Determine if vector space.

[karush]

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 $k=2$

 

[topsquark]

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

 

[karush]

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$

 

[topsquark]

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

 

[HOI]

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".

 

[karush]

That was a great help ..
Much Mahalo

It hard to find really good help with these