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#1 streek405  Icon User is offline

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Linear Algebra: Prove/Disprove that this is a vector space

Posted 10 March 2015 - 07:17 PM

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

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Replies To: Linear Algebra: Prove/Disprove that this is a vector space

#2 macosxnerd101  Icon User is online

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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 row-vector, use the transpose operator.
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#3 streek405  Icon User is offline

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Re: Linear Algebra: Prove/Disprove that this is a vector space

Posted 11 March 2015 - 12:36 AM

View Postmacosxnerd101, 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 row-vector, 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?
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#4 macosxnerd101  Icon User is online

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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.
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