Serlo: EN: Sum of subspaces

{{#invoke:Mathe für Nicht-Freaks/Seite|oben}} In this article, we define the sum of two subspaces. This sum will again be a subspace, containing the two initial subspaces. We can think of the sum as a structure-preserving union.

What is the sum of subspaces?

Consider two subspaces $U$ and $W$ of a vector space $V$. Now we want to combine these subspaces into a larger subspace that contains $U$ and $W$. A first approach could be to consider $U\cup W$. However, we have already seen in the article union and intersection of vector spaces that the union is generally not a subvector space.

Why is that the case? For $u\in U$ and $w\in W$, the vector $u+w$ is not always in $U\cup W$, as you can see from this example.

Union of two lines in two-dimensional space

In order to solve the problem, we add all sums of the form $u+w$ with $u\in U$ and $w\in W$ to the union of the two subspaces $U$ and $W$. That means, we consider $U\cup W\cup \{u+w\mid u\in U,w\in W\}$. This expression still seems very complicated, but we can simplify it to $\{u+w\mid u\in U,w\in W\}$. Mathe für Nicht-Freaks: Vorlage:Frage We call this set the sum of $U$ and $W$ because it consists of the sums of vectors from $U$ and $W$. Later we will show that this is a subspace.

The sum is a subspace

We still have to prove that $U+W$ is a subspace. Mathe für Nicht-Freaks: Vorlage:Satz

Examples

Sum of two lines in ℝ² Vorlage:Anker

The lines $U$ and $W$

We consider the following two lines in ${ℝ}^2$: Vorlage:Einrücken So $U$ is the $x$-axis and $W$ is the line that runs through the origin and the point $(1,1)$. What is the sum $U+W$?

Using the definition $U+W=\{u+w\mid u\in U,w\in W\}$ we can calculate a convenient set description for $U+W$: Vorlage:Einrücken

We can write each vector in ${ℝ}^2$ as $(x+y,y{)}^T$ with matching $x,y\in ℝ$. Specifically, for each vector $(a,b{)}^T\in {ℝ}^2$ we can find scalars $x$ and $y\in ℝ$ such that $(a,b)=(x+y,y)$, namely $x:=a-b$ and $y:=b$. Therefore, $U+W={ℝ}^2$ holds.

Intuitively, you can immediately see that $U+W={ℝ}^2$. This is because $U+W$ is a subspace of ${ℝ}^2$, which contains the straight lines $U$ and $W$. The only subspaces of ${ℝ}^2$ are the null space, lines that run through the origin and ${ℝ}^2$. As the straight lines $U$ and $W$ do not coincide but are different, $U+W$ cannot be a line. Therefore, we must have $U+W={ℝ}^2$.

Sum of two lines in ℝ³

The lines $U$ and $W$

Consider the following lines in ${ℝ}^3$: Vorlage:Einrücken Here $U$ is the line in ${ℝ}^3$ that runs through the origin and the point $(1,1,2)$ and $W$ is the line that runs through the origin and $(3,0,5)$. We want to determine the sum $U+W=\{u+w\mid u\in U,w\in W\}$.

So $U+W$ is the plane that is spanned by the vectors $(1,1,2{)}^T$ and $(3,0,5{)}^T$.

Sum of two planes in ℝ³ Vorlage:Anker

The planes $U_1$ and $W$

Consider the following two planes: Vorlage:Einrücken

The planes are not equal. We can see this, for example, from the fact that the vector $(2,1,0{)}^T$ lies in $U_1$, but not in $W$. Therefore, the two planes should intuitively span the entire space ${ℝ}^3$. So we can initially assume that $U_1+W={ℝ}^3$.

We now try to prove this assumption. To do so, we have to show that each vector $(a,b,c{)}^T\in {ℝ}^3$ lies in the sum $U_1+W=\{u+w\mid u\in U_1,w\in W\}$. We must therefore find vectors $(a,b,c{)}^T$ for $u\in U_1$ and $w\in W$ such that $(a,b,c{)}^T=u+w$. Then $(a,b,c{)}^T\in U_1+W$ applies. Here we can use the definitions of $U_1$ and $W$: Each vector $u\in U_1$ can be written as $(2x_1,x_1,y_1{)}^T$ with $x_1,y_1\in ℝ$. Similarly, each vector $w\in W$ can be written as $(0,2x_2,-y_2{)}^T$ with $x_2,y_2\in ℝ$. So we want to find numbers $x_1,y_1,x_2,y_2\in ℝ$ for the vector $(a,b,c{)}^T\in {ℝ}^3$ satisfying Vorlage:Einrücken We can re-write this as Vorlage:Einrücken How can we choose $x_1,y_1,x_2,y_2\in ℝ$ such that the above equation is satisfied? For instance, Vorlage:Einrücken will do this job.

