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Show how the axioms for a vector space $V$ can be used to prove the elementary properties described after the definition of a vector space. Fill in the blanks with the appropriate axiom numbers. Because of Axiom $2,$ Axioms 4 and 5 imply, respectively, that $\mathbf{0}+\mathbf{u}=\mathbf{u}$ and $-\mathbf{u}+\mathbf{u}=\mathbf{0}$ for all $\mathbf{u}$.Fill in the missing axiom numbers in the following proof that $0 \mathbf{u}=\mathbf{0}$ for every $\mathbf{u}$ in $V .$$0 \mathbf{u}=(0+0) \mathbf{u}=0 \mathbf{u}+0 \mathbf{u}$ $\qquad$ by Axiom ______ (a)Add the negative of 0 $\mathbf{u}$ to both sides:$\begin{aligned} 0 \mathbf{u}+(-0 \mathbf{u}) &=[0 \mathbf{u}+0 \mathbf{u}]+(-0 \mathbf{u}) \\ 0 \mathbf{u}+(-0 \mathbf{u}) &=0 \mathbf{u}+[0 \mathbf{u}+(-0 \mathbf{u})] \\ \mathbf{0} &=0 \mathbf{u}+\mathbf{0} \\ \mathbf{0} &=0 \mathbf{u} \end{aligned}$by Axiom ____ (a)by Axiom ____ (b)by Axiom ____ (c)

Given $0 \mathbf{u}=\mathbf{0},$ for every $\mathbf{u}$ in $V$$0 \mathbf{u}=(0+0) \mathbf{u}=0 \mathbf{u}+0 \mathbf{u}[\text { By Axiom } 8,(c+d) \mathbf{u}=c \mathbf{u}+d \mathbf{u}]-(a)$Add the negative of 0 $\mathbf{u}$ to both sides$0 \mathbf{u}+(0 \mathbf{u})=[0 \mathbf{u}+0 \mathbf{u}]+(-0 \mathbf{u})$$0 \mathbf{u}+(-0 \mathbf{u})=0 \mathbf{u}+[0 \mathbf{u}+(-0 \mathbf{u})]^{[\mathrm{By} \text { axiom } 3,(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})] \rightarrow(b)}$$\mathbf{0}=0 \mathbf{u}+\mathbf{0}[\text { By axiom } 5, \mathbf{u}+(-\mathbf{u})=\mathbf{0}] \rightarrow(c)$$\mathbf{0}=0 \mathbf{u}[\text { By axiom } 4, \mathbf{u}+\mathbf{0}=\mathbf{u}] \rightarrow(d)$

Calculus 3

Chapter 4

Vector Spaces

Section 1

Vector Spaces and Subspaces

Vectors

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02:56

In mathematics, a vector (…

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Let $\mathbf{u}=\langle a,…

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in this video song, Call number 28 of Section four. Bone, which gives us two steps of a proven tells us what axioms. Uh, these two steps are possible by and the first one's asked him for, as it says, that any vector is added to the zero vector. So any vector you in, uh, vector space would be equal to you plus zero or vice versa. And you is also here. Zero is the vector that is being attitude. So you can also be zero in this. Um, in this case, so zero vector. It also applies for the zero vector. If you were to be zero vector. So that's why we're were able to weaken. Read Dr Step One and Step two is given to us by axiom seven, which states that, uh, any scaler value see times two vectors u plus V would equal to see you plus CV and they're interchangeable. So the scaler value just distributes to you to the Becker's

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