Serlo: EN: Complements of vector spaces

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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.

Existence and Uniqueness

Existence Vorlage:Anker

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 $V={ℝ}^2$ 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 Vorlage:Einrücken According to the basis completion theorem, we can add a vector from ${ℝ}^2$ to a basis $B_V$ of $V$ by adding a vector that does not lie on the line $U$: Vorlage:Einrücken If we define $B_W=B_V\setminus B_U$ as the set of newly added basis vectors and $W=\mathrm{span}(B_W)$, then $V=U\oplus W$ should hold. In our example, we obtain the $y$-axis for $W$: Vorlage:Einrücken 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:

Complements are not unique Vorlage:Anker

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 $V={ℝ}^2$ and the first angle bisector $U$. We already know that Vorlage:Einrücken 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 $W=\mathrm{span}\{(0,1{)}^T\}$ 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 Vorlage:Einrücken and therefore, $W^{\prime }=\mathrm{span}\{(1,0{)}^T\}$ is also a complement of $U$ in $V$. We have thus found two complements: $W$ and $W^{\prime }$. These vector spaces are the coordinate axes of $V={ℝ}^2$ and therefore $W\ne W^{\prime }$ holds. This means that $U$ has no unique complement in $V$ and complements are not unique.

Examples and exercises

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