Starting from:
$30

$24

Homework 7 Solution




In each of the following you are given a vector space (you do not need to prove that it is a vector space). Find a basis for this space and prove that
the system you found is indeed a basis.

0 1







t + 2s + r








B
t s r
C










i.
f
B
t + 3r
C
: t; r; s 2 Rg


5r + 5s




B


C






@


A






B
2t + 2s + r
C


fA 2 M4(R) : (A)ij = (A)ji 81 i; j 4g



(Remark: This is the collection of all 4 4 symmetric matrices).




iii. fA 2 M4(R) : (A)ij = (A)ji 81 i; j 4g




(Remark: This is the collection of all 4 4 anti symmetric matrices). iv. The set of solutions of the homogeneous system




8
x + y
z = 0


x + 2y
5z = 0
:
2x + 5y
8z =
0
<
fp(x) 2 R2[x] : p(1) = p(2)g.



fp(x) 2 R3[x] : p(1) = 0 and p′ (1) = 0g.
( )

vii. fA 2 M2(R) : A
1
= 0g.
2



a Consider the space R2[x]. Recall that we mentioned in class that B = (1; x; x2) and C = (x; x2; 1) are both ordered bases for R2[x].



Prove that D = (1; 1 + x; (1 + x)2) is also an ordered basis for for
R2[x].

ii. Compute [3
2x + x2]B, [3
2x + x2]C and [3
2x + x2]D.
iii. Find polynomials p1(x); p2(x); p3(x) 2 R2[x] which satisfy:
@
1
A
@
1
A
@
1
A
1
1
1
[p1(x)]B = 0
3
1
; [p2(x)]C = 0
3
1; [p3(x)]D =
0
3
1:
b. Consider the space R3.




1






2




i. Find all values of k for which B = (
0
3
1
;
0
1
1
;
0
7
1
) is






@
1
A @
2
A @
k
A


a basis for R3.


1
1
3
1


0
1
























For k = 2 nd [@ 14 A]B.
8

0 1

1

iii. For k = 2 nd b 2 R3 such that [b]B = @ 1 A.

2




c. Consider the vector space in Q1(vii) and the basis you found for this space. Give this basis an order and denote this ordered basis B. For




each of the following vectors determine whether it belongs to the space and if so nd its coordinates with respect to the ordered basis B:

( )

3
4 2
)
ii.
0
0
.
2
1







Let V be a vector space and let v1; :::; vn 2 V . Prove the following claims.



If fv1; :::; vng is a basis for V then it is a maximal linearly independent set in V .



If fv1; :::; vng is a maximal linearly independent set in V then it is a basis in V .



If v1; :::; vn is linearly independent then v1; :::; vn is a basis for spanfv1; :::; vng.



If V is a subspace of some vector space W , and v1; :::; vn is a basis for V , then for w 2 W we have: spanfv1; :::; vn; wg = V if and only if fv1; ::; vn; wg is linearly dependent.



Let A 2 Mn(R). Prove that the following are equivalent.



A is invertible.
The columns of A form a basis for Rn.
The rows of A form a basis for Rn.



(Hint: You might want to use Q6 from HW6).

More products