Is this matrix a vector space?

  • Thread starter mikephy
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  • #1
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Homework Statement



a b c
0 b 8
0 0 c

Homework Equations


10 axioms to determine vector space:

1. If u and v are objects in V, then u + v is in V.
2. u + v = v + u
3. u + (v + w) = (u + v) + w
4. There is an object 0 in V, called a zero vector for V, such that 0 + u = u+ 0 = for all u in V
5. For each u in V, there is an object -u in V, called a negative of u, such that u + (-u) = (-u) + u = 0.
6. If k is any scalar and u is any object in V, then ku is in V.
7. k(u + v) =ku + kv
8. (k + m)u = ku + mu
9. k(mu) = (km)(u)
10. 1u=u

The Attempt at a Solution


I am brand new to this, but this is what I got so far.

u =
a b c
0 b 8
0 0 c

v =
a' b' c'
0 b' 8
0 0 c'

I got that it isn't a vector space because
u + v =
a + a' b + b' c + c'
0 b + b' 16 <----- Because this should be 8
0 0 c + c'

Is this correct and it would fail the 1st axiom?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Yes. Your set of matrices isn't closed under matrix addition.
 
  • #3
421
1
Hi mikephy,
a simpler way to check whether the given matrix would satisfy the conditions for a vector space would be check whether the vector

C = A + [tex]\lambda[/tex]B

lies in the vector space if A and B are vectors in your vector space and [tex]\lambda[/tex] is a scalar from the appropriate field. Using just this, you can see yourself that if this condition is satisfied, all 10 of your axioms will be satisfied.
 
  • #4
34,934
6,698
Hi mikephy,
a simpler way to check whether the given matrix would satisfy the conditions for a vector space would be check whether the vector

C = A + [tex]\lambda[/tex]B

lies in the vector space if A and B are vectors in your vector space and [tex]\lambda[/tex] is a scalar from the appropriate field.
This is not simpler than what the OP already has done. He zeroed in on the fact that addition is not closed for these matrices, so he's done. And besides, your formula doesn't confirm that scalar multiples of A are in the space.
Using just this, you can see yourself that if this condition is satisfied, all 10 of your axioms will be satisfied.
Not true. If you know that your vectors belong to a vector space over some field of scalars, then by definition all 10 vector space axioms are satisfied, but to check that a set of vectors from that space forms a subspace of the vector space, all you need to check are these three axioms:
  • The set contains the zero vector, 0.
  • Addition is closed for vectors in the set; i.e., u + v is in the set whenever u and v are in the set.
  • Scalar multiplication is closed for any scalar in the field and for any vector in the set; i.e., cu is in the set whenever u is in the set, and c is any scalar in the field.
You can take a shortcut by showing that au + bv is in the set whenever u and v are in the set, and a and b are scalars. When a = b = 0, the first axiom is covered. When a = b = 1, the second axiom is covered. When a is any scalar and b = 0, the third axiom is covered.
 
Last edited:
  • #5
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Thanks a lot guys what Mark44 said agrees with my textbook, to verify if a set of vectors forms a subspace it says you only need to verify axioms 1, 4, 5 and 6. And then it says that even axiom 4 and 5 can be omitted.
 
  • #6
421
1
This is not simpler than what the OP already has done. He zeroed in on the fact that addition is not closed for these matrices, so he's done. And besides, your formula doesn't confirm that scalar multiples of A are in the space.

Eh? You are talking about A as though I have mentioned some specific matrix. A is ANY matrix from my space, so technically, scalar multiples of A fall under that.

Not true.

I am quite sure what I have stated is true. Maybe I should have included some more detail though. The formula that I quoted will work so long as he is checking for subspaces under a vector space. For instance, the above example was checking whether the given set of matrices formed a subspace over the space of all 3x3 matrices (which itself could be considered as a subspace). If the rule what I stated is applied, the condition will not be satisfied. I personally don't see any difference in difficulty.
 

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