# Thread: [SOLVED] Proving k*0=0 using linear algebra

1. ## [SOLVED] Proving k*0=0 using linear algebra

Prove that in any vector space k × 0 = 0, where k is any real number and 0 is the zero vector.

Do I use the distributivity of scalar multiplication with respect to field addition ??

(a+b)v= av+bv

let a=b=0

(0+0)v=0v+oV
?????

little confused since it's pretty much obvious!

Thanks

2. Originally Posted by Khonics89
Prove that in any vector space k × 0 = 0, where k is any real number and 0 is the zero vector.

I know I have to use the vector axioms..

AxiomSignificationAssociativity of additionu + (v + w) = (u + v) + w.Commutativity of additionv + w = w + v.Identity element of additionThere exists an element 0V, called the zero vector, such that v + 0 = v for all vV.Inverse elements of additionFor all v ∈ V, there exists an element wV, called the additive inverse of v, such that v + w = 0. The additive inverse is denoted −v.Distributivity of scalar multiplication with respect to vector addition  a(v + w) = av + aw.Distributivity of scalar multiplication with respect to field addition(a + b)v = av + bv.Compatibility of scalar multiplication with field multiplicationa(bv) = (ab)v [nb 3]Identity element of scalar multiplication1v = v, where 1 denotes the multiplicative identity in F.

Do I use the distributivity of scalar multiplication with respect to field addition ??

(a+b)v= av+bv

let a=b=0

(0+0)v=0v+oV
?????

little confused since it's pretty much obvious!

Thanks
$\alpha\bold{0}=\alpha(\bold{0}+\bold{0})=\alpha\bo ld{0}+\alpha\bold{0}\implies \alpha\bold{0}=\bold{0}$

3. Originally Posted by Khonics89
Prove that in any vector space k × 0 = 0, where k is any real number and 0 is the zero vector.

I know I have to use the vector axioms..

AxiomSignificationAssociativity of additionu + (v + w) = (u + v) + w.Commutativity of additionv + w = w + v.Identity element of additionThere exists an element 0V, called the zero vector, such that v + 0 = v for all vV.Inverse elements of additionFor all v ∈ V, there exists an element wV, called the additive inverse of v, such that v + w = 0. The additive inverse is denoted −v.Distributivity of scalar multiplication with respect to vector addition  a(v + w) = av + aw.Distributivity of scalar multiplication with respect to field addition(a + b)v = av + bv.Compatibility of scalar multiplication with field multiplicationa(bv) = (ab)v [nb 3]Identity element of scalar multiplication1v = v, where 1 denotes the multiplicative identity in F.

Do I use the distributivity of scalar multiplication with respect to field addition ??

(a+b)v= av+bv

let a=b=0

(0+0)v=0v+oV
?????

little confused since it's pretty much obvious!

Thanks
hate these little proofs! always make an error
I would say

V+0=V
k.(V+0)=k.V
Kv+K0=KV

but kV is a vector in so additive inverse vector exists ,-kV
so

-kV+kV+k0=-KV+kV
0+k0=0
kO=0