Inner direct sum – "Math for Non-Geeks"

Derivation and definition

We have already learned about sums of two subspaces. If $U$ and $W$ are two subspaces, then the sum of $U$ and $W$ is again a subspace $Z=U+W$. So for each vector $v\in Z$ we may find two vectors $u\in U$ and $w\in W$, such that $v=u+w$. Now the question arises: Are there several ways to write $v$ as such a combination?

The answer is yes, there can be several possibilities. As an example, let's look at the vector space ${ℝ}^3$. This space can be viewed as the sum of the $xy$-plane and the $yz$-plane. This means that if $v\in {ℝ}^3$, then there are actually several ways to represent $v$ as the sum of vectors from the $xy$-plane and the $yz$-plane. For the vector $(2,3,5)$, for example, we have $(2,3,5)=(2,2,0)+(0,1,5)=(2,1,0)+(0,2,5)$.

Such representations are therefore generally not unique. We now want to find a criterion for uniqueness.

Suppose we have two different representations of $v$, i.e. $v=u+w$ and $v=u^{\prime }+w^{\prime }$ with $u\ne u^{\prime }$ and $w\ne w^{\prime }$ (if one of them is the same, then so is the other). In particular, we know that $u-u^{\prime }\ne 0$ and $w-w^{\prime }\ne 0$. If we now rearrange the equation $u+w=v=u^{\prime }+w^{\prime }$, we get $\underset{\in U}{\underbrace{u-u^{\prime }}}=\underset{\in W}{\underbrace{w^{\prime }-w}}$. Because the left-hand side is in $U$ and the right-hand side is in $W$, this is an element in $U\cap W$ which is not a zero vector at the same time. So $U\cap W$ is not just $\{0\}$. (The zero vector is in the intersection because $U$ and $W$ are both subspaces). This means that if the representation is not unique, then the intersection $U\cap W$ does not only contain the zero vector.

Conversely, if the intersection is not $\{0\}$, we do not have a unique representation: Let $v\in U\cap W$ with $v\ne 0$. Then there are two representations of $v$, namely $v=v+0=0+v$ (on the one hand $v=u+w$ with $u=v$ and $w=0$ and on the other hand $v=u^{\prime }+w^{\prime }$ with $u^{\prime }=0$ and $w^{\prime }=v$). Because of $v\ne 0$, these representations are different from each other.

We can therefore conclude an equivalence: The intersection $U\cap W$ is exactly $\{0\}$ if the representation of all vectors in $V$ is unique.

In this case, we give the sum a special name: We call the sum of $U$ and $W$, in the case $U\cap W=\{0\}$, the direct sum of $U$ and $W$ and write $U\oplus W=U+W$.

Definition (Direct sum)

Let $U$ and $W$ be two subspaces of a vector space $V$. We call the sum $U+W$ direct if $U\cap W=\{0\}$ holds. The subspace $Z=U+W$ is called the direct sum of $U$ and $W$ and we write $Z=U\oplus W$.

Examples

Sum of two lines in R2

The lines $U$ and $W$

We consider the following two lines in ${ℝ}^2$:

$\begin{array}{c}U:=\{\left(\begin{array}{c}x\\ 0\end{array}\right)\mid x\in ℝ\}\text{ and }W:=\{\left(\begin{array}{c}x\\ x\end{array}\right)\mid x\in ℝ\}\end{array}$

So $U$ is the $x$-axis and $W$ is the line that runs through the origin and the point $(1,1)$. Their sum is $U+W={ℝ}^2$

Question: Why do wa have $U+W={ℝ}^2$?

By the definition $U+W=\{u+w\mid u\in U,w\in W\}$ we can describe the set $U+W$ as

$\begin{array}{cc}& U+W\\ =& \{\left(\begin{array}{c}x\\ 0\end{array}\right)\mid x\in ℝ\}+\{\left(\begin{array}{c}x\\ x\end{array}\right)\mid x\in ℝ\}\\ =& \{\left(\begin{array}{c}x\\ 0\end{array}\right)\mid x\in ℝ\}+\{\left(\begin{array}{c}y\\ y\end{array}\right)\mid y\in ℝ\}\\ =& \{\left(\begin{array}{c}x\\ 0\end{array}\right)+\left(\begin{array}{c}y\\ y\end{array}\right)\mid x,y\in ℝ\}\\ =& \{\left(\begin{array}{c}x+y\\ y\end{array}\right)\mid x,y\in ℝ\}\end{array}$

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 $y:=b$. We conclude $U+W={ℝ}^2$.

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

Let us now investigate whether this sum is direct. To do so, we need to determine $U\cap W$. If $v=(x,y{)}^T\in U\cap W$, then we know the following: Because $v\in U$, we have $y=0$. And because $v\in W$, we have $x=y$. Therefore $x=y=0$ and we get $v=0$. Because $U\cap W$ also contains $0$, we get $U\cap W=\{0\}$. This means that the sum of $U$ and $W$ is direct, and we can write $U\oplus W$.

