How can a subspace have a lower dimension than its parent space?

Question

If $V$ is a vector subspace of $W$, then

$$\dim(V) \le \dim(W)$$

Why? Does that mean that for

$$W = \mathbb{R}^3\\ V = \{(0,0)\}$$

$V$ is a valid subspace of $W$? But $V$ only has two coordinates, and $W$ has $3$...

I've always been under the impression that a subspace's dimension should be equal to its parent space's dimension.

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EDIT

I made a mistake with my example - the answer is no (as you all pointed out), however, the real core question is, how can a subspace have a lower dimension than its parent space?

Answer 1

No. $\{(0,0)\}$ is not a subspace of $\mathbb{R}^3$.

A subspace must be a subset of its parent vector space.

Now, $\{(0,0,0)\}$ is a subspace of $\mathbb{R}^3$ and $\{(0,0)\}$ is a subspace of $\mathbb{R}^2$, but $\{(0,0)\}$ is not a subspace of $\mathbb{R}^3$.

As for your other question, a subspace's dimension cannot exceed its parent's dimension, but it by no means must be equal to it.

For example: The subspaces of $\mathbb{R}^3$ are...

$\mathbb{R}^3$ itself (every vector space is a subspace of itself).

Any plane through the origin is a 2-dimensional subspace of $\mathbb{R}^3$.

Any line through the origin is a 1-dimensional subspace of $\mathbb{R}^3$.

The origin itself, $\{(0,0,0)\}$ is the 0-dimensional subspace of $\mathbb{R}^3$.

Why must the dimension of a subspace not exceed the dimension of it's parent? It's essentially because any linearly independent subset (like a basis for a subspace) can be extended to a basis for the whole vector space.

Edit: Coordinates written down to represent a subspace $\not=$ the dimension of the subspace.

Example: $W = \{(c,2c,3c)\;|\; c \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^3$. Now, yes, elements of $W$ are 3-tuples, but this does not make $W$ itself 3-dimensional.

Think of "dimension" as meaning the minimum number of parameters needed to describe the subspace, so for $W$ this is "1". Notice that $W$ consists of multiples of $(1,2,3)$. This means that $\{(1,2,3)\}$ is a basis for $W$ and thus (since the basis has only 1 element), $W$ is a 1-dimensional subspace of $\mathbb{R}^3$.

Now what if all you know is the world of $W$? Then you really don't need a 3-tuple to know who you are. Instead of saying, "Hey I'm $(2,4,6)$." You could just say, "Hey I'm 2." (as long as everyone knows that's shorthand for $2(1,2,3)=(2,4,6)$.

Another analogy is like driving on a highway. If someone asks you where you are on I-40 in North Carolina, you don't need to give them your latitude and longitude. You could just say, I'm at mile marker 319. Even though the highway is a 3-dimensional thing (going different directions at different altitudes) in some sense (internally) it's really 1-dimensional. That internal kind of dimension is the one being discussed here.

I hope this helps!

Answer 2

Sub- just means within.

-space means when viewed in isolation from the parent space, it is a vector space in its own right.

In using the term "subspace", there is no implication that the subspace has to have the same dimension as the parent space.

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Also, you are confusing what dimension means. The set $\{(0,0,0)\}$ has dimension zero, because it is just a point. The fact that we are discussing an element that is expressed as an ordered triple is irrelevant to discussing dimension. At best, that just tells us our space is a subspace of $\mathbb{R}^3$.

Answer 3

I've found it useful not to picture a space in its totality. Try to understand what a space is by its bases or spanning sets. Suppose $V$ is a vector space with dimension $n$. That is, any $n$ linearly independent vectors in $V$ will span it. Now consider a linearly independent set of vectors in $V$ with less than $n$ vectors. Clearly the subspace spanned by this set has a smaller dimension, by definition.

Now for your question. Consider any set of vectors in $V$ with more than $n$ vectors. You might have learnt that this set cannot be linearly independent and cannot be a basis for $V$. Furthermore every subspace in $V$ will have a spanning set and hence a basis. This discussion should tell you that the dimension of such subspaces must be less than that of the whole space since a basis for such a set will contain $n$ or less vectors.