Home page for accesible maths 2 Vector spaces 2.A Vector spaces 2.C Linear independence

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2.B Subspaces and spanning sequences

To check whether something is a vector space, one method is to check all of the vector space axioms. But there is usually a shorter way. Usually the vector spaces we encounter naturally sit inside a larger vector space; in the same way that ${\mathbb{Q}}$ sits inside ${\mathbb{R}}$ , which sits inside ${\mathbb{C}}$ . We formalize this idea as follows.

Definition 2.7:

If $V$ is a vector space over the field $F$ , then any subset $W\subset V$ that also forms a vector space over $F$ , (with the same addition and scalar multiplication operations as in $V$ ) is called a subspace of $V$ .

There is an easy check to test whether a subset is a subspace:

Theorem 2.8.

Let $V$ be a vector space over the field $F$ , and let $W$ be a subset of $V$ . Then $W$ is a subspace of $V$ if and only if all of the following conditions hold:

S1.

$\vec{0}\in W$ ,
S2.

If $\vec{v},\vec{w}\in W$ then $\vec{v}+\vec{w}\in W$ ,
S3.

If $\alpha\in F$ and $\vec{v}\in W$ then $\alpha\vec{v}\in W$ .

Proof.

If $W$ is a vector space, then the conditions follow from V4, V1, and V6.

Conversely, if $W$ satisfies the above three conditions, then we need to prove all 10 vector space axioms. Since $V$ is a vector space, and $W\subset V$ , this automatically gives us V2,V3,V7,V8,V9, and V10. Clearly S2 $\Rightarrow$ V1; S1 $\Rightarrow$ V4; and S3 $\Rightarrow$ V6. Finally, S3 together with Lemma 2.5(iv) implies V5. Therefore $W$ is a vector space over $F$ . ∎

Example 2.9.

i.

Prove $W:=\{(x,y)\in{\mathbb{R}}^{2}\;|\;x+y=0\}\subset\mathbb{R}^{2}$ is a subspace.

Solution: Since $(0,0)\in W$ , S1 is satisfied. To check S2, assume we have $(x_{1},y_{1}),(x_{2},y_{2})\in W$ . This means $x_{1}+y_{1}=0$ and $x_{2}+y_{2}=0$ . Therefore, $(x_{1}+x_{2})+(y_{1}+y_{2})=0$ . So the vector $(x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})\in W$ . This verifies S2. Finally, we use that $x_{1}+y_{1}=0$ implies that $\alpha x_{1}+\alpha y_{1}=0$ is true for any $\alpha\in\mathbb{R}$ , and hence $\alpha(x_{1},y_{1})\in W$ , which proves S3 is satisfied.
ii.

The subset $\{\vec{0}\}$ is always a subspace. This is because $\vec{0}+\vec{0}=\vec{0}$ , and $\alpha\vec{0}=\vec{0}$ , for any $\alpha\in F$ by Lemma 2.5.
iii.

In the vector space $\operatorname{M}_{{3}}({{\mathbb{R}}})$ over ${\mathbb{R}}$ , the set of diagonal matrices forms a subspace:

$W:=\{\begin{bmatrix}a_{11}&0&0\\ 0&a_{22}&0\\ 0&0&a_{33}\end{bmatrix}\;|\;a_{11},a_{22},a_{33}\in{\mathbb{R}}\}.$

We will write diagonal matrices as $\operatorname{diag}(a_{11},a_{22},a_{33})$ . To prove this is a subspace, we use that $\operatorname{M}_{{3}}({{\mathbb{R}}})$ is a vector space, together with Theorem 2.8. The zero matrix $\vec{0}=\operatorname{diag}(0,0,0)\in W$ , so S1 is satisfied. The sum of any two diagonal matrices is again diagonal, so S2 is satisfied. Finally, for any $\alpha\in{\mathbb{R}}$ , we have that

$\alpha\operatorname{diag}(a_{11},a_{22},a_{33})=\operatorname{diag}(\alpha a_{% 11},\alpha a_{22},\alpha a_{33})\in W.$

So the scalar multiple of any diagonal matrix is diagonal; S3 is verified.

Exercise 2.10:

Consider the set $W$ of real polynomials of degree equal to 3, which is a subset of the vector space $\mathcal{P}_{3}({\mathbb{R}})$ , from Example 2.1(v). So

W:=\{c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}\;|\;c_{i}\in{\mathbb{R}},\textup{ and % }c_{3}\neq 0\}.

Recall the degree of a polynomial is the largest $r$ such that the coefficient of $x^{r}$ is non-zero. Prove that $W$ satisfies none of S1, S2, nor S3.

[End of Exercise]

Exercise 2.11:

Each of the following are subsets of a given vector space. Determine which are subspaces. Justify your answer.

i.

$W:=\{(a,2a+b,b)\;|\;a,b\in\mathbb{R}\}\subset\mathbb{R}^{3}$ .
ii.

$W:=\{A\in\operatorname{M}_{{3}}({{\mathbb{R}}})\;|\;A^{T}=A\}\subset% \operatorname{M}_{{3}}({{\mathbb{R}}})$ . A matrix is symmetric when $A=A^{T}$ .
iii.

$W:=\{A\in\operatorname{M}_{{3}}({{\mathbb{R}}})\;|\;A^{T}=-A\}\subset% \operatorname{M}_{{3}}({{\mathbb{R}}})$ . A matrix is skew-symmetric when $A=-A^{T}$ .
iv.

