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Why are the following two subspaces are not the same ?

  1. all of $R^{2}$
  2. All lines through vector \begin{bmatrix} 0 \\0 \end{bmatrix}

REFERENCE: Lecture 5

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    A line does not contain all the points of $\Bbb{R}^2$. IOW *any line* refers to a single, unspecified line. – 2017-01-05
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    I have made changed the word any to all ? Are they then the same ? – 2017-01-05
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    "all lines through $0$" is not a subspace (it is not even a subset). All lines are subspaces, (any of those is a subspace). – 2017-01-05
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    Depends on your definition of "the same". Both have the same dimension, hence there is a linear isomorphism between them. – 2017-01-05
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    I agree with Crostul that when you talk about several lines through the origin, then you are talking about a collection of 1-dimensional subspaces. You can say that *the union of all the lines through the origin is the same as $\Bbb{R}^2$*. – 2017-01-05
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    The question does not correspond to the image. The lecturer says that any line through the origin is a (proper) subspace of $\Bbb R^{2}$. – 2017-01-05
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    Why wont "all" of the "any lines" NOT be the same as $R^{2}$ ? – 2017-01-05
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    You need to be careful when you describe sets. A line through the origin is a set that has individual points as elements. All the lines (a very hazy description if you ask me) describes a set that has *lines* (as opposed to points) as elements. And, this is not just splitting hairs. We absolutely need to be careful in making the distinction, because otherwise we would not be able to communicate in precise terms. – 2017-01-05
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    And, your teacher is clearly saying that any single one of those lines is a subspace. That picture describes a list of subsets of $\Bbb{R}^2$. The first and the last items on the list describe only a single subset. The second item describes infinitely many subsets. Any of them a subspace by itself. Your teacher has a reason for doing it this way, grouping those infinitely many lines. It will become clear later. – 2017-01-05

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I think you are a little bit confused. I don't know how the lecturer defined a vector space but I'll try to explain what seems to bother you. I won't get into the details but I shall try to remain intuitive.

A line through the origin is a vector space. If you give one non-zero vector of the line, you can determine any other vector of the line as a multiplication by a real number of this vector:

For example, take the line through $(0,0)$ and $(1,1)$. This contains the vector $(1,1)$ and any vector of this line is of the form $(x,x) = x(1,1)$ where $x\in\Bbb R$. Such a vector $(1,1)$ is called a basis of the line through $(0,0)$ and $(1,1)$.

The dimension of a vector space is the number of elements in its basis. This is intuitive: the basis is the smallest set of vectors you have to specify to determine any other vector of the vector space.

For example, the space $\Bbb R^{2}$ can be seen as the set $\{(x,y)\vert x\in\Bbb R, y\in\Bbb R\}$, that is, the set of all ordered pairs of real numbers. To specify a vector, you have to give two numbers: $x$ and $y$. From a vector space perspective, you can see $\Bbb R^{2}$ as the space of all combinations of $(1,0)$ and $(0,1)$: $$\Bbb R^{2} = \{x(1,0) + y(0,1)\vert x\in\Bbb R, y\in\Bbb R\}$$ Beware that this representation is not unique, you could take $(-1,0)$ and $(1,1)$ as a basis of $\Bbb R^{2}$. Indeed, we have $$(x,y) = x(-1,0) + (2x+y)(1,1)$$ Hence, we can generate all the vectors $(x,y)$ of $\Bbb R^{2}$ as a combination of $(-1,0)$ and $(1,1)$.

Such combinations where you multiply by a number and add vectors are called "linear combinations". A vector space is a space containing the zero vector and that is stable under linear combinations: adding the zero vector to an initial vector does not change the initial vector, adding vectors yields a new vector and multiplying a vector by a number yields a vector.

The importance is that the vectors of a basis cannot be expressed as a combinations of the other vectors of this basis.

For example, $(-1,0)$ and $(1,0)$ cannot be a basis for $\Bbb R^{2}$ since $(-1,0) = (-1) \cdot (1,0)$. Similarly, $\{(1,0),(0,2),(1,-1)\}$ is not à basis because it is not minimal: this is expressed by the fact $(1,1)=(1,0) + \tfrac{-1}{2}(0,2)$.

$\Bbb R^{2}$ is of dimension $2$: you have to specify two "parameters" to obtain a vector of $\Bbb R^{2}$. Hence, it is distinct from any line through $(0,0)$, which is of dimension $1$.

There is a big difference between "any line through $(0,0)$" and "all lines through $(0,0)$".

"Any line through $(0,0)$" means "one arbitrary line through $(0,0)$".

"All lines through $(0,0)$" means "the collection of all the lines through $(0,0)$"

These two things are distinct since there exists (at least) two distincts line through $(0,0)$. For example, the line through $(0,0)$ and $(1,0)$ is distinct from the line through $(0,0)$ and $(0,1)$.