Subspaces — Linear Algebra

Subspaces and subspace operations. Within a vector space, we often focus on elements that satisfy a particular condition. The important question is whether the condition survives the operations of the larger space: do addition and scalar multiplication keep us among elements satisfying that condition? If they do, and the zero vector also satisfies the condition, the collection forms a vector space on its own. Such a collection is called a subspace.

Definition 1.3.1. Let $$V$$ be a vector space over a field $$F$.$ A subset $$W$$ of $$V$$ is called a subspace if $$W$$ is a vector space under the operations inherited from $$V$.$ Note that $\{0_V\}$ is a subspace; it is called the zero subspace of $$V$.$

Since $$W$$ uses the operations of $$V$,$ the algebraic laws governing those operations are inherited from $$V$.$ Thus, to decide whether $$W$$ is a subspace, it is enough to check that $$W$$ contains $$0_V$$ and is closed under the two operations of addition and scalar multiplication. The following theorem makes this criterion precise.

Theorem 1.3.2. Let $$V$$ be a vector space over $$F,$$ and $$W$$ a subset of $$V.$$ Then $$W$$ is a subspace of $$V$$ if and only if $$W$$ has the following three properties:

We have $$0_V\in W$.$
For all $$u,v\in W$,$ we have $$u+v\in W$.$
For all $a\in F$ and $$v\in W$,$ we have $$av\in W$.$

Proof. Suppose that $$W$$ is a subspace of $$V.$$ By Definition 1.3.1, (2) and (3) hold. Since there exists $0_W\in W$ such that $$v+0_W = v$$ for all $v\in W$ and $$0_W = 0_V$,$ (1) holds. Conversely, suppose that (1), (2), and (3) hold. Then, $$W$$ is closed under scalar multiplication and vector addition. Since $-v=(-1)v\in W$ for all $$v\in W$,$ each element of $$W$$ has its additive inverse in $$W$.$ Therefore $$W$$ is a subspace of $$V$.$ $\square$

Example 1.3.3. Let $$A$$ be an $m\times n$ matrix with entries from a field $$F$.$ The transpose of $$A$,$ denoted by $$A^{\!\top}$,$ is the unique matrix whose $$i$-$ th row is the $$i$-$ th column of $$A$,$ for each $$i=1,\dots,n$,$ where a row and a column are identified with the same coordinate tuple. Then, $(A^{\!\top})_{ij}$ is the $$j$-$ th entry of the $$i$-$ th column of $$A$,$ namely, $$A_{ji}$.$ Consequently, one may think of $A^{\!\top}$ as being obtained from $$A$$ by interchanging the rows with the columns. If $$A^{\!\top}=A$,$ then $$A$$ is called a symmetric matrix. In this case, $$A$$ is a square matrix with $$A_{ij} = A_{ji}$.$ The set $$W$$ of all symmetric matrices in $\operatorname{Mat}_{n\times n}(F)$ is a subspace of $$\operatorname{Mat}_{n\times n}(F)$.$ Indeed, $O\in W$ since $$O^{\!\top}=O$.$ Let $A,B\in W.$ We have $A_{ij}=A_{ji}$ and $$B_{ij}=B_{ji}$;$ thus $(A+B)_{ij}=A_{ij}+B_{ij}=A_{ji}+B_{ji}=(A+B)_{ji}.$ Therefore $(A+B)\in W.$ Let $A\in W$ and $c\in F.$ We have $$A_{ij}=A_{ji}$;$ thus $(cA)_{ij}=cA_{ij}=cA_{ji}=(cA)_{ji}.$ Therefore $cA\in W.$

Example 1.3.4. Let $$n$$ be a nonnegative integer, and let ${F}_n[x]$ consist of all polynomials in $${F}[x]$$ having degree less than or equal to $$n.$$ Since the zero polynomial $\mathbf 0\in {F}[x]$ has degree $-\infty,$ it is in ${F}_n[x].$ Let $P, Q \in {F}_n[x].$ Since $\text{deg}(P+Q)\leq \max{\lbrace \text{deg}(P),\text{deg}(Q)\rbrace }\leq n,\ (P+Q)\in {F}_n[x].$ Let $P \in {F}_n[x]$ and $c\in F.$ Since $$\text{deg}(cP)\leq n$,$ we have $cP\in {F}_n[x].$ It follows from Theorem 1.3.2 that ${F}_n[x]$ is a subspace of $${F}[x].$$

