Did I do this correctly?
*note: square brackets on top of each other represent a single matrix so,
[a]
[b]
is actually a 2x1*
Let w = {
[a] : a, b are elements in all reals, a + b = 1
[b]}
define:
[a] + [c] = [a + c  1]
[b] + [d] = [ b + d ]
and
r*[a] = [ra  r + 1]
[b] = [ rb ]
Begin:
[a] + [c] = [a + c  1]
[b] + [d] = [ b + d ]
=
[a + (c  1)]
[ b + d ]
=
[(c  1) + a]
[ d + a ]
=
[c] + [a]
[d] + [b]
thus, closed under addition
r*[a] = [ra  r + 1]
[b] = [ rb ]
=
[r(a 1)+ 1]
[ rb ]
=
r[a  1] + [1]
[b] [0]
but this cannot simplify down to:
r[a]
[b]
this fails scaler multiplication and is NOT a vector space
3 Replies  322 Views  Last Post: 11 March 2015  07:55 AM
#1
Linear Algebra: Prove/Disprove that this is a vector space
Posted 10 March 2015  07:17 PM
Replies To: Linear Algebra: Prove/Disprove that this is a vector space
#2
Re: Linear Algebra: Prove/Disprove that this is a vector space
Posted 10 March 2015  07:22 PM
You didn't prove closure under addition. You proved commutativity under addition.
Just show it fails on scalar multiplication. 2 * (0, 1) = (0, 2).
By the way your vector notation is a pain to read. Just use standard tuple notation. If you want a column vector rather than a rowvector, use the transpose operator.
Just show it fails on scalar multiplication. 2 * (0, 1) = (0, 2).
By the way your vector notation is a pain to read. Just use standard tuple notation. If you want a column vector rather than a rowvector, use the transpose operator.
#3
Re: Linear Algebra: Prove/Disprove that this is a vector space
Posted 11 March 2015  12:36 AM
macosxnerd101, on 10 March 2015  07:22 PM, said:
You didn't prove closure under addition. You proved commutativity under addition.
Just show it fails on scalar multiplication. 2 * (0, 1) = (0, 2).
By the way your vector notation is a pain to read. Just use standard tuple notation. If you want a column vector rather than a rowvector, use the transpose operator.
Just show it fails on scalar multiplication. 2 * (0, 1) = (0, 2).
By the way your vector notation is a pain to read. Just use standard tuple notation. If you want a column vector rather than a rowvector, use the transpose operator.
OK thanks but when I do 2 (0,1) I get
(0  2 +1, 2)=(1,2)...what did I do wrong?
#4
Re: Linear Algebra: Prove/Disprove that this is a vector space
Posted 11 March 2015  07:55 AM
Nope. I'm the one who screwed up. It looks like this might be a vector space. I'll think more about it and get back to you if it isn't.
Yeah it looks like a vector space to me. At the very least, closure is satisfied.
Yeah it looks like a vector space to me. At the very least, closure is satisfied.
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