The extension lemma has all sorts of consequences that are very useful for making arguments about the dimension of a vector space. In this section we’ll write down the most common ones.
As soon as you see $k$ linearly independent elements in a vector space, you know its dimension is at least $k$.
Let $V$ be a vector space and let ${\mathbf{v}}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{\mathbf{v}}_{k}$ be linearly independent elements of $V$. Then $\mathrm{dim}V\mathrm{\u2a7e}k$.
You can extend these elements to a basis of $V$ having size at least $k$, and the size of that basis is the dimension of $V$. ∎
Any sequence of at least $n\mathrm{+}\mathrm{1}$ elements in a vector space of dimension $n$ is linearly dependent.
If they were linearly independent, we could extend them to a basis of size larger than $n$ using Proposition 4.12.2 contradicting that every basis has size $n$. ∎
For example, if you have 4 vectors in ${\mathbb{R}}^{3}$ you know they must be linearly dependent, no matter what they are.
If $U\mathrm{\u2a7d}V$ then
$dimU\u2a7ddimV$, and
if $dimU=dimV$ then $U=V$.
A basis of $U$ is a linearly independent sequence in $V$, so we can extend it to a basis of $V$. So its size is less than or equal to the size of a basis of $V$.
Let $dimV=n$ and let ${\mathbf{u}}_{1},\mathrm{\dots},{\mathbf{u}}_{n}$ be a basis of $U$, so $U=\mathrm{span}({\mathbf{u}}_{1},\mathrm{\dots},{\mathbf{u}}_{n})$. Suppose for a contradiction that $U\ne V$, and let $\mathbf{v}$ be an element of $V$ not in $U$. Then ${\mathbf{u}}_{1},\mathrm{\dots},{\mathbf{u}}_{n},\mathbf{v}$ is linearly independent (by the extension lemma, Lemma 4.12.1), which contradicts Theorem 4.13.2. ∎
As soon as you have $n$ linearly independent elements in a vector space of dimension $n$, they must be a basis.
Let $V$ be a vector space of dimension $n$. Any sequence of $n$ linearly independent elements of $V$ are a basis of $V$.
Let $U$ be the span of this sequence. This length $n$ sequence spans $U$ by definition, and it is linearly dependent, so it is a basis of $U$ and $dimU=n$. The proposition tells us $U=V$, so in fact the sequence is a basis of $V$. ∎
Consider two sets $X$ and $Y$. What’s the size of $X\cup Y$ in terms of the size of $X$ and the size of $Y$? It isn’t $|X|+|Y|$, in general, because elements belonging to $X$ and $Y$ get counted twice when you add the sizes like this. The correct answer is $|X|+|Y|-|X\cup Y|$. We would like a similar result for sums of subspaces.
Let $V$ be a vector space and $X\mathrm{,}Y\mathrm{\u2a7d}V$. Then
$$dim(X+Y)=dimX+dimY-dimX\cap Y.$$ |
Take a basis $\mathcal{I}={\mathbf{i}}_{1},\mathrm{\dots},{\mathbf{i}}_{k}$ of $X\cap Y$. Extend $\mathcal{I}$ to a basis $\mathcal{X}={\mathbf{i}}_{1},\mathrm{\dots},{\mathbf{i}}_{k},{\mathbf{x}}_{1},\mathrm{\dots},{\mathbf{x}}_{n}$ of $X$, using 4.12.2. Extend $\mathcal{I}$ to a basis $\mathcal{Y}={\mathbf{i}}_{1},\mathrm{\dots},{\mathbf{i}}_{k},{\mathbf{y}}_{1},\mathrm{\dots},{\mathbf{y}}_{m}$ of $Y$. It’s now enough to prove that $\mathcal{J}={\mathbf{i}}_{1},\mathrm{\dots},{\mathbf{i}}_{k},{\mathbf{x}}_{1},\mathrm{\dots},{\mathbf{x}}_{n},{\mathbf{y}}_{1},\mathrm{\dots},{\mathbf{y}}_{m}$ is a basis of $X+Y$, because if we do that then we will know the size of $\mathcal{J}$, which is $k+n+m$, equals the size of a basis of $\mathcal{I}$ (which is $k+n$) plus the size of a basis of $Y$ (which is $k+m$) minus the size of a basis of $X\cap Y$ (which is $k$).
To check something is a basis for $X+Y$, as always, we must check that it is a spanning sequence for $X+Y$ and that is it linearly independent.
Spanning: let $\mathbf{x}+\mathbf{y}\in X+Y$, where $\mathbf{x}\in X,\mathbf{y}\in Y$. Then there are scalars such that
$\mathbf{x}$ | $={\displaystyle \sum _{j=1}^{k}}{a}_{j}{\mathbf{i}}_{j}+{\displaystyle \sum _{j=1}^{n}}{c}_{j}{\mathbf{x}}_{j}$ | ||
$y$ | $={\displaystyle \sum _{j=1}^{k}}{b}_{j}{\mathbf{i}}_{j}+{\displaystyle \sum _{j=1}^{m}}{d}_{j}{\mathbf{y}}_{j}$ |
and so
$$\mathbf{x}+\mathbf{y}=\sum _{j=1}^{k}({a}_{j}+{b}_{j}){\mathbf{i}}_{j}+\sum _{j=1}^{n}{c}_{j}{\mathbf{x}}_{j}+\sum _{j=1}^{m}{d}_{j}{\mathbf{y}}_{j}$$ |
Linear independence: suppose
$$\sum _{j=1}^{k}{a}_{j}{\mathbf{i}}_{j}+\sum _{j=1}^{n}{c}_{j}{\mathbf{x}}_{j}+\sum _{j=1}^{m}{d}_{j}{\mathbf{y}}_{j}=0.$$ |
Rearrange it:
$$\sum _{j=1}^{k}{a}_{j}{\mathbf{i}}_{j}+\sum _{j=1}^{n}{c}_{j}{\mathbf{x}}_{j}=-\sum _{j=1}^{m}{d}_{j}{\mathbf{y}}_{j}.$$ |
The left hand side is in $X$ and the right hand side is in $Y$. So both sides are in $X\cap Y$, in particular, the right hand side is in $X\cap Y$. Since $\mathcal{I}$ is a basis of $X\cap Y$, there are scalars ${e}_{j}$ such that
$$\sum _{j=1}^{k}{e}_{j}{\mathbf{i}}_{j}=-\sum _{j=1}^{m}{d}_{j}{\mathbf{y}}_{j}$$ |
This is a linear dependence on $\mathcal{Y}$ which is linearly independent, so all the ${d}_{j}$ are 0. Similarly all the ${c}_{j}$ are 0. So the ${a}_{j}$ are 0 too, and we have linear independence. ∎