..READ THIS
FIRST..Problems
about Vector spaces
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about Vector spaces
Level 1 problems
-
Show that the set
V = {(x,y,z) | x,y,z in R and x+y = 11}
is not a subspace of RxRxR .
Since (0,0,0) is not in V, V is not a subspace of RxRxR.
-
Show that the set
V = {(x,y,z) | x,y,z in R and x.x = z.z }
is not a subspace of RxRxR.
(1,1,1) and (1,1,-1) are in V but the sum (1,2,0) is not!
-
Show that the set
V = {(x,y,z) | x,y,z in R and x + 2y + z = 0}
is a subspace of RxRxR.
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' = 0 -
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.
-
S = {(2,5,3)} and T = {(2,0,5)}
The intersection of span(S) and span(T) is a vector space.
Find this space.
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. -
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! -
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! -
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.
-
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 -
Find the solution space of the linear system 3x + 2y + 6z = 0 -x - y + 2z = 0 2x + y + 8z = 0The 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) -
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).
Level 2 problems
-
S = {(2,5,3),(1,0,2)} and T = {(2,0,5),(3,5,5)}
The intersection of span(S) and span(T) is a vector space.
What is the dimension of that space.
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. -
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. -
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.
(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 -
Find ,for each m, the dimension of the row space of the matrix [2 m m-1] [3 m 5 ] [1 0 m+1]
-
Find ,for each m, the solution space of the linear system 3x + 2y + mz = 0 mx - y + 4z = 0 2x + 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} -
In the vector space V = RxRxR, we take a set
S = {(4,5,6) , (r,5,1) , (4,3,2)}
Find the values of r such that the vector space spanned by S is not V.
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 -
In RxRxR we have basis B = ((1,0,1) , (0,2,0) , (1,2,3))
and a basis C = ((1,0,0) , (2,0,1) , (0,0,3))
The coordinates of a vector v with respect to B are (x,y,z).
The coordinates of a vector v with respect to C are (x',y',z').
Write the relation between these coordinates in matrix notation.
-
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 )
Information Provided by Johan.Claeys@ping.be