I think I am a little bit confused with the terms in the title, so I hope you can correct me if I got it wrong...

$$ \left\lbrace \begin{bmatrix} x_{1}\\0\\0 \end{bmatrix} : x_{1} \in \mathbb{R} \right\rbrace $$ is a vector space. So far so good.

The dimension of the vector space is the number of vectors in the basis.

In class, I wrote down that a basis is

$$ B=(\begin{bmatrix}{1}\\{0}\\{0}\end{bmatrix}, \begin{bmatrix}{0}\\{0}\\{0}\end{bmatrix}, \begin{bmatrix}{0}\\{0}\\{0}\end{bmatrix}) $$

But this does not seem logical to me now, as it should be the smallest number of independent vectors that span the vector space. And shouldn’t that be just $ (\begin{bmatrix}{1}\\{0}\\{0}\end{bmatrix}) $

So this is the place where I'm not sure: what is the dimension of the vector space? It is a line in ${\mathbb R}^3$, and it would seem logical that the dimension of a line is $1$.

And what would the rank be? I totally confused myself.

Or is it like the dimension is $3$ and the rank is $1$. so the above solution for the basis would be correct...

Many thanks for your help

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    $\begingroup$ A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice). $\endgroup$ – Bungo Nov 7 '18 at 17:24
  • $\begingroup$ Yes sorry i will add it $\endgroup$ – Lillys Nov 7 '18 at 17:28
  • $\begingroup$ "it should be the smallest number...shouldn't that just be $\left(\left[\begin{smallmatrix}1\\0\\0\end{smallmatrix}\right]\right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space. $\endgroup$ – JMoravitz Nov 7 '18 at 17:30

The differences:

A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.

The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.

The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)

The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:V\to W$, the rank of $f$ is $\dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.

So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.

So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.

The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($\mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $\mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $W\subseteq V$ is that $\dim(W)\leq \dim(V)$, which your example demonstrates.

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  • $\begingroup$ I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer $\endgroup$ – Lillys Nov 7 '18 at 18:19
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    $\begingroup$ @Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say. $\endgroup$ – rschwieb Nov 7 '18 at 18:21
  • $\begingroup$ Im sorry im bothering you again, but i have one more (last question) if A is a linear transformation than the rank(A) =dim(img(A)), n = dim(img(A)) + dim(ker(A)) and n-rank(A)=dim(kern(A)) is this correct??? You cant believe how thankful i am i confuse these terms all the time $\endgroup$ – Lillys Nov 8 '18 at 10:10
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    $\begingroup$ @Lillys Yes, those all say the same thing, the so called 'rank-nullity' theorem. $\endgroup$ – rschwieb Nov 8 '18 at 10:47

You are correct in finding just one vector in your basis.

The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.

Usually the rank is defined if you have a linear transformation or a matrix.

I did not see a matrix or a linear transformation in your statement to find the rank of it.

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You don't want to think of $$\begin{bmatrix} x_{1}\\0\\0 \end{bmatrix}$$ being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want $$\left\lbrace \begin{bmatrix} x_{1}\\0\\0 \end{bmatrix} : x_{1} \in \mathbb{R} \right\rbrace$$

Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.

Then your basis for this vector space will contain a single vector, and so is one-dimensional.

As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.

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