + s(x' + 2y' + z') = 0
and this is true since x + 2y + z = 0 and x' + 2y' + z' = 0Go to the top of the page
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Examine if M = {(r,r+2,0) | r in R} is a subspace of RxRxR. |
Since (0,0,0) is not in M, M is not a subspace of RxRxR.
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S = {(2,5,3)} and T = {(2,0,5)} |
· span(S) = {r.(2,5,3) | r in R}
· span(T) = {r'.(2,0,5) | r'in R}
· The only common vector of span(S) and span(T) is (0,0,0).
· The intersection of span(S) and span(T) is a vector space {(0,0,0)}.
· This space is called the zero space or the null space.
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Show that {(1,2,3) , (2,3,4) , (3,4,5) } is not a basis of RxRxR. |
· The three vectors are linear dependent
·
· <=> There is a k,l,m not all zero such that
· l(1,2,3)+m(2,3,4)+n(3,4,5)=(0,0,0)
·
· <=> The following system has a non trivial solution
· 1.l + 2.m + 3.n = 0
· 2.l + 3.m + 4.n = 0
· 3.l + 4.m + 5.n = 0
·
· <=> | 1 2 3|
· | 2 3 4| = 0
· | 3 4 5|
· And this is true!
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Find a basis of RxRxR containing the vectors (1,2,5) and (0,1,2). |
· Since the dimension is 3, we have to find 1 vector (a,b,c) such that (1,2,5) ;(0,1,2) and (a,b,c) are linear independent.
·
· The three vectors are linear dependent
·
· <=> There is a k,l,m not all zero such that
· l(1,2,5)+m(0,1,2)+n(a,b,c)=(0,0,0)
·
· <=> The following system has a non trivial solution
· 1.l + 2.m + 5.n = 0
· 0.l + 1.m + 2.n = 0
· a.l + b.m + c.n = 0
·
· <=> | 1 2 5|
· | 0 1 2| = 0
· | a b c|
· There are many choices for a b and c , such that this is true!
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Assume that v and w are linear independent vectors. Prove that v , w and (v + w) are linear dependent vectors. |
· Since v + w = 1.v + 1.w , the vector (v + w) is a linear combination of v and w.
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Find the coordinates of the vector (3,2,1) with respect to the basis ((1,0,2),(2,1,0),(0,3,5)) in RxRxR. |
· Say (x,y,z) are those coordinates, then
· (3,2,1) = x(1,0,2)+y(2,1,0)+z(0,3,5)
· <=>
· / x+2y+0z = 3
· | 0x+y+3z = 2
· \ 2x+0y+5z= 1
· <=>
· x = 1/17 ; y = 25/17 ; z = 3/17
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Find the solution space of the linear system 3x + 2y + 6z = 0 -x - y + 2z = 02x + y + 8z = 0 |
· The third equation is a linear combination of the previous equations.
·
· So, the system is equivalent with
·
· 3x + 2y + 6z = 0
· -x - y + 2z = 0
·
· For each value of z, we have just one solution of the system.
·
· x = -10z ; y = 12z ; z = z
·
· The solutions are {(-10z,12z,z) | with z in R}
·
· These solutions form a one dimensional vector space with basis (-10,12,1)
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All polynomials p(x) with degree not greater than 2 constitute a vector space V. Replace in (1, 1 + x2 , b(x) ) the polynomial b(x) such that it becomes an ordered basis for that vector space. Calculate the coordinates of (2x2 - 7x) with respect to the chosen basis. |
· The vectors 1 and (1 + x2) are linear independent. It is easy to see that x can't be written as a linear combination of the given vectors. So, (1, 1 + x2 , x ) is an ordered basis for that vector space.
· Now we write (2x2 - 7x) as a linear combination of the vectors of the ordered basis.
· We find -2.(1) + 2.(1 + x2) - 7.(x) = (2x2 - 7x)
· The coordinates of (2x2 - 7x) are (-2,2,-7).
