I Dimension of a vector space and its subspaces

[Txema]

TL;DR Summary

Can a vector subspace have the same dimension as the space it is part of?

Can a vector subspace have the same dimension as the space it is part of?
If so, can such a subspace have a Cartesian equation?
if so, can you give an example.

Thanks in advance;

 

[PeroK]

TL;DR Summary: Can a vector subspace have the same dimension as the space it is part of?

Can a vector subspace have the same dimension as the space it is part of?
If so, can such a subspace have a Cartesian equation?
if so, can you give an example.

Thanks in advance;

Is this homework?

Technically, a vector space is a subspace of itself. This is sometimes called a trivial subspace or not a proper subspace. See, for example:

https://en.wikipedia.org/wiki/Linear_subspace

 

[fresh_42]

TL;DR Summary: Can a vector subspace have the same dimension as the space it is part of?

Can a vector subspace have the same dimension as the space it is part of?
If so, can such a subspace have a Cartesian equation?
if so, can you give an example.

Thanks in advance;

If the dimension is finite, then inclusion $U\subseteq V$ AND equal (finite) dimension implies equality $U=V.$ Both properties are required for this conclusion.

If the dimension is infinite, then things are more complicated. For example $\mathbb{Q}\subseteq \mathbb{R}$ has "the same dimension over $\mathbb{Q}$" as $\mathbb{Q}(\sqrt{2})\subseteq \mathbb{R}$ and $\mathbb{Q}\subsetneq \mathbb{Q}(\sqrt{2})$ but they are obviously not equal. Infinite cardinalities behave different than ordinary numbers.

 

[Bosko]

TL;DR Summary: Can a vector subspace have the same dimension as the space it is part of?

Can a vector subspace have the same dimension as the space it is part of?
If so, can such a subspace have a Cartesian equation?
if so, can you give an example.

Thanks in advance;

The dimension of a vector space is the number of vectors in its basis.
Each basis of a vector space has the same number of vectors and $dim(Ssub) \leq dim(S)$.

Technically, a vector space is a subspace of itself. This is sometimes called a trivial subspace or not a proper subspace. See, for example:

The Cartesian equation (If it is a scalar product that determines the subspace) is also trivial.
The scalar product with zero vector
$\langle 0,x \rangle=0 \cdot x_1+0 \cdot x_2+...+0 \cdot x_n=0$

That is it in finite dimensional spaces .

When the vector space has infinite dimensions, you can use the property that $\infty+1$ is still $\infty$.

You can choose the set of all infinite sequences as a vector space $x=(x_1, x_2, x_3, ... ) \in S$
and all sequences which sum is zero as a subspace $Ssub$ . Its dimension is $\infty$.

The Cartesian equation in that situation is
$\langle 1,x \rangle=1 \cdot x_1+1 \cdot x_2+...+1 \cdot x_n +...=0$
where 1 is infinite series $(1, 1, 1, ... )$

All constant sequences ( which all members are the same number ) is a subspace . Its dimension is 1.
e.g. $(1, 1, 1, ...)$, $(-4, -4, -4, ... )$, ...

A proper, non-trivial, subspace with the same dimension as the space can only be obtained if the vector space is infinite-dimensional.

 

[Txema]

Is this homework?

No, I wish, I'm 60 years old.
Thank you Perok for your answer.

 

[Txema]

fresh_42, Bosko, my question is perfectly answered, thank you very much.

 

[martinbn]

If the dimension is infinite, then things are more complicated. For example $\mathbb{Q}\subseteq \mathbb{R}$ has "the same dimension over $\mathbb{Q}$" as $\mathbb{Q}(\sqrt{2})\subseteq \mathbb{R}$ and $\mathbb{Q}\subsetneq \mathbb{Q}(\sqrt{2})$ but they are obviously not equal. Infinite cardinalities behave different than ordinary numbers.

This doesn't fit in the question, does it? What are the space and the subspace with the same dimensions?

 

[fresh_42]

This doesn't fit in the question, does it? What are the space and the subspace with the same dimensions?

You are right, I should have written ...

The subspaces are $\mathbb{Q}(\pi)$ and $\mathbb{Q}(\pi,\sqrt{2})$ contained in the rational vector space $\mathbb{R}.$ The dimensions of both subspaces are then equal but the subspaces are not.

 

[martinbn]

It fits perfectly.

The subspaces are $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ contained in the rational vector space $\mathbb{R}.$ The dimensions of both subspaces are countably infinite, the dimension of $\mathbb{R}$ is uncountable infinite.

I don't understand. What do you mean the dimensions are countably infinite? The dimensions of $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ over the field $\mathbb{Q}$ are $1$ and $2$.

But the question ask for a subspace of a space such that their dimensions are the same. So is it possible to have two spaces (over a field $F$) say $W\subset V$ such that $dim_FW=dim_FV$? What are the $W$ and $V$ in your example?

 

[fresh_42]

I don't understand. What do you mean the dimensions are countably infinite? The dimensions of $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ over the field $\mathbb{Q}$ are $1$ and $2$.

But the question ask for a subspace of a space such that their dimensions are the same. So is it possible to have two spaces (over a field $F$) say $W\subset V$ such that $dim_FW=dim_FV$? What are the $W$ and $V$ in your example?

I corrected it.

 

[martinbn]

I corrected it.

Still, what are the $W$ and $V$ in your example?

 

[fresh_42]

Still, what are the $W$ and $V$ in your example?

\begin{align*}
\mathrm{F}&=\mathbb{Q}\\
\mathrm{V}&=\mathbb{R}\\
\mathrm{W_1}&=\mathbb{Q}(\pi)\\
\mathrm{W_2}&=\mathbb{Q}(\pi,\sqrt{2})\\
\end{align*}
\begin{align*}
\dim_\mathrm{F} \mathrm{W_1}=\infty = \dim_\mathrm{F} \mathrm{W_2} \wedge \mathrm{W_1}\subsetneq \mathrm{W_2}
\end{align*}

 

[martinbn]

\begin{align*}
\mathrm{F}&=\mathbb{Q}\\
\mathrm{V}&=\mathbb{R}\\
\mathrm{W_1}&=\mathbb{Q}(\pi)\\
\mathrm{W_2}&=\mathbb{Q}(\pi,\sqrt{2})\\
\end{align*}
\begin{align*}
\dim_\mathrm{F} \mathrm{W_1}=\infty = \dim_\mathrm{F} \mathrm{W_2} \wedge \mathrm{W_1}\subsetneq \mathrm{W_2}
\end{align*}

Then there is a missing $\pi$ in your first post. Why do you need to include $\mathbb R$?

 

[fresh_42]

Then there is a missing $\pi$ in your first post. Why do you need to include $\mathbb R$?

I was mistaken. My thought was that $\mathbb{Q}\subseteq \mathbb{R}$ was a vector space and a subspace with an uncountably infinite index. I then confused index and dimension instead of building a correct counterexample.

 

[martinbn]

I was mistaken. My thought was that $\mathbb{Q}\subseteq \mathbb{R}$ was a vector space and a subspace with an uncountably infinite index. I then confused index and dimension instead of building a correct counterexample.

I know. I am not picking on you. I just thought it would be usfull to the OP if you clarify it. By the way the real numbers as a subspace of the complex numbers, both over the rationals may be an easier example for the OP.

 

[mathwonk]

since the dimension of a vector space is the cardinality of a basis, and every basis of a subspace extends to a basis of the containing space, and every set is the index set for a basis of a vector space, a simpler version of this question is whether a proper subset of a set can have the same cardinality.