Example 1.3.5. Let $C(\mathbb R, \mathbb R)$ denote the set of all continuous real-valued functions defined on $\mathbb R.$ Since constant functions are continuous functions, $\mathbf {0} \in C(\mathbb R, \mathbb R).$ Let $f,g\in C(\mathbb R, \mathbb R).$ Since $$f+g$$ is continuous function, $(f+g)\in C(\mathbb R, \mathbb R).$ Let $f\in C(\mathbb R, \mathbb R)$ and $c\in \mathbb R.$ Since $$cf$$ is a continuous function, $cf\in C(\mathbb R, \mathbb R).$ Theorem 1.3.2 shows $C(\mathbb R, \mathbb R)$ is a subspace of ${\mathbb R}^{\mathbb R}.$

Example 1.3.6. An $n\times n$ matrix $$A$$ is called a diagonal matrix if $A_{ij}=0$ whenever $$i\neq j$.$ Let $$W$$ be the set of all $n\times n$ diagonal matrices. Since $O_{ij}=0$ for all $$i$$ and $j,\ O\in W.$ Let $A,B\in W.$ If $i\neq j$ then $(A+B)_{ij}=A_{ij}+B_{ij}=0+0=0.$ Thus $(A+B)\in W.$ Let $A\in W$ and $c\in F.$ If $i\neq j$ then $(cA)_{ij}=cA_{ij}=c0=0c=0.$ Thus $cA\in W.$ Theorem 1.3.2 shows that $$W$$ is a subspace of $$\operatorname{Mat}_{n\times n}(F)$.$

Example 1.3.7. Let $$A$$ be a $n\times n$ matrix with the entries $$A_{ij}\in F$.$ The trace of $$A$$ is the sum of all the entries of the main diagonal, denoted by $$\operatorname{tr}(A)$.$ That is,

\[\operatorname{tr}(A) = A_{11} + A_{22} + \cdots + A_{nn}.\]

The set of $n\times n$ matrices with their traces equal to zero is a subspace of $$\operatorname{Mat}_{n\times n}(F)$.$

Definition 1.3.8. Let $$X$$ be a set. A collection $\mathcal C$ of subsets of $$X$$ is called a Moore¹ collection if $\mathcal C$ is closed under intersection, that is, if any intersection of elements of $\mathcal C$ is also in $$\mathcal C$.$

Theorem 1.3.9. The collection of all subspaces of a vector space $$V$$ is a Moore collection.

Proof. Let $\{W_i\}_{i\in I}$ be any collection of subspaces of $$V$,$ and let $$W$$ denote the intersection of $$\{W_i\}_{i\in I}$.$ Since every subspace contains the zero vector, $0_V\in W.$ Let $a\in F$ and $u,v\in W.$ Then $$u$$ and $$v$$ are elements of every subspace $$W_i$.$ Since each subspace $$W_i$$ is closed under addition and scalar multiplication, $$u+v$$ and $$av$$ are elements of every subspace $$W_i$.$ Hence $$u+v$$ and $$av$$ are also elements of $$W$,$ so $$W$$ is a subspace of $$V$$ by Theorem 1.3.2. $\square$

We now ask whether the union of subspaces of $$V$$ is a subspace of $$V$.$ It is easily seen that the union contains the zero vector and is closed under scalar multiplication. But the union need not be closed under addition in general. In fact, the union of two subspaces becomes a subspace if and only if one of the subspaces contains the other.

Proposition 1.3.10. Let $$W_1$$ and $$W_2$$ be subspaces of a vector space $$V$.$ Then, $W_1\cup W_2$ is a subspace of $$V$$ if and only if $W_1\subseteq W_2$ or $$W_2\subseteq W_1$.$

Proof. Suppose $W_1\cup W_2$ is a subspace of $$V$,$ and suppose $x\in W_1\setminus W_2$ and $$y\in W_2\setminus W_1$.$ Then $x+y\in W_1\cup W_2$ and, without loss of generality, $$x+y\in W_1$.$ Since $$W_1$$ is a subspace of $$V$,$ we have $y=(x+y)+(-x)\in W_1.$ This contradicts $$y\in W_2\setminus W_1$.$ Therefore if $W_1\cup W_2$ is a subspace of $$V$$ then $W_1\subseteq W_2$ or $$W_2\subseteq W_1$.$ Since $$W_1$$ and $$W_2$$ are subspaces of $$V$,$ the converse automatically holds. $\square$

We have seen that the union itself is not an operation that generates a subspace. There is, however, a natural way to combine two subspaces $$W_1$$ and $$W_2$$ to obtain a subspace containing both $$W_1$$ and $$W_2$.$ The key to finding such an operation is to ensure that the result must be closed under addition.

