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Additive Identity

by tc_11
Tags: additive, identity
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Feb1-10, 11:50 PM
P: 8

I am new with linear algebra, and I'm having a hard time wrapping my mind around the 0 vector and the additive identity v + 0 = v, where 0 is the 0 vector.
If I had a 2x2 matrix, and v + w = C + (C^T)*D ... (where (C^T) is the transpose, v & w are vectors, and C & D are matrices)... would the additive identity hold? I feel like it wouldn't, because I don't see how it would be unique... but I think I may be wrong.. can someone please help?
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Lord Crc
Feb2-10, 02:35 AM
P: 88
I can't make sense of your expression, how does adding two vectors get you a 2x2 matrix? Are you confusing something here, or am I the confused one?

Anyway, if the additive identity does not hold, you're not dealing with a vector space and all bets are off (as far as linear algebra is concerned). One of the requirements of a vector space [tex]V[/tex] is that there exists an element [tex]\mathbf{0} \in V[/tex] such that [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] for all [tex]\mathbf{v} \in V[/tex].
Feb2-10, 07:27 AM
P: 8
That's what I'm trying to prove though, that the additive identity v + 0 = v does in fact hold, and if not it's not a vector space, but we have to test the axioms anyway to see which ones do hold.
Addition of 2 vectors in this problem translates to:
vector v := C (where C is a 2x2 matrix)
vector w:= D (where D is a 2x2 matix)
(v+w):= C + (C^T)D
so i would set up my equation as v + 0 =? v
C + (C^T)D =? C

Feb2-10, 07:40 AM
HW Helper
radou's Avatar
P: 3,220
Additive Identity

If 0 is a 2x2 zero-matrix (the 0-vector you were referring to), then v + 0 = v, but 0 + v = 0, which can't hold for a vector space.

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