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

• MHB
• karush
In summary, the given set of vectors $\begin{bmatrix}x\\y\\3x+2y\end{bmatrix}$ $\in \Bbb{R}^3$ forms a vector space with the usual addition and scalar multiplication for vectors in $\Bbb{R}^3$, as it satisfies the criteria of closure under scalar multiplication and vector addition. This can be seen by representing the vectors in the form of a matrix and observing that the characteristic of the third element being equal to $3$ times the first element plus $2$ times the second is preserved under these operations.
karush
Gold Member
MHB
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

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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.

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

this was the given question

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.

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

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.

karush said:
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!

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

karush said:
Mahalo

that helped a lot

Excellent!

karush said:
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.

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 $$\{\begin {bmatrix}1 \\ 0 \\ 3\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}\}$$.

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

Why the capitals X and Y

karush said:
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

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, such as addition and scalar multiplication. In order for a set of vectors to form a vector space, it must satisfy certain properties, such as closure under addition and scalar multiplication, and the existence of a zero vector and additive inverses.

2. How do you determine if a set of vectors forms a vector space?

To determine if a set of vectors forms a vector space, you must check if it satisfies the properties of a vector space. This includes checking for closure under addition and scalar multiplication, the existence of a zero vector and additive inverses, and the distributive and associative properties.

3. What is R^3 form?

R^3 form refers to the three-dimensional coordinate system in which a vector is represented by three coordinates (x, y, z). This is also known as the Cartesian coordinate system.

4. How do you determine if a set of vectors in R^3 form a vector space?

In order for a set of vectors in R^3 to form a vector space, they must satisfy the properties of a vector space, as well as the additional property of being linearly independent. This means that none of the vectors in the set can be written as a linear combination of the other vectors in the set.

5. Can a set of vectors in R^3 form a vector space if they are not linearly independent?

No, a set of vectors in R^3 cannot form a vector space if they are not linearly independent. This is because a vector space must satisfy the property of linear independence, meaning that none of the vectors in the set can be written as a linear combination of the other vectors in the set. If this property is not satisfied, the set of vectors cannot form a vector space.

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