Sum of two lines in R3

The lines $U$ and $W$

We have the following lines in ${ℝ}^3$:

$\begin{array}{c}U:=\{\left(\begin{array}{c}x\\ x\\ 2x\end{array}\right)\mid x\in ℝ\}\text{ and }W:=\{\left(\begin{array}{c}3x\\ 0\\ 5x\end{array}\right)\mid x\in ℝ\}\end{array}$

Then $U$ is a line in ${ℝ}^3$ that runs through the origin and the point $(1,1,2)$, and $W$ is a line that runs through the origin and $(3,0,5)$. The sum $U+W$ is a plane spanned by the vectors $(1,1,2{)}^T$ and $(3,0,5{)}^T$, i.e.

$U+W=\{x\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)+y\left(\begin{array}{c}3\\ 0\\ 5\end{array}\right)\mid x,y\in ℝ\}.$

Question: Why is this the sum?

$\begin{array}{cc}& U+W\\ =& \{\left(\begin{array}{c}x\\ x\\ 2x\end{array}\right)\mid x\in ℝ\}+\{\left(\begin{array}{c}3y\\ 0\\ 5y\end{array}\right)\mid y\in ℝ\}\\ =& \{\left(\begin{array}{c}x\\ x\\ 2x\end{array}\right)+\left(\begin{array}{c}3y\\ 0\\ 5y\end{array}\right)\mid x,y\in ℝ\}\\ =& \{x\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)+y\left(\begin{array}{c}3\\ 0\\ 5\end{array}\right)\mid x,y\in ℝ\}\end{array}$

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

Also here, we want to determine whether the sum is direct. To do so, we consider a vector $v=(x,y,z{)}^T\in U\cap W$. Then, because $v\in U$, we have $x=y=2z$. And because $v\in W$, we obtain $y=0$. Therefore, $x=z=y=0$ and the sum is direct. This means that we can write $U\oplus W$.

Sum of a line and a plane in R3

The line $U_3$ and the plane $W$

We consider the subspaces $U_3$ and $W$ of ${ℝ}^3$.

$\begin{array}{cc}{U}_3& =\{\left(\begin{array}{c}x\\ x\\ x\end{array}\right)\mid x\in ℝ\}=\mathrm{span}\left\{\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\right\}\\ W& =\{\left(\begin{array}{c}0\\ y\\ z\end{array}\right)\mid y,z\in ℝ\}\end{array}$

The subspace $U_3$ is the line through the origin and the point $(1,1,1)$, while $W$ represents the y-z-plane. Together, $U_3$ and $W$ span the entire ${ℝ}^3$, i.e. $U_3+W={ℝ}^3$.

Question: Why is the sum of $U_3$ and $W$ the entire space ${ℝ}^3$?

Since $U_3$ and $W$ are subspaces of ${ℝ}^3$, the sum $U_3+W$ is also a subspace of ${ℝ}^3$. We still have to show that ${ℝ}^3$ is contained in $U_3+W$. To do so, we prove that any vector $(a,b,c{)}^T\in {ℝ}^3$ lies in $U_3+W$. To accomplish this, we show that there is a $u\in U_3$ and a $w\in W$ with $(a,b,c{)}^T=u+w$.

We choose $u:=(a,a,a{)}^T$ and $w:=(0,b-a,c-a{)}^T$. Then $u+w=(a,a,a{)}^T+(0,b-a,c-a{)}^T=(a,b,c{)}^T$. In addition, $u\in U_3$ and $w\in W$ apply.

Therefore, the entire ${ℝ}^3$ is contained in $U_3+W$. Thus ${ℝ}^3=U_3+W$.

One may now ask whether the sum $U_3+W$ is direct. To check this, we need to analyze the intersection $U_3\cap W$. If $U_3\cap W$ only contains the zero vector $(0,0,0{)}^T$, then the sum is direct.

Let $(a,b,c{)}^T$ be a vector in $U_3\cap W$. Since $(a,b,c{)}^T\in U_3$, we have $a=b=c$. Consequently, we can write $(a,b,c{)}^T$ as $(a,a,a{)}^T$. Furthermore, $(a,a,a{)}^T\in W$, which implies $a=0$. We have therefore shown that $(a,b,c{)}^T=(0,0,0{)}^T$.

It follows that $U_3\cap W=\{(0,0,0{)}^T\}$. Since the intersection only contains the zero vector, the sum $U_3+W$ is direct. Therefore, we can conclude ${ℝ}^3=U_3\oplus W$.

