Subspaces of Rn and Linear Independence

Subspaces

For any vector space, a subspace is a subset that is itself a vector space, under the inherited operations.

Important Lemma on Subspaces

For a nonempty subset $S$ of a vector space, under the inherited operations, the following are equivalent statements.

1. $S$ is a subspace of that vector space
2. $S$ is closed under linear combinations of pairs of vectors: for any vectors ${\vec {s}}_{1},{\vec {s}}_{2}\in S$ and scalars $r_{1},r_{2}$ the vector $r_{1}{\vec {s}}_{1}+r_{2}{\vec {s}}_{2}$ is in $S$
3. $S$ is closed under linear combinations of any number of vectors: for any vectors ${\vec {s}}_{1},\ldots ,{\vec {s}}_{n}\in S$ and scalars $r_{1},\ldots ,r_{n}$ the vector $r_{1}{\vec {s}}_{1}+\cdots +r_{n}{\vec {s}}_{n}$ is in $S$.

Briefly, the way that a subset gets to be a subspace is by being closed under linear combinations.

Proof:

"The following are equivalent" means that each pair of statements are equivalent.

$(1)\!\iff \!(2)\qquad (2)\!\iff \!(3)\qquad (3)\!\iff \!(1)$

We will show this equivalence by establishing that $(1)\implies (3)\implies (2)\implies (1)$. This strategy is suggested by noticing that $(1)\implies (3)$ and $(3)\implies (2)$ are easy and so we need only argue the single implication $(2)\implies (1)$.

For that argument, assume that $S$ is a nonempty subset of a vector space $V$ and that $S$ is closed under combinations of pairs of vectors. We will show that $S$ is a vector space by checking the conditions.

The first item in the vector space definition has five conditions. First, for closure under addition, if ${\vec {s}}_{1},{\vec {s}}_{2}\in S$ then ${\vec {s}}_{1}+{\vec {s}}_{2}\in S$, as ${\vec {s}}_{1}+{\vec {s}}_{2}=1\cdot {\vec {s}}_{1}+1\cdot {\vec {s}}_{2}$.

Second, for any ${\vec {s}}_{1},{\vec {s}}_{2}\in S$, because addition is inherited from $V$, the sum ${\vec {s}}_{1}+{\vec {s}}_{2}$ in $S$ equals the sum ${\vec {s}}_{1}+{\vec {s}}_{2}$ in $V$, and that equals the sum ${\vec {s}}_{2}+{\vec {s}}_{1}$ in $V$ (because $V$ is a vector space, its addition is commutative), and that in turn equals the sum ${\vec {s}}_{2}+{\vec {s}}_{1}$ in $S$. The argument for the third condition is similar to that for the second.

For the fourth, consider the zero vector of $V$ and note that closure of $S$ under linear combinations of pairs of vectors gives that (where ${\vec {s}}$ is any member of the nonempty set $S$) $0\cdot {\vec {s}}+0\cdot {\vec {s}}={\vec {0}}$ is in $S$; showing that ${\vec {0}}$ acts under the inherited operations as the additive identity of $S$ is easy.

The fifth condition is satisfied because for any ${\vec {s}}\in S$, closure under linear combinations shows that the vector $0\cdot {\vec {0}}+(-1)\cdot {\vec {s}}$ is in $S$; showing that it is the additive inverse of ${\vec {s}}$ under the inherited operations is routine.

We usually show that a subset is a subspace with $(2)\implies (1)$.

Example 1

The plane $P=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+y+z=0\}$ is a subspace of $\mathbb {R} ^{3}$. As specified in the definition, the operations are the ones inherited from the larger space, that is, vectors add in $P$ as they add in $\mathbb {R} ^{3}$

${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}$

and scalar multiplication is also the same as it is in $\mathbb {R} ^{3}$. To show that $P$ is a subspace, we need only note that it is a subset and then verify that it is a space. Checking that $P$ satisfies the conditions in the definition of a vector space is routine. For instance, for closure under addition, just note that if the summands satisfy that $x_{1}+y_{1}+z_{1}=0$ and $x_{2}+y_{2}+z_{2}=0$ then the sum satisfies that $(x_{1}+x_{2})+(y_{1}+y_{2})+(z_{1}+z_{2})=(x_{1}+y_{1}+z_{1})+(x_{2}+y_{2}+z_{2})=0$.

Example 2

The $x$-axis in $\mathbb {R} ^{2}$ is a subspace where the addition and scalar multiplication operations are the inherited ones.

${\begin{pmatrix}x_{1}\\0\end{pmatrix}}+{\begin{pmatrix}x_{2}\\0\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\0\end{pmatrix}}\qquad r\cdot {\begin{pmatrix}x\\0\end{pmatrix}}={\begin{pmatrix}rx\\0\end{pmatrix}}$

As above, to verify that this is a subspace, we simply note that it is a subset and then check that it satisfies the conditions in definition of a vector space. For instance, the two closure conditions are satisfied: (1) adding two vectors with a second component of zero results in a vector with a second component of zero, and (2) multiplying a scalar times a vector with a second component of zero results in a vector with a second component of zero.