The set of invertible matrices in $\operatorname{M}_{{n}}({{\mathbb{C}}})$ .
v.

The set of non-invertible matrices in $\operatorname{M}_{{2}}({{\mathbb{R}}})$ .

[End of Exercise]

Exercise 2.12:

Let $V$ be a vector space over a field $F$ .

i.

Prove that if a non-empty subset $W\subset V$ satisfies S3 of Theorem 2.8, then it also must satisfy S1.
ii.

Is it true that if $W\subset V$ satisfies S2 of Theorem 2.8, then $W$ must also satisfy S1?

[End of Exercise]

It will be desirable to express our subspaces as the set of all linear combinations of some finite set of vectors (when it is possible to do so!). For example, the subspace from Example 2.9(i) is equal to $\{\alpha\vec{v}\;|\;\alpha\in{\mathbb{R}}\}$ , where $\vec{v}=(1,-1)$ ; so it equals the set of all linear combinations of $\{\vec{v}\}$ . In this case, we say $\{\vec{v}\}$ spans the subspace.

Definition 2.13:

Let $\vec{v_{1}},\cdots,\vec{v_{r}}\in V$ be a collection of vectors. We define the span of these vectors to be:

\operatorname{span}_{F}\{\vec{v_{1}},\cdots,\vec{v_{r}}\}:=\{\alpha_{1}\vec{v_% {1}}+\alpha_{2}\vec{v_{2}}+\cdots+\alpha_{r}\vec{v_{r}}\;|\;\alpha_{i}\in F\}.

In other words, the it is the set of all possible linear combinations of the given vectors. More generally, if $S\subset V$ is any subset of vectors, possibly infinitely many, then the span of $S$ is the smallest subspace of $V$ that contains $S$ .

[Technical note: A particularly careful reader should question whether such a subspace always exists, whether it is unique, and whether it agrees with the apparently different definition immediately above.]

Theorem 2.14.

The span of a set of vectors is always a subspace.

If $W\subset V$ is a subspace, and $W=\operatorname{span}_{F}\{\vec{v_{1}},\cdots,\vec{v_{r}}\}$ , then we say $W$ is spanned by the sequence $\vec{v_{1}},\cdots,\vec{v_{r}}\in V$ . We also say the sequence $\vec{v_{1}},\cdots,\vec{v_{r}}$ spans $W$ .

Notice how the definition of span depends on the field, which is the purpose of the subscript $F$ in the notation. If $F={\mathbb{R}}$ , then we take all linear combinations with scalars in ${\mathbb{R}}$ . But if the field is ${\mathbb{C}}$ , then there are more linear combinations. If there is no doubt about what the field is, then one may omit the subscript $F$ from the notation.

Example 2.15.

i.

In the vector space $\operatorname{M}_{{2}}({{\mathbb{C}}})$ over the field ${\mathbb{C}}$ , consider the subspace spanned by the following set of matrices:

$W:=\operatorname{span}_{{\mathbb{C}}}\{\begin{bmatrix}1&0\\ 0&0\end{bmatrix},\begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 0&1\end{bmatrix}\}=\{\begin{bmatrix}a&b\\ 0&c\end{bmatrix}\;|\;a,b,c\in{\mathbb{C}}\}.$

So, the span is all complex upper-triangular matrices.
ii.
Consider the following system of equations in ${\mathbb{R}}^{4}$ :
1. $x+y=0$
2. $z-w=0$
There are several ways of writing the set of all solutions:

$S=\{(x,y,z,w)=(x,-x,z,z)\;|\;x,z\in{\mathbb{R}}\}=\{s(1,-1,0,0)+t(0,0,1,1)\;|% \;s,t\in{\mathbb{R}}\}$

Here $S$ is called the solution space. The right-most expression says that $S=\operatorname{span}_{{\mathbb{R}}}\{(1,-1,0,0),(0,0,1,1)\}$ ; so in this case, the solution set is a subspace.

Exercise 2.16:

i.

Give an example of a pair of vectors in ${\mathbb{R}}^{2}$ whose span is ${\mathbb{R}}^{2}$ , and whose coordinates are all non-zero.
ii.

Give an example of three vectors in ${\mathbb{R}}^{3}$ whose span is ${\mathbb{R}}^{3}$ , and whose coordinates are all not rational.
iii.

Can you find a sequence of 100 vectors in ${\mathbb{R}}^{2}$ whose span is ${\mathbb{R}}^{2}$ ?

Exercise 2.17:

True or false: $\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}\in\operatorname{span}_{\mathbb{R}}\{\begin{bmatrix}2\\ -1\\ 1\end{bmatrix},\begin{bmatrix}3\\ -4\\ -1\end{bmatrix}\}$ .

Exercise 2.18:

For each of the following systems of linear equations, find a finite set of vectors which spans the solution space.

i.
The set of $(x,y,z,w)\in{\mathbb{R}}^{4}$ such that
1. $x+z=2w$ ,
2. $x-3y=0$
ii.

$\displaystyle\begin{bmatrix}3&1&0\\ 0&2&i\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}$ , $x,y,z\in{\mathbb{C}}$ .
iii.

$x+\sqrt{2}y=0$ , $x,y\in{\mathbb{Q}}$ .

[End of Exercise]