Complements of vector spaces – Serlo

Introduction

We consider a vector space $V$ with some subspace $U$ of $V$. Can we then find a subspace $W$ of $V$ that complements $U$ to $V$? "That is, if we add $W$ to $U$, we would like to get all of $V$. But at the same time, what we add shall not already have be in $U$.

We have already seen earlier how to add two vector spaces, and in this context we would like $U+W=V$ to hold. Further, $W$ shall not contain anything from $U$. We have already learned about this concept in the article on inner direct sum: We want $U$ and $W$ to form an inner direct sum. So $U\oplus W=V$ should apply.

To summarize, we are looking for a subspace $W$ of $V$ for which $U\oplus W=V$ holds. If $V$ is written as a direct sum of subspaces, this is also called a decomposition of $V$. This is because we decompose $V$ into "smaller" parts using the direct sum.

Definition

Definition (Complement)

Let $K$ be a field and $V$ a $K$ vector space. Let $U$ be a subspace of $V$. Then a complement $W$ of $U$ in $V$ is defined as a subspace of $V$ such that $U\oplus W=V$ holds. This means $U+W=V$ holds, and this sum is an inner direct sum.

Existence and Uniqueness

Existence

Suppose we have given $V$ and a subspace $U$. How do we find a subspace $W$ of $V$ so that $U\oplus W=V$ holds? For example, let $V=\mathbb {R} ^{2}$ and let the subspace $U$ be the diagonal line through the origin with slope 1. According to the theorem on the basis of a direct sum, the following applies: If $U\oplus W=V$ holds, then a basis $B_{U}$ of $U$ together with a basis $B_{W}$ of $W$ will form a basis of $V$. So we first choose a basis $B_{U}$ of $U$: For example, we can choose

$B_{U}=\left\{{\begin{pmatrix}1\\1\end{pmatrix}}\right\}$

According to the basis completion theorem, we can add a vector from $\mathbb {R} ^{2}$ to a basis $B_{V}$ of $V$ by adding a vector that does not lie on the line $U$:

$B_{V}=\left\{{\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\right\}$

If we define $B_{W}=B_{V}\setminus B_{U}$ as the set of newly added basis vectors and $W=\operatorname {span} (B_{W})$, then $V=U\oplus W$ should hold. In our example, we obtain the $y$-axis for $W$:

$W=\operatorname {span} \left\{{\begin{pmatrix}0\\1\end{pmatrix}}\right\}.$

We can see that the sum is direct because the intersection of the two lines is the set $\{0\}$, while togwther, they span the entire vector space.

We may even prove that that this kind of construction always provides a complement of a given subspace of a vector space via the basis completion theorem:

Theorem (Complements always exist)

Let $V$ be a $K$-vector space over a field $K$. Let further $U\subseteq V$ be a subspace. Then there is a subspace $W\subseteq V$ such that $U\oplus W=V$, i.e. $W$ is a complement of $U$ in $V$.

How to get to the proof? (Complements always exist)

We know from the theorem about the basis of a direct sum that a basis of $U$ together with a basis of $W$ must result in a basis of $V$. Since $U$ and $V$ are given, we first choose a basis of $U$ and add this to a basis of $V$. The span of the newly added basis vectors is then a canidate for the required subspace $W$. We only have to check that the sum of $U$ and $W$ is direct and results in $V$.

Proof (Complements always exist)

In this proof, we will use bases. These will be defined later in this series, but they are unavoidable here. There is no circular reasoning because we have not used complements in the articles on bases.

Let $U\subseteq V$ be a subspace. We choose a basis $B$ of $U$. According to the basis completion theorem, we can add $B$ to a basis $B'$ of $V$. Let then $W=\operatorname {span} (B\setminus B')$. This is by definition a subspace of $V$.

$U+W=V$ holds, since $U\cup W$ already contains the basis $B'$ of $V$.

It remains to show that $U\cap W=0$. Let $x\in U\cap W$. Then $x$ has representations as a linear combination of vectors in $B$ on the one hand, and of vectors in $B'\setminus B$ on the other. However, since $B'=B\uplus (B'\setminus B)$ forms a basis of $V$ and is therefore linearly independent, only $x=0$ remains as an option.

Warning

Complements always exist in our setting. However, in your further studies, it may happen that the term "complement" is defined somewhat differently, e.g. in functional analysis. Then there are examples of subspaces that have no complement.

Hint

Strictly speaking, we have only shown the existence of complements for finite-dimensional $V$, because we have only proved the basis completion theorem in the finite-dimensional case. However, there is a more general version of the basis completion theorem that works for all vector spaces. This can be used to prove the above theorem in exactly the same way and obtain the existence of complements in infinite-dimensional vector spaces as well.