To summarise, the following applies to any vector $(a,b,c{)}^T\in {ℝ}^3$: Vorlage:Einrücken

Therefore, $U_1+W={ℝ}^3$ indeed holds, i.e. the two planes together span the entire ${ℝ}^3$.

Absorption property of the sum

The line $U_2$ and the plane $W$

We have already looked at a few examples of sums in the space ${ℝ}^3$. Now let's look at another example in ${ℝ}^3$. Let Vorlage:Einrücken Then $U_2$ is the line that runs through the origin and through the point $(0,1,2)$. The subspace $W$ is the $y,z$-plane.

What is the sum of the subspaces $U_2+W$? The line $U_2$ lies in the $y,z$-plane, i.e. in $W$. The sum is intuitively the subspace consisting of $U_2$ and $W$. Since $U_2$ is already contained in $W$, the sum should simply be $W$, i.e. $U_2+W=W$. This is indeed the case, as the exercise below shows.

Intuitively, this should also apply more generally: Let $U$ and $W$ be two subspaces of an arbitrary vector space $V$. If $U$ lies in $W$, i.e. $U\subseteq W$, then the sum $U+W$ should simply result in $W$. This is called the absorption property, as $U$ is absorbed by $W$ when taking the sum. We prove it in the following exercise.

Alternative definitions

Using the intersection

We have constructed a subspace $U+W$ of $V$, which contains the two subspaces $U$ and $W$. Since we have included only "necessary" vectors in our construction of $U+W$, this sum $U+W$ should be the smallest subspace that contains both $U$ and $W$.

We can also describe the smallest subspace containing $U$ and $W$ differently: We first consider all subspaces that contain $U$ and $W$ and then take the intersection of these subspaces. This intersection still contains $U$ and $W$ and is also a subspace, since the intersection of any number of subspaces is again a subspace. Intuitively, there should be no smaller subspace with this property. Thus, we also obtain the smallest subspace that contains both $U$ and $W$. According to these considerations, it should therefore be the case that $U+W$ is equal to the intersection of all subspaces containing $U$ and $W$. We now want to prove this: Mathe für Nicht-Freaks: Vorlage:Satz This renders us the two alternative definitions:

Using the span

We can describe the smallest subspace containing $U$ and $W$ or $U\cup W$ in yet a third way. In the article "span", we saw that for a given subset $M$ of $V$, the span of $M$ is the smallest subspace containing $M$. Therefore, $\mathrm{span}(U\cup W)$ is the smallest subspace that contains $U$ and $W$. So it must also be equal to the sum $U+W$.

Dimension formula Vorlage:Anker

Now that we know what the sum of two subspaces $U$ and $W$ of a vector space $V$ is, we can ask ourselves how large the sum $U+W$ is. The sum of subspaces is the vector space analogue of the union of sets. For two sets $X$ and $Y$, the union $X\cup Y$ has a maximum of $|X|+|Y|$ elements. If $X$ and $Y$ share elements, i.e. have a non-empty intersection, then $X\cup Y$ has fewer than $|X|+|Y|$ elements, because we count the elements from $X\cap Y$ twice. This gives us the formula Vorlage:Einrücken In order to transfer this formula to vector spaces, we need the correct concept of the size of a vector space, i.e. the analogue for the cardinality of a set for vector spaces. This is exactly the idea of the dimension of a vector space. Therefore, if an analogue formula holds for vector spaces, the following should be true: Vorlage:Einrücken If $\dim (U\cap W)$ is finite, we can convert this formula to a formula for $\dim (U+W)$, namely Vorlage:Einrücken

Before we prove our assumption, we will test it with a few examples:

The lines $U$ and $W$

Let us reconsider the two lines from the example above: Vorlage:Einrücken We have already calculated above that $U+W={ℝ}^2$. This fits our assumption: ${ℝ}^2$ is two-dimensional, $U$ and $W$ are one-dimensional and the intersection $U\cap W=\{0\}$ is zero-dimensional.

The planes $U_1$ and $W$

Let us look again at the example above with the two planes: Vorlage:Einrücken We have already calculated above that $U_1+W={ℝ}^3$ and the figure shows that $U_1$ and $W$ intersect in a straight line. This means that the dimension of $U_1+W$ is three, the dimension of $U_1$ and $W$ are both two and the dimension of $U_1\cap W$ is just one. So the dimension formula also holds in this case.

As a final example, we consider the subspace $U={ℝ}^3$ in $V={ℝ}^3$ and Vorlage:Einrücken The subspace $W$ is a line through the origin, i.e. $\dim (W)=1$ and we have $\dim (U)=\dim ({ℝ}^3==3$. Because $U\subseteq W$, the Absorption property of the sum tells us that $U+W=U={ℝ}^3$. For the same reason, we have $U\cap W=W$. Thus, Vorlage:Einrücken So the dimension formula is also valid in this case.

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