Sum of even and odd polynomials

We will now look at an example of a direct sum in the vector space of real polynomials $ℝ[x]$. Let us consider the following subspaces $U$ and $W$ of $ℝ[x]$: $U$ consists of all odd polynomials over $ℝ$, while $W$ is the space of the even polynomials over $ℝ$. In formulas, this is

$\begin{array}{cc}U& =\{\sum _{i=0}^n{a}_i{x}^{2i+1}\mid n\in \mathbb{N},a_i\in ℝ\}=\mathrm{span}\{x,x^3,x^5,\dots \}\\ W& =\{\sum _{i=0}^n{a}_i{x}^{2i}\mid n\in \mathbb{N},a_i\in ℝ\}=\mathrm{span}\{1,x^2,x^4,\dots \}\end{array}$

The odd polynomials $\sum _{i=0}^n{a}_i{x}^{2i+1}$ only contain monomials with odd exponents, while the even polynomials $\sum _{i=0}^n{a}_i{x}^{2i}$ only contain monomials with even exponents. For example, $3x^{18}+19x^{12}-x^4+8x^2$ is an even polynomial, while $x^{10}-x$ is neither even nor odd. We now show that the even and odd polynomials together generate the entire polynomial space $ℝ[x]$. Expressed in formulas: $U+W=ℝ[x]$.

To show this, we need to prove that every polynomial in $ℝ[x]$ can be written as the sum of an odd and an even polynomial. To do so, we consider any polynomial $p=\sum _{i=0}^n{a}_i{x}^i$ from $ℝ[x]$. We must write $p$ as the sum of an even and an odd polynomial.

$p=\sum _{i=0}^n{a}_i{x}^i=\underset{\in U}{\underbrace{\sum _{i=0}^{\lfloor \frac{n-1}{2}\rfloor }{a}_{2i+1}{x}^{2i+1}}}+\underset{\in W}{\underbrace{\sum _{i=0}^{\lfloor \frac{n}{2}\rfloor }{a}_{2i}{x}^{2i}}}$

Therefore, $p$ is contained in the sum $U+W$.

Now we want to check whether the sum $U+W$ is direct. That is, we need to check whether the intersection of the two subspaces $U\cap W$ only contains the zero vector, i.e. the zero polynomial. Let $p$ be a polynomial in the intersection $U\cap W$. Then $p$ lies both in $U$ and in $W$. We can write $p$ as $\sum _{i=0}^n{a}_i{x}^i$. Since $p$ lies in $U$, $p$ only consists of odd monomials. Therefore, the prefactors of the even monomials must be equal to $0$. So $a_i=0$ for all even $i$. Since $p$ lies in $W$, $p$ only consists of even monomials. So $a_i=0$ for all odd $i$. This means that all coefficients $a_i$ are equal to zero and $p$ is therefore the zero polynomial. Thus $U\cap W=0$, and the sum of $U$ and $W$ is direct.

We have seen that $ℝ[x]=U\oplus W$. In other words, the polynomial space $ℝ[x]$ can be written as the direct sum of the subspaces $U$ and $W$, where $U$ is the subspace of odd polynomials and $W$ is the subspace of even polynomials.

Counterexamples

Two planes in R3

The planes $U$ and $W$

We consider the following two planes:

$U=\{\left(\begin{array}{c}2x\\ x\\ y\end{array}\right)\mid x,y\in ℝ\}\text{ and }W=\{\left(\begin{array}{c}0\\ 2x\\ -y\end{array}\right)\mid x,y\in ℝ\}$

The two planes together span all of ${ℝ}^3$. However, the sum is not direct, as the intersection is a line and therefore does not only contain the zero vector. That is, $U\cap W\ne \{(0,0,0{)}^T\}$.

We want to check this mathematically. This requires looking for a vector in the intersection of $U$ and $W$ which is not zero. We consider a vector $(a,b,c{)}^T$ that lies in the intersection $U\cap W$. Because this vector lies in $U$, we have $x,y\in ℝ$ so $(a,b,c{)}^T=(2x,x,y{)}^T$. In addition, there must be $v,w\in ℝ$ so $(a,b,c{)}^T=(0,2v,-w{)}^T$, since $(a,b,c{)}^T\in W$.

We now look for suitable values for $x,y,v,w$ to fulfill both conditions. From $a=2x$ and $a=0$, we get $x=0$. Because $2v=b=x$, we also have $v=0$. Furthermore, $b=0$ results from $b=x$. Finally, we conclude $y=c=-w$.

One possible solution is $x=v=0$, $y=1$ and $w=-1$. The vector $(a,b,c{)}^T=(0,0,1{)}^T$ therefore lies in the intersection of $U$ and $W$. Hence, $U\cap W\ne \{(0,0,0{)}^T\}$.

