Question

1. Let V = set of all ordered pairs in R^2 and define the operations on V
(x,y) + (x1,y1) = (x+x1,y_y1)
k(x,y) = (kx,y)
Determine whether V is a vector space.

2. Which of the following are linear combination of u = (1,-3,2) and v = (1,0,-4)
a. (0,-3,6) c. (0,0,0)
b. (3,-9,-2) d. (1,6,-16)

Determine whether the vectors
V1 = (1,-2,3) V2 = (5,6-1) V3 = (3,2,1)
are linear independent or linear dependent

Answer 1

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Answer 2

3. Determine whether the vectors
V1 = (1,-2,3) V2 = (5,6-1) V3 = (3,2,1)
are linear independent or linear dependent

Answer 3

Help me T-T

Answer 4

Yes please :(
T---------------T

Answer 5

i think you should take a look at this
http://openstudy.com/study#/updates/565bceb0e4b0b09ed5de4edc

Answer 6

Let \(V(R)=\left\{ (x_1,x_2)/x_1,x_2 \in \mathbb{R} \right\}\)
With this, we are able to construct a vectoria lspace in two dimensions that will allows us to define three parameters in which we will be able to study if this is a valid Vectorial space.
therefore, let's define \(\alpha , \beta , \theta \in V(R)\) as three vectors whose tail belong in the reference system, since for this vectors to form a vectorial space means they are linearlly independant, we deduce:
\[\alpha (x_1,x_2), \beta (y_1, y_2), \theta (z_1,z_2)\]
Let \(c_1,c_2,c_3 \in F \iff (V_1)(V,+) \) is an abelian group.
Now, with all this defined we have to study if all the Real axioms are satisfied by the vectors we defined, thus, proving that this vectorial space is indeed a Vectorial space over the field of the real numbers.
I'll show you a couple, we can begin with the addition one:
\[(i) \alpha + \beta=(x_1,x_2)+(y_1,y_2) \iff \alpha + \beta=(x_1+y_1, x_2 + y_2)\]
As you can see, the addition property is satisfied, we can move on to associative maybe:
\[\left[ ii \right]\alpha + (\beta+\theta)=(\alpha+\beta)+\theta\]
starting from the left side:
\[\alpha + (\beta+\theta)=(x_1,x_2)+((y_1,y_2)+(z_1,z_2))\]
\[(x_1,x_2)+(y_1+z_1,y_2+z_2) = [(x_1+y_1+z_1),(x_2+y_2+z_2)]\]
The right side:
\[(\alpha+\beta)+ \theta=[(x_1,x_2)+(y_1,y_2)]+(z_1,z_2)\]
\[[(x_1+y_1,x_2+y_2)]+(z_1,z_2) = [(x_1+y_1+z_1),(x_2+y_2+z_2)]\]
Thus proving that the associative property s satisfied in this vectorial space.
As a final example, I'll try to prove the existence of a neutral element 0:
\[\ [iii] \exists 0=(0,0) \in V \iff 0+\alpha = \alpha \]
\[(0,0)+(x_1,x_2)=[(x_1+0),(x_2+0)]=(x_1,x_2)=\alpha\]

Answer 7

Now, you have to prove that:
there exists an opposite element,
that the commutative property is satisfied,
that the distributive property is satisfied,
that there exists a neutral element of multiplication
That a vector multiplied an ordered pair yield: \((c_1+c_2) \alpha= (c_1\alpha+c_2 \alpha)\).