Complements are not unique

Is the complement $W$ from the last section unique? To define the complement, we used the basis addition theorem. Now we know that bases are in general not unique. Therefore, we could also complete a basis of $U$ to another basis of $V$, which would lead to another subspace $W$ as the complement. We will now try this out using an example:

Let's look at the example from the last section again: We consider $V=\mathbb {R} ^{2}$ and the first angle bisector $U$. We already know that

$B_{U}=\left\{{\begin{pmatrix}1\\1\end{pmatrix}}\right\}$

is a basis of $U$ and that we can add $B_{U}$ to a basis of $V$ by adding the vector $(0,1)^{T}$. We have thus seen that $W=\operatorname {span} \{(0,1)^{T}\}$ is a complement of $U$ in $V$. Another vector that is not in $U$ is $(1,0)^{T}$. This means that we can also add $B_{U}$ to the basis

$\left\{{\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\0\end{pmatrix}}\right\}$

and therefore, $W'=\operatorname {span} \{(1,0)^{T}\}$ is also a complement of $U$ in $V$. We have thus found two complements: $W$ and $W'$. These vector spaces are the coordinate axes of $V=\mathbb {R} ^{2}$ and therefore $W\neq W'$ holds. This means that $U$ has no unique complement in $V$ and complements are not unique.

Examples and exercises

Example (Trivial complements)

Let $V$ be a vector space. We always have $0\oplus V=V$. Therefore, $0$ is a complement of $V$ in $V$.

The construction from the proof of the theorem on the existence of complements also works in this case: If $U=V$, then we do not need to add any vectors to the basis $B_{U}$ of $U$. Then $W=\operatorname {span} (\emptyset )=\{0\}$ is a complement, because the span of the empty set is the null space. It works in the same way in the case $U=\{0\}$: Then $B_{U}=\emptyset$ and we may complete it to a basis of $V$.

Example (Complement of a plane in space)

We consider the plane $U$, which is spanned by the vectors $(1,1,0)^{T}$ and $(0,1,1)^{T}$, i.e.

$U=\operatorname {span} \left\{{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\1\end{pmatrix}}\right\}.$

Our aim is to find a complement of $U$. We can proceed in a similar way to the theorem on the existence of complements. First, we choose a basis of $U$, then we complement it to a basis of the entire $\mathbb {R} ^{3}$. The two vectors that span $U$, namely $(1,1,0)^{T}$ and $(0,1,1)^{T}$, are already linearly independent. Therefore, they already form a basis of $U$. To construct a complement of $U$, we only need one more vector, because $\mathbb {R} ^{3}$ is a $3$-dimensional vector space. We therefore require a vector that is linearly independent of the vectors $(1,1,0)^{T}$ and $(0,1,1)^{T}$. For instance, we may choose the vector $(1,1,1)^{T}$. It is easy to check that the three vectors are indeed linearly independent.

Question: Are the three vectors really linearly independent?

Let $\lambda _{1},\lambda _{2},\lambda _{3}\in \mathbb {R}$ with

${\begin{pmatrix}0\\0\\0\end{pmatrix}}=\lambda _{1}{\begin{pmatrix}1\\1\\0\end{pmatrix}}+\lambda _{2}{\begin{pmatrix}0\\1\\1\end{pmatrix}}+\lambda _{3}{\begin{pmatrix}1\\1\\1\end{pmatrix}}.$

We have to show that $\lambda _{1}=\lambda _{2}=\lambda _{3}=0$ has to hold. If we look at the vectors line by line, we get a system of equations with three equations:

${\begin{aligned}0&=\lambda _{1}+\lambda _{3}\\0&=\lambda _{1}+\lambda _{2}+\lambda _{3}\\0&=\lambda _{2}+\lambda _{3}\end{aligned}}$

From the first and third equation, we deduce $\lambda _{1}=-\lambda _{3}=\lambda _{2}$. Substituting this into the second equation, we obtain $0=\lambda _{1}+\lambda _{1}-\lambda _{1}$. So $\lambda _{1}=0$ and therefore also $\lambda _{2}=\lambda _{3}=0$.

The three vectors $(1,1,0)^{T},(0,1,1)^{T}$ and (1,1,1) therefore form a basis of $\mathbb {R} ^{3}$. The new vector $(1,1,1)^{T}$ spans a possible complement $W$:

$W:=\operatorname {span} \left\{{\begin{pmatrix}1\\1\\1\end{pmatrix}}\right\}$

Example (Decomposition of polynomials)

We consider the vector space $V=K[X]$ of the polynomials over $K$. Then, $U=\{f\in K[X]\mid f(0)=0\}$ is a subspace of $V$. We want to find a complement of $U$ in $V$.

We can also write the condition that $f(0)=0$ differently: We can write $f=a_{n}X^{n}+\dots a_{1}X+a_{0}$. Then $f(0)=a_{0}$ and such a polynomial $f$ lies in $U$ if $a_{0}=0$. In order to construct a complement of $U$, we must therefore find enough polynomials with $a_{0}\neq 0$. One such polynomial is the constant polynomial $f=1$.