Definition 1.3.11. Let $$W_1$$ and $$W_2$$ be subspaces of a vector space $$V$.$ We define the sum of $$W_1$$ and $$W_2$$ by

\[W_1 + W_2 = \{x+y: x\in W_1 \text{ and } y\in W_2\}.\]

Theorem 1.3.12. Let $$W_1$$ and $$W_2$$ be subspaces of a vector space $$V$.$

The sum $$W_1+W_2$$ is a subspace of $$V$$ that contains both $$W_1$$ and $$W_2$.$
Every subspace of $$V$$ that contains both $$W_1$$ and $$W_2$$ contains $$W_1+W_2$.$

Proof. (1) Since $0_V\in W_1$ and $$0_V\in W_2$,$ we have $W_1\subseteq W_1+W_2$ and $$W_2\subseteq W_1+W_2$;$ we also have $$0_V\in W_1 + W_2$.$ If $$v\in W_1 + W_2$,$ then there exist $x\in W_1$ and $y\in W_2$ such that $$v=x+y$.$ Hence for any scalar $$a$,$

\[av = ax + ay\in W_1+W_2.\]

If $$v_1,v_2\in W_1+W_2$,$ then there exist $x_1,x_2\in W_1$ and $y_1,y_2\in W_2$ such that $$v_1=x_1+y_1$$ and $$v_2=x_2+y_2$.$ Hence,

\[v_1+v_2=(x_1+x_2)+(y_1+y_2)\in W_1+W_2.\]

(2) Let $$U$$ be a subspace of $$V$$ that contains both $$W_1$$ and $$W_2$.$ If $$v\in W_1+W_2$,$ then there exist $x\in W_1$ and $y\in W_2$ such that $$v=x+y$.$ Since $x\in U$ and $$y\in U$,$ closedness of $$U$$ under addition gives $$x+y\in U$.$ Therefore $$W_1 + W_2\subseteq U$.$ $\square$

The preceding theorem shows that $$W_1+W_2$$ is the minimal subspace containing both $$W_1$$ and $$W_2$.$ Every vector in this sum can be written as $$x+y$,$ where $x\in W_1$ and $$y\in W_2$,$ but this representation need not be unique. For a nonzero vector $$z\in W_1\cap W_2$,$ we have $$x+y=(x+z)+(y-z)$:$ a common vector can be moved from one summand to the other. The following definition describes the case in which this ambiguity disappears.

Exercises

1. Prove that

\[(aA+bB)^{\!\top} = aA^{\!\top} + bB^{\!\top}\]

for any $A,B\in \operatorname{Mat}_{m\times n}(F)$ and any $a,b\in F.$

Proof. Let $1\leq i \leq n$ and $1\leq j \leq m.$ Then we have

\[\begin{aligned} \left( (aA+bB)^{\!\top}\right)_{ij} &= (aA+bB)_{ji}\\ &= (aA)_{ji} + (bB)_{ji}\\ &= aA_{ji} + bB_{ji}\\ &= a\left( A^{\!\top}\right)_{ij} + b\left( B^{\!\top}\right)_{ij}\\ &= \left( aA^{\!\top}\right)_{ij} + \left( bB^{\!\top}\right)_{ij}\\ &= \left( aA^{\!\top} + bB^{\!\top}\right)_{ij}. \end{aligned}\]

Therefore $(aA+bB)^{\!\top} = aA^{\!\top} + bB^{\!\top}.$ $\square$

2. Prove that $A+A^{\!\top}$ is symmetric for any square matrix $$A$.$

In the term Moore collection, “Moore” refers to the American mathematician Eliakim Hastings Moore (1862–1932). He was an important figure in the development of abstract algebra and mathematical foundations, who played a major role in establishing the mathematics department at the University of Chicago. His surname is pronounced /mʊr/. ↩