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Solutions – Assignment 1
- The system gives the matrix A which row reduces as follows:
2 1 3 1
1 2
1 3
3
3
R R R R
R R
R R
A −−
−
−
= ^ ^ →^
→ ^
→ ^
So we can conclude that x 1 (^) = −1, x 2 (^) = 2, and x 3 = −1. Furthermore, since the row reduced echelon form of A is the identity matrix, this solution is unique.
- The plane z = 5 is not a vector space since it does not have a zero vector. Also, addition of two vectors in the plane gives a vector outside the plane. For
example:
−^ −
- Yes, the space P ( ) n of polynomials of degree at most n is a vector space. Proof: Let p = anxn + ...+ a x 1 1 + a 0 , q = b xn n + ...+ b x 1 1 + b 0 , and r = c xn n + ...+ c x 1 + c 0 be arbitrary polynomials in P ( ) n , and let r and s be arbitrary real numbers. a. Commutativity:
1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 0
n n n n n n n n n n n n^ n n
a
a
p q x a x a b x b x b a b x a b x a b b a x b a x b a b x b x b x a x a q p
b. Associativity:
[ ] ( ) ( ) ( )
1 0 1 0 1 0 1 1 1 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0
n n^ n n^ n n n n n n n n n n n n^ n n^ n n
a
a
p q r x a x a b x b x b c x c x c a b c x a b c x a b c a b c x a b c x a b c x a x a b x b x b c x c x c
+ + = + + + + ^ + + + + + + +
= ^ + + + + + + + + + + +
= [ p + q ] + r
c. Existence of a zero: 0 = 0 x n + ... + 0 x + 0
. Note that
1 0 1 0 1 0
0
n (^) n n n n n n
a a a
p x x x a x a x a x a x a x a p
d. Existence of negatives: − p = ( − an ) x n + ...+ −( a 1 ) x + −( a 0 ). We see
easily that
1 0 1 0 1 1 0 0
n n n n n n n
a a a a
p p x a x a x a x a x a a x a a
e. Distributivity of scalar multiplication over vector addition:
1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 0
n n^ n n n n n n n n n^ n n
r p q r a b x a b x a b r a b x r a b x r a b ra rb x ra rb x ra rb ra x ra x ra rb x rb x rb rp rq
f. Distributivity of scalar multiplication over scalar addition:
[ ] [ ] [ ]
1 0 1 0 1 1 0 0 1 0 1 0
n n n n n n n n n n n
a ra a
r s p r s a x a x a r s x r s a x r s a s x ra sa x ra sa ra x ra x ra sa x sa x sa rp rs
- The most efficient way to describe a subspace is by giving a basis. To do this, I first row reduce A:
2 1 4 1
2 3 2 4 2
1 2
2
1 (^92)
4
R R R R
R R R R R
R R
A −−
− −
−
= ^ ^ →^
→ ^
→ ^
This gives a row reduced echelon form with two pivot columns and two free columns. We deduce that a basis of the column space is given by the first two column vectors of the original matrix: 1 4 2 1 col( ) span , 0 2 1 3
A
= ^ ^ ^
Similarly, a basis for the row space is given by the first two row vectors of the
row reduced matrix: row( A ) = span {( 1 0 1 2 , 0) ( 1 1 1 )}. Finally,
the null space can be found by translating the row reduced matrix back into a system of linear equations: 1 2 3 4 1 2 3 4
x x x x x x x x
^ +^ +^ +^ =
Since x (^) 3 and x 4 are free variables, we can assign them the variables s and t respectively. So our system reduces to 1 2
x s 2 t x s t
^ = −^ −
We construct the basis from this information by creating two vectors: one where ( , ) s t = (1, 0)and the second where ( , ) s t = (0,1). This gives us the basis
null( ) span , 1 0 0 1
A
= ^
^
^
^
