Determine vec {{x},{y},{3x+2y}} in R^3 form a vec space

[karush]

Determine if the set of vectors
$\begin{bmatrix}
x\\y\\3x+2y
\end{bmatrix}$ $\in \Bbb{R}^3$
form a vector space
(with the usual addition and scalar multiplication for vectors in $\Bbb{R}^3$).OK first of all this doesn't have z in it.
So I don't know if this meets the requirement of
whether number of elements in the set are equal to the dimension of given vector spaceOk I assume a matrix can be formed of this as albeit assuming
$x_1,x_2,x_3 \textit{ and } y_1,y_2,y_3$
$\begin{bmatrix}
1&0\\0&1\\3&2
\end{bmatrix}$
I don't see how this would be linearly independent

View attachment 8767​

 

[Ackbach]

Forms a basis? Or forms a vector space? The question title indicates you want to know if these vectors form a vector space, but your comments seem to indicate that you're trying to see if they form a basis. They'll be quite different criteria you'd use for one versus the other.

 

[karush]

https://www.physicsforums.com/attachments/8768

this was the given question

 

[Ackbach]

So number of dimensions, and linear dependence are irrelevant for determining if it's a vector space. Those are criteria for a basis. What you need to check is closure under scalar multiplication and vector addition. My guess is that a number of the other properties you get to inherit from the parent vector space.

 

[karush]

$\lambda \cdot (a,b) = (\lambda a, \lambda b)$
are you suggesting this for closure under scalar multiplication
thus$\lambda\cdot\begin{bmatrix}
1&0\\0&1\\3&2
\end{bmatrix}=
\begin{bmatrix}
\lambda &\\&\lambda \\3\lambda &2\lambda
\end{bmatrix}$

 

[Ackbach]

I was thinking something more along the lines of
$$\lambda\left[\begin{matrix}x\\y\\3x+2y\end{matrix}\right]=\left[\begin{matrix}\lambda x\\ \lambda y\\ \lambda(3x+2y)\end{matrix}\right]=\left[\begin{matrix}\lambda x\\ \lambda y\\ 3\lambda x+2\lambda y\end{matrix}\right],$$
which we can see is another vector in the space. Same idea with vector addition. Basic idea: everything in the vector space looks like triples, where the third element always looks like $3$ times the first element plus $2$ times the second. If that characteristic is preserved under scalar multiplication and vector addition, you're probably home-free.

 

[topsquark]

Determine if the set of vectors
$\begin{bmatrix}
x\\y\\3x+2y
\end{bmatrix}$ $\in \Bbb{R}^3$
form a vector space
(with the usual addition and scalar multiplication for vectors in $\Bbb{R}^3$).OK first of all this doesn't have z in it.

I think you are making difficulties for yourself and didn't understand the nature of the vector given.
x = x, y = y, z = 3x + 2y

So there is a z after all.

Let's look at closure, for example. So say we have two supposed vectors and we add them together.
\(\displaystyle \left [ \begin{matrix} x_1 \\ y_1 \\ 3x_1 + 2y_1 \end{matrix} \right ] \oplus \left [ \begin{matrix} x_2 \\ y_2 \\ 3x_2 + 2y_2 \end{matrix} \right ] = \left [ \begin{matrix} x_1 + x_2 \\ y_1 + y_2 \\ 3(x_1 + x_2) + 2(y_1 + y_2) \end{matrix} \right ] = \left [ \begin{matrix} X \\ Y \\ 3X + 2Y \end{matrix} \right ] \)

After the addition we get a vector of the same form. So addition is closed.

See what you can do with the rest of it.

-Dan

PS Looks like Ackbach beat me to the punch!

 

[karush]

Mahalo (thanks all)

that helped a lot

I tried to see what stack exchange had on this
but that site is very unfriendly and very little examples

ok I have a few more, but will try first

 

[Ackbach]

Mahalo

that helped a lot

Excellent!

I tried to see what stack exchange had on this
but that site is very unfriendly and very little examples

ok I have a few more, but will try first

M.SE is good for what it's good for: questions that are well-defined, and not too high-level (or you go to Math Overflow). What it's emphatically not good at is the back-and-forth required for teaching. Forum software like vBulletin is much better at this than the SE suite of sites.

 

[HOI]

The first thing I observe is that $$\begin{bmatrix} x \\ y \\ 3x+ 2y\end{bmatrix}= \begin {bmatrix}1 \\ 0 \\ 3\end{bmatrix}x+ \begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}y$$. This the two dimensional vector space having basis $$\{\begin {bmatrix}1 \\ 0 \\ 3\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}\}$$.

 

[karush]

\(\displaystyle \left [ \begin{matrix} x_1 \\ y_1 \\ 3x_1 + 2y_1 \end{matrix} \right ] \oplus \left [ \begin{matrix} x_2 \\ y_2 \\ 3x_2 + 2y_2 \end{matrix} \right ] = \left [ \begin{matrix} x_1 + x_2 \\ y_1 + y_2 \\ 3(x_1 + x_2) + 2(y_1 + y_2) \end{matrix} \right ] = \left [ \begin{matrix} X \\ Y \\ 3X + 2Y \end{matrix} \right ] \)

Why the capitals X and Y

 

[topsquark]

Why the capitals X and Y

I'm just defining \(\displaystyle X = x_1 + x_2\) (similar for Y) as a shorthand. If \(\displaystyle x_1\) and \(\displaystyle x_2\) are x components in some "subspace" then X is also in there. It would be the same result if I used \(\displaystyle x_3\), etc. I just picked the capitals in this case. No real reason.

-Dan