# Is This a Valid Vector Space with Unusual Operations?

• MHB
• karush
In summary, this vector space has a zero vector, additive inverses, associativity, and closure. However, it does not obey the distributive law.
karush
Gold Member
MHB
On the set of vectors
$\begin{bmatrix} x_1 \\ y_1 \end{bmatrix}\in \Bbb{R}^2$
with $x_1 \in \Bbb{R}$, and $y_1$ in $\Bbb{R}^{+}$ (meaning $y_1 >0$) define an addition by
$$\begin{bmatrix} x_1 \\ y_1 \end{bmatrix} \oplus \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \\ y_1y_2 \end{bmatrix}$$
and a scalar multiplication by
$$k \odot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} k x \\ y^{k} \end{bmatrix}.$$
Determine if this is a vector space.
If it is, make sure to explicitly state what the $0$ vector is.
OK the only the only thing I could come up with was $2+2=4$ and $2\cdot 2=4$
and zero vectors are orthogonal with $k=2$

Last edited:
karush said:
On the set of vectors
$\begin{bmatrix} x_1 \\ y_1 \end{bmatrix}\in \Bbb{R}^2$
with $x_1 \in \Bbb{R}$, and $y_1$ in $\Bbb{R}^{+}$ (meaning $y_1 >0$) define an addition by
$$\begin{bmatrix} x_1 \\ y_1 \end{bmatrix} \oplus \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \\ y_1y_2 \end{bmatrix}$$
and a scalar multiplication by
$$k \odot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} k x \\ y^{k} \end{bmatrix}.$$
Determine if this is a vector space.
If it is, make sure to explicitly state what the $0$ vector is.
OK the only the only thing I could come up with was $2+2=4$ and $2\cdot 2=4$
and zero vectors are orthogonal with $k=2$
"zero vectors?" There's only one.

I've got closure, associativity, a zero vector, additive inverses, and it's even commutative. However it doesn't obey the distributive law.

Can you get these?

-Dan

ok I don't know how you would try the distributive property since the scalar was different
Distributive law: For all real numbers c and all vectors $u, v \in V$, $c\cdot(u + v) = c\cdot u + c\cdot v$

karush said:
ok I don't know how you would try the distributive property since the scalar was different
Distributive law: For all real numbers c and all vectors $u, v \in V$, $c\cdot(u + v) = c\cdot u + c\cdot v$
I had this whole blasted thing written out in LaTeX just to find out I made an error. The distributive law also works.

Here it is anyway.

$$\displaystyle k \odot \left ( \left [ \begin{matrix} x_1 \\ y_1 \end{matrix} \right ] \oplus \left [ \begin{matrix} x_2 \\ y_2 \end{matrix} \right ] \right ) = \left ( k \odot \left [ \begin{matrix} x_1 \\ y_1 \end{matrix} \right ] \right ) \oplus \left ( k \odot \left [ \begin{matrix} x_2 \\ y_2 \end{matrix} \right ] \right ) = \left [ \begin{matrix} kx_1 \\ y_1^k \end{matrix} \right ] \oplus \left [ \begin{matrix} kx_2 \\ y_2^k \end{matrix} \right ] = \left [ \begin{matrix} kx_1 + kx_2 \\ y_1^k y_2^k \end{matrix} \right ]$$

$$\displaystyle k \odot \left ( \left [ \begin{matrix} x_1 \\ y_1 \end{matrix} \right ] \oplus \left [ \begin{matrix} x_2 \\ y_2 \end{matrix} \right ] \right ) = k \odot \left [ \begin{matrix} x_1 + x_2 \\ y_1 y_2 \end{matrix} \right ] = \left [ \begin{matrix} k(x_1 + x_2 ) \\ (y_1 y_2)^k \end{matrix} \right ]$$

So they are the same.

-Dan

The 0 vector (additive identity) is $\begin{bmatrix}0 \\ 1\end{bmatrix}$: for any vector $v= \begin{bmatrix}a \\ b\end{bmatrix}$, $v+ 0= 0+ v= \begin{bmatrix}a+ 0 \\ b(1)\end{bmatrix}= \begin{bmatrix}a \\ b\end{bmatrix}= v$.

What about the additive inverse of $\begin{bmatrix}a \\ b\end{bmatrix}$? Calling that $\begin{bmatrix}p \\ q\end{bmatrix}$, We must have $\begin{bmatrix}a \\ b\end{bmatrix}+ \begin{bmatrix}p \\ q \end{bmatrix}= \begin{bmatrix}a+ p \\ bq \end{bmatrix}= \begin{bmatrix} 0 \\ 1\end{pmatrix}$ so we have a+ p= 0 and bq= 1 so we must have p= -a and q= 1/b. That is the reason for the condition "y> 0".

That was a great help ..
Much Mahalo

It hard to find really good help with these

## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and operations that can be performed on those vectors. These operations include addition and scalar multiplication, and they must follow certain properties to be considered a vector space.

## 2. How do you determine if a set is a vector space?

To determine if a set is a vector space, you must check if it satisfies the 10 axioms or properties of a vector space. These properties include the existence of a zero vector, closure under addition and scalar multiplication, and the existence of additive and multiplicative inverses for each vector.

## 3. What are the properties of a vector space?

The properties of a vector space include the existence of a zero vector, closure under addition and scalar multiplication, associativity and commutativity of addition, distributivity of scalar multiplication over addition, and the existence of additive and multiplicative inverses for each vector.

## 4. Can a set be a vector space if it does not have a zero vector?

No, a set cannot be a vector space if it does not have a zero vector. The zero vector is a necessary property of a vector space and without it, the set would not satisfy all of the axioms of a vector space.

## 5. Is every set of vectors a vector space?

No, not every set of vectors is a vector space. To be considered a vector space, a set must satisfy all of the axioms or properties of a vector space. If even one of these properties is not satisfied, the set cannot be considered a vector space.

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