Example 3

Another subspace of $\mathbb {R} ^{2}$ is

$\{{\begin{pmatrix}0\\0\end{pmatrix}}\}$

which is its trivial subspace.

Any vector space has a trivial subspace $\{{\vec {0}}\,\}$.

At the opposite extreme, any vector space has itself for a subspace. Template:AnchorThese two are the improper subspaces. Template:AnchorOther subspaces are proper.

Example 4

The condition in the definition requiring that the addition and scalar multiplication operations must be the ones inherited from the larger space is important. Consider the subset $\{1\}$ of the vector space $\mathbb {R} ^{1}$. Under the operations $1+1=1$ and $r\cdot 1=1$ that set is a vector space, specifically, a trivial space. But it is not a subspace of $\mathbb {R} ^{1}$ because those aren't the inherited operations, since of course $\mathbb {R} ^{1}$ has $1+1=2$.

Example 5

All kinds of vector spaces, not just $\mathbb {R} ^{n}$'s, have subspaces. The vector space of cubic polynomials $\{a+bx+cx^{2}+dx^{3}\,{\big |}\,a,b,c,d\in \mathbb {R} \}$ has a subspace comprised of all linear polynomials $\{m+nx\,{\big |}\,m,n\in \mathbb {R} \}$.

Example 6

This is a subspace of the $2\!\times \!2$ matrices

$L=\{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a+b+c=0\}$

(checking that it is nonempty and closed under linear combinations is easy).

To parametrize, express the condition as $a=-b-c$.

$L=\{{\begin{pmatrix}-b-c&0\\b&c\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}=\{b{\begin{pmatrix}-1&0\\1&0\end{pmatrix}}+c{\begin{pmatrix}-1&0\\0&1\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}$

As above, we've described the subspace as a collection of unrestricted linear combinations (by coincidence, also of two elements).

Span

The span(or linear closure) of a nonempty subset $S$ of a vector space is the set of all linear combinations of vectors from $S$.

$[S]=\{c_{1}{\vec {s}}_{1}+\cdots +c_{n}{\vec {s}}_{n}\,{\big |}\,c_{1},\ldots ,c_{n}\in \mathbb {R} {\text{ and }}{\vec {s}}_{1},\ldots ,{\vec {s}}_{n}\in S\}$

The span of the empty subset of a vector space is the trivial subspace. No notation for the span is completely standard. The square brackets used here are common, but so are "${\mbox{span}}(S)$" and "${\mbox{sp}}(S)$".

Lemma

In a vector space, the span of any subset is a subspace.

Proof:

Call the subset $S$. If $S$ is empty then by definition its span is the trivial subspace. If $S$ is not empty, then we need only check that the span $[S]$ is closed under linear combinations. For a pair of vectors from that span, ${\vec {v}}=c_{1}{\vec {s}}_{1}+\cdots +c_{n}{\vec {s}}_{n}$ and ${\vec {w}}=c_{n+1}{\vec {s}}_{n+1}+\cdots +c_{m}{\vec {s}}_{m}$, a linear combination

$p\cdot (c_{1}{\vec {s}}_{1}+\cdots +c_{n}{\vec {s}}_{n})+r\cdot (c_{n+1}{\vec {s}}_{n+1}+\cdots +c_{m}{\vec {s}}_{m})$

$=pc_{1}{\vec {s}}_{1}+\cdots +pc_{n}{\vec {s}}_{n}+rc_{n+1}{\vec {s}}_{n+1}+\cdots +rc_{m}{\vec {s}}_{m}$

($p$, $r$ scalars) is a linear combination of elements of $S$ and so is in $[S]$ (possibly some of the ${\vec {s}}_{i}$'s forming ${\vec {v}}$ equal some of the ${\vec {s}}_{j}$'s from ${\vec {w}}$, but it does not matter).

Example 1

The span of this set is all of $\mathbb {R} ^{2}$.

$\{{\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\-1\end{pmatrix}}\}$

To check this we must show that any member of $\mathbb {R} ^{2}$ is a linear combination of these two vectors. So we ask: for which vectors (with real components $x$ and $y$) are there scalars $c_{1}$ and $c_{2}$ such that this holds?

$c_{1}{\begin{pmatrix}1\\1\end{pmatrix}}+c_{2}{\begin{pmatrix}1\\-1\end{pmatrix}}={\begin{pmatrix}x\\y\end{pmatrix}}$

Gauss' method

${\begin{array}{rcl}{\begin{array}{*{2}{rc}r}c_{1}&+&c_{2}&=&x\\c_{1}&-&c_{2}&=&y\end{array}}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}c_{1}&+&c_{2}&=&x\\&&-2c_{2}&=&-x+y\end{array}}\end{array}}$

with back substitution gives $c_{2}=(x-y)/2$ and $c_{1}=(x+y)/2$. These two equations show that for any $x$ and $y$ that we start with, there are appropriate coefficients $c_{1}$ and $c_{2}$ making the above vector equation true. For instance, for $x=1$ and $y=2$ the coefficients $c_{2}=-1/2$ and $c_{1}=3/2$ will do. That is, any vector in $\mathbb {R} ^{2}$ can be written as a linear combination of the two given vectors.