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VECTOR SPACES AND LINEAR MAPS
1. Show that there no numbers x and y such that
x y
2. Express the vector (4,-11) as a linear combination of (2,-1) and (1,4). 3. If
a and b ; compute: a. at ; a. bt; a.(a+b)t
4. When is
u t w
t v t u
t 1
5. Construct the matrix
ij 3 x 3 A a where aij=2i-j.
6. Show that A.A=A for
A
7. Let us consider subspace S x , y , z R^3 | x y z k 1 ; k^2 x y z 0
a) Find k in such a way that subspace S has dimension 2. b) Find the dimension and a basis for S when K 1. (Solutions: a) k = - 1; b) dim(S) = 2, basis={(1,0,-1), (0,1,-1)})
8. Let us consider the vectors: v 1 (^) (1, 2, ,1), a v 2 (^) ( ,1, 2,3), a v 3 (0,1, ,0) b
a) Find a and b in such a way that the vectors are linearly dependent. b) By using the a and b values found, find the equations of the subspace generated by the three vectors. (Solutions: a) a = 3, b = 7/5; b) x 1 + 7x 2 – 5x 3 = 0, x 1 – x 4 = 0)
9. Find the dimension and a basis for subspace S given by the following system of equations 0 2 0
x y y z
(Solution: dim(S) = 1, basis={(-1,1,2)})
10. Let v 1 (^) (1, 4, 5, 2), v 2 (^) (1, 2, 3,1), v 3 (3, 2, , x y )be vectors of the linear space R^4
a) Find x and y so that v 3 belongs to the subspace generated by v 1 and v 2 b) Calculate the equations of the resulting subspace (Solutions: a) x = -5, y = 1; b) x 1 + x 2 + x 3 = 0, x 2 - 2x 4 = 0)
11. Given the linear subspace S of R^4 described by the following equations:
x 1^ ^ a x ,^2^ ^ a^ ^ b x ,^3^ ^ c x ,^4^ ^ b con a b c ;^ , , R ^ , find a basis for the subspace and
determine the coordinates of the vector (1, 2,0,1)with respect to it. (Solutions: basis={(1,1,0,0), (0,1,0,1), (0,0,1,0)}; x 1 = 1, x 2 = 1, x 3 = 0)
12. Find a basis for the linear subspace of 4 R determined by the equation: 2 x 1 (^) 3 x 3 (^) x 4 0 (Solution: basis={(1,0,0,-2), (0,1,0,0), (0,0,1,3)}
13. Given the following vectors: (0, 1, 2, 1), ( , a 1,0,1), (1, 1, ,0) b , find a and b so
that they generate a dimension 2 subspace and find its equations.
VECTOR SPACES AND LINEAR MAPS
(Solution: a = 2, b = 1; x 1 + 2x 2 + x 3 = 0, x 1 + x 2 – x 4 = 0)
14. Let us consider the subset of R^3 identified by the following equations:
2 0 2
ax y z x ay z b
^ ^ ^
^ ^
. Calculate the values of a and b so that the subset is a dimension 2
linear subspace. (Solution: b = 0, a = 1)
15. Find the dimension, a basis, and the equations of the subspace identified by the vectors
whose components are ( x 3 x 2 , x 2 , x 3 , 2 x 3 )
(Solution: dim(S) = 2, basis = {(1,0,1,2), (1,1,0,0)}, x 3 + x 2 - x 1 = 0, x 4 + 2x 2 - 2x 1 = 0).
16. Given the following vectors:
v 1 (^) (2,0, 2,9), v 2 (^) (1, 2,1,3), v 3 (^) (1,0,1,3), v 4 (2, 4, 2,6)
a) Is v v 1 , 2 , v 3 , v 4 a basis for the linear space R^4? Why? Similarly, is it a linear span
for 4 R. b) Compute a and b so that the vector (0,1, a b , )belongs to the subspace identified by
the vectors v v 1 , 2 , v 3 , v 4 .
(Solutions: a) No, because v 2 and v 4 are L.D. It can’t be a linear span because rank({v 1 ,v 2 ,v 3 ,v 4 }) < dim( 4 R ) = 4; b) a = 0, b in 4 R )
17. Answer the following questions:
a) Is it possible to have a dimension 3 subspace inside of the linear space R^3? Motivate your answer. If yes, find a base for such a subspace and its equations. b) Given the following subspace:
1 1 (1, 5,0) (0,0,1) (0,0,0) ( 1,5, 2) ( , 1, ) 5 5
S
find its dimension and its parametric and Cartesian equations. c) Find a vector that belongs to the previous subspace, verifying that it is linear combination of the base of S. (Solutions: a) No, it’s not possible because a subspace must be dimensionally smaller than the original vector space. ; b) dim(S) = 2, {(a, -5a, b)}, 5x 1 + x 2 = 0; c) Any vector that is linear combination of (1,-5,0) and (0,0,1) would make the job. For example: (1,-5,1), with coordinates λ 1 =λ 2 =1)
18. Given the following linear span: (2,1,0,1),( ,0,1, a 1),(1,1, 1, 2)
a) Find the dimension of the subspace generated by the vectors with respect to the value taken by parameter a.
b) Find the equations of the subspace generated by the vectors when a 1 and when
a 0. (Solutions: a) dim(S) = 2 when a=1, and dim(S) = 3 otherwise; b) For a = 1 the equations are: x 1 + 2x 2 + 3x 3 = 0 and x 1 + 5x 2 – 3x 4 = 0, for a = 0 the equation is: x 2 – x 3 – x 4 = 0)