Have we already found enough polynomials with $f$ to have a complement? To answer this question, we may check whether $U+\operatorname {span} \{f\}=V$. Let $g=b_{n}X^{n}+\dots +b_{1}X+b_{0}\in V$ be an arbitrary polynomial. Then $g_{1}=b_{0}$ is contained in the span of $1$. Again, $g_{2}=g-g_{1}$ is a polynomial with $g_{2}(0)=0$. This means $g=g_{1}+g_{2}$ with $g_{1}\in \operatorname {span} \{f\}$ and $g_{2}\in U$. Therefore, $U+\operatorname {span} \{f\}=V$.

Furthermore, this sum is direct because we know that $f\not \in U$. Thus, we have found a complement of $U$ in $V$. The subspace $W=\operatorname {span} \{f\}$ is exactly the subspace of constant polynomials. That means, we have just proved that every polynomial $f$ can be decomposed into a polynomial $f_{0}$ with $f(0)=0$ and a constant polynomial. The constant part is sometimes also also called the $y$-intercept.

Of course, we could also have generated a complement of $U$ using any other polynomial $g$ with $g(0)\neq 0$.

Exercise (Uniqueness of complements)

Let $V$ be a $K$-vector space. Show that a subspace $U$ has a unique complement in $V$ if and only if either $U=0$ or $U=V$.

Solution (Uniqueness of complements)

Proof step: $\Rightarrow$

Let $U\subseteq V$ be a subspace with $\{0\}\subsetneq U\subsetneq V$. Note that this implies in particular $\dim(V)\geq 2$. (The only subspaces of a one-dimensional vector space are $\{0\}$ and the space itself). Let $W$ be a complement of $U$ in $V$. We show that $W$ is not unique by constructing another complement $W'\neq W$ of $U$.

Neither $U\subseteq W$ nor $W\subseteq U$ applies: In the first case, $U+W=W$, but this cannot be all of $V$, as otherwise $U\cap W=U\neq \{0\}$. In the second case, $U+W=U\neq V$ would also be true. It therefore follows from the theorem on the union of subvector spaces that $U\cup W\subsetneq V$. Hence there exist vectors in $V$ that lie neither in $U$ nor in $W$. Choose such a vector $w\in V\setminus (U\cup W)$. Because $w$ is not in $W$, $w$ is linearly independent of all vectors in $W$. Because $w$ is not in $U$, $w$ is also linearly independent of all vectors in $U$.

Now, choose a basis of $W$, replace one of the basis vectors with $w$ and define $W'$ as the span of the new basis. Because $w\in W'$ but $w\notin W$, we have $W'\neq W$. In addition, $W'$ is also a complement to $U$: To show $W'\cap U=\{0\}$, let $v\in W'\cap U$ be arbitrary. Let $B_{W'}=\{w_{1},\ldots ,w_{k}\}$ be the basis over which we have defined $W'$. By construction, each of the vectors in $B_{W'}$ is linearly independent of all vectors in $U$. Let $B_{U}=\{u_{1},\ldots ,u_{s}\}$ be a basis of $U$. Then the following applies

$v=\sum _{i=1}^{k}\lambda _{i}w_{i}=\sum _{j=1}^{s}\mu _{j}u_{j}$

for certain $\lambda _{i},\mu _{j}\in K$. Rearranging the equation results in

$0=\sum _{i=1}^{k}\lambda _{i}w_{i}+\sum _{j=1}^{s}(-\mu _{j})u_{j}$

and because the $w_{i},u_{j}$ are linearly independent, $\lambda _{i}=-\mu _{j}=0$ follows for all $i,j$. Therefore, $v=0$ and the sum is direct. The sum results in the whole $V$, because by construction we have $\dim(W')=\dim(W)$: Using the dimension formula for subspaces, we get

$\dim(U+W')=\dim(U)+\dim(W')-\underbrace {\dim(U\cap W')} _{=0}=\dim(U)+\dim(W)=\dim(V),$

where the last equality holds because $U\oplus W=V$. Since $U+W'\subseteq V$, and the dimensions are equal, we must therefore have $U+W'=V$.

Proof step: $\Leftarrow$

Suppose $U=0$. We know that $V$ is a complement of $U$ in $V$. Let $W\subseteq V$ be another subspace with $\{0\}\oplus W=V$. Since the sum of the two subspaces in particular results in $V$, we get $V=W+\{0\}=W$.

Assuming $U=V$. Then $\{0\}$ is a complement of $U$ in $V$. Let $W\subseteq V$ be another subspace with $V\oplus W=V$. Because the sum is direct, we have in particular $W\cap V=\{0\}$. However, since $W\subseteq V$, we conclude $W\cap V=W=\{0\}$.