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Level 2 problems
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S = {(2,5,3),(1,0,2)} and T =
{(2,0,5),(3,5,5)} |
· span(S) = { r(2,5,3) + s(1,0,2) | r,s in R}
· span(T) = { r'(2,0,5) + s'(3,5,5) | r',s' in R}
· (x,y,z) is an element of the intersection if and only if
· there is an r, s , r', s' such that
· (x,y,z) = r(2,5,3) + s(1,0,2) = r'(2,0,5) + s'(3,5,5)
·
· <=> there is an r, s , r', s' such that
· 2r + s = 2r' + 3s'
· 5r = 5s'
· 3r +2s = 5r' + 5s'
·
· <=> there is an r, s , r', s' such that
· 2r + s - 2r' - 3s' = 0
· 5r - 5s' = 0
· 3r +2s - 5r' - 5s' = 0
· This is a homogeneous system of the second kind.
· The rank of the coefficient matrix is 3.
· We choose a main matrix
· [2 1 -2]
· [5 0 0]
· [3 2 -5]
· s' is the side unknown and s' is an arbitrary number.
· Hence, for each choice of s' there is a vector of the intersection.
· The dimension of the the intersection space is 1.
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Assume that v and w are linear independent vectors. Prove that v and (v + w) are linear independent vectors. |
· rv + s(v + w) = 0
· => (r+s)v + sw = 0
· since v en w are linear independent
· => r+s = 0 and s = 0
·
· => r = s = 0
·
· => v and (v + w) are linear independent vectors.
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Find the condition for r and s
such that the vectors |
· (r,2,s) , (r+1,2,1) and (3,s,1) are linear dependent.
· <=>
· There is a (k,l,m) not (0,0,0) such that
· k(r,2,s) + l(r+1,2,1) + m(3,s,1) = 0
· <=>
· The system
· / r.k + (r+1).l + 3.m = 0
· | 2.k + 2. l + s.m = 0
· \ s.k + 1. l + 1.m = 0
· has a solution different from (0,0,0).
· <=>
· | r r+1 3|
· | 2 2 s| = 0
· | s 1 1|
· <=>
· rs2 +s2 - rs -6s + 4 = 0
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Find ,for each m, the dimension of the row space of the matrix [2 m m-1] [3 m 5 ] [1 0 m+1] |
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Find ,for each m, the solution space of the linear system 3x + 2y + mz = 0 mx - y + 4z = 02x + y + 3z = 0 |
· |3 2 m|
· |m -1 4| = ...=(m-5)(m+1)
· |2 1 3|
· Case 1 : m is different from 5 and -1
· The only solution is x = y = z = 0.
· The solution space is {(0,0,0)}
· Case 2 : m = -1 ; The system becomes
·
· 3x + 2y - z = 0
· -x - y + 4z = 0
· 2x + y + 3z = 0
·
· The third equation is a linear combination of the previous ones.
· The system is equivalent to
·
· 3x + 2y - z = 0
· -x - y + 4z = 0
·
· and this system is equivalent to
·
· x = -7z
· y = 11z
·
· The solution space is {(-7,11,1).z | with z in R}
·
· Case 3 : m = 5 ; The system becomes
·
· 3x + 2y + 5z = 0
· 5x - y + 4z = 0
· 2x + y + 3z = 0
·
· The third equation is a linear combination of the previous ones.
· The system is equivalent to
·
· 3x + 2y + 5z = 0
· 5x - y + 4z = 0
·
· and this system is equivalent to
·
· x = -z
· y = -z
·
· The solution space is {(-1,-1,1).z | with z in R}
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In the vector space V = RxRxR,
we take a set |
· If the three vectors of S are linear independent, the vector space spanned by S is V. If this is not the case, then (r,5,1) has to be a linear combination of (4,5,6) and (4,3,2).
· This is equivalent to:
· |r 5 1|
· |4 5 6| =0
· |4 3 2|
·
· <=> r = 9
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In RxRxR we
have basis B = ((1,0,1) , (0,2,0) , (1,2,3)) |
M = span { (1+m, 4, 2); (5,6,-1-m) }N = span { (5+2m, 10,0) }
Show that (1+m, 4, 2) and (5,6,-1-m) are linear independent for all real m values. Calculate m such that M+N is a direct sum. |
(Ans: m = 1 or m = 15 )