Hi sarah2728!!
1) Let W be the set of all n X n real matrices of rank 1. Is W a subspace
of Mn,n? Please justify your answer.
The answer is NO, and there are many reasons that justified this
answer, one of the strongest and simpliest is that the not a subspace
of W because it does not contain the neutral element of W (the zero
matrix is not of rank 1).
This subset is not closed under addition, see for example:
| 1 0 | | 0 0 | | 1 0 |
| | + | | = | |
| 0 0 | | 0 1 | | 0 1 |
-----------------------------
2) Determine whether the set of all positive real numbers together
with the operations of addition and scalar multiplication, is a vector
space. Let x and y be real numbers. Let c be a real number too.
Addition: x + y = xy, and scalar multiplication: c * x = x^c
Let see if all of the definition axioms are satisfied:
A- V is closed under vector addition: x + y ? R+.
x + y = xy ? R+ ---> ok
B- Vector addition is associative: u + (v + w) = (u + v) + w.
u + (v + w) = u + vw = uvw = uv + w = (u + v) + w ---> ok
C- There exists a neutral element 0 in R+, such that for all x in R+:
0 + x = x.
The number 1 satisfies 1x = x for all x in R+, then for all x in R+:
1 + x = 1x = x , then 1 is the neutral element ---> ok
D- For all v in R+, there exists an element w in R+, such that v + w = 0
If 0 = v + w ==> vw = 1 ==> w = 1/v (remember that the zero number is
not part of the set of all real positive numbers R+) ---> ok.
E- Vector addition is commutative: v + w = w + v.
v + w = vw = wv = w + v ---> ok.
F- V is closed under scalar multiplication: c*x ? R+.
c*x = x^c ? R+ ---> ok.
G- Scalar multiplication is associative: a*(b*v) = (ab)*v.
a*(b*v) = a*(v^b) = (v^b)^a = v^(ba) = (ba)*v = (ab)*v ---> ok.
H- 1*v = v, where 1 denotes the multiplicative identity in R (the number one).
1*v = v^1 = v ---> ok.
I- Scalar multiplication distributes over vector addition:
a*(v + w) = a*v + a*w .
a*(v + w) = (v + w)^a = (vw)^a = (v^a)(w^a) = a*v + a*w ---> ok.
J- Scalar multiplication distributes over scalar addition:
(a + b)*v = a*v + b*v.
(a + b)*v = v^(a + b) = (v^a)(v^b) = (v^a) + (v^b) = a*v + b*v ---> ok.
The set of all positive real numbers together with the following
operations of addition and scalar multiplication:
Let x and y be real numbers. Let c be a real number too.
Addition: x + y = xy, and scalar multiplication: c * x = x^c
is a vector space.
-----------------------
3) Suppose that the n vectors v1, v2, ... , vn are linearly
independent. Show that any collection of n-1 of them are also linearly
independent. Hint: for simplicity, show that the n-1 vectors v1, v2,
... , vn-1 are linearly independent.
Supose that the n-1 vectors V1, V2, ... , Vn-1 are linearly dependent:
The above means that:
c1V1 + c2V2 + ... + c(n-1)V(n-1) = 0, where not all c1, c2, ...,
c(n-1) are zero, then we can write:
0 = c1V1 + c2V2 + ... + c(n-1)V(n-1) + 0Vn , where not all c1, c2,
..., c(n-1) are zero, this contradicts the assumtion that the n
vectors v1, v2, ... , vn are linearly independent. Then c1V1 + c2V2 +
... + c(n-1)V(n-1) = 0 is satisfied only if all c1, c2, ..., c(n-1)
are zero, that is only if all the vectors are linear independent.
--------------------------------------------------------
I hope that this helps you. Feel free to request for a clarification
if you need it.
Regards.
livioflores-ga |
,
0 Comments
)