Various polynomials in polynomial space

Let $K$ be a field. We consider two subspaces in the polynomial space $K[X]$: Let $U=\{f\in K[X]\mid \mathrm{deg}f\le 2\}$ be the space of polynomials of degree less or equal to two, and let $V=\{f=a_0+a_{1}X+\dots a_n{X}^n\mid a_0+\dots +a_n=0\}$ be the space of polynomials whose sum of coefficients is $0$. We want to investigate whether the sum $U+V$ is direct. To find this out, we need to decide whether $U\cap V=0$.

An element $f\in U\cap V$ is a polynomial $f=a_0+a_{1}X+\dots +a_n{X}^n$, which has a maximum degree of $2$ and for which $a_0+\dots ,a_n=0$ applies. Because the polynomial has degree two, we have $a_3=\dots =a_n=0$. Therefore, we get $a_0+a_1+a_2=0$. This means $U\cap V$ consists of all polynomials $f=a_0+a_{1}X+a_2{X}^2$ for which $a_0+a_1+a_2=0$. Thus, we can find a non-zero element of $U\cap V$ if we use the equation $a_0+a_1+a_2=0$ with non-trivial $a_0,a_1,a_2$. One possibility for this is $a_0=1,a_1=-1,a_2=0$, i.e. $f=1-X\in U\cap V$. So the intersection of $U$ and $V$ is not zero, and the sum $U+V$ is therefore not direct.

Unique decomposition of vectors

We have already considered in the derivation that the decomposition of vectors is unique for the direct sum. We will prove this result here rigorously.

Theorem (Equivalent characterizations of the direct sum)

Let $U_1,U_2$ be subspaces of $V$. Then the following statements are equivalent:

1. The sum of $U_1$ and $U_2$ is direct (that is, $U_1+U_2=U_1\oplus U_2$).
2. $U_1$ and $U_2$ have trivial intersection (that is, $U_1\cap U_2=\{0\}$ is the trivial subspace).
3. The representation of all elements of $U_1+U_2$ is unique (that is, if $u=u_1+u_2={u_1}^{\prime }+{u_2}^{\prime }$ with $u_1,{u_1}^{\prime }\in U_1$ and $u_2,{u_2}^{\prime }\in U_2$, then we already have $u_1={u_1}^{\prime }$ and $u_2={u_2}^{\prime }$).
4. The representation of the zero is unique (that is, if $u_1+u_2=0$ with $u_1\in U_1$ and $u_2\in U_2$, then we already have $0=u_1=u_2$).

Proof (Equivalent characterizations of the direct sum)

The definition of the inner direct sum is just $1⟺2$. We now show the implications $2⟹3⟹4⟹2$. The statement then readily follows.

Proof step: $2⟹3$

Let $u\in U_1+U_2$. We must prove that $u$ can be written uniquely as the sum of two elements of $U_1$ and $U_2$.

Let $u_1,{u_1}^{\prime }\in U_1$ and $u_2,{u_2}^{\prime }\in U_2$ with the property that $u_1+u_2=u={u_1}^{\prime }+{u_2}^{\prime }$. In order to prove uniqueness, we must show that these two representations of $u$ are equal. "Equal" means that $u_1={u_1}^{\prime }$ and $u_2={u_2}^{\prime }$.

Because $u_1+u_2={u_1}^{\prime }+{u_2}^{\prime }$, we have $u_1-{u_1}^{\prime }={u_2}^{\prime }-u_2$. This element lies in $U_1$ (because of the representation on the left of "$=$") and in $U_2$ (because of the representation on the right of "$=$"). So $u_1+u_2={u_1}^{\prime }+{u_2}^{\prime }$ lies in the intersection $U_1\cap U_2$. According to the prerequisite, $U_1\cap U_2=\{0\}$. This means $0=u_1-{u_1}^{\prime }={u_2}^{\prime }-u_2$. So $u_1={u_1}^{\prime }$ and $u_2={u_2}^{\prime }$. This is exactly what we wanted to show.

Proof step: $3⟹4$

Let $u_1\in U_1$ and $u_2\in U_2$ with $u_1+u_2=0\in U_1+U_2$. This is a representation of $0\in U_1+U_2$.

On the other hand, $0=0+0\in U_1+U_2$ is also a representation of $0$.

Since representations are unique according to the requirements, we conclude $u_1=0$ and $u_2=0$.

Proof step: $4⟹2$

Let $u\in U_1\cap U_2$. Then, of course, $u\in U_1$ and $u\in U_2$. Since $U_2$ is a subspace, for each element $x\in U_2$ also its additive inverse element must be in this space, i.e., $-x\in U_2$. Therefore, $-u\in U_2$.

This gives us $\underset{\in U_1}{\underbrace{u}}+\underset{\in U_2}{\underbrace{(-u)}}=0=\underset{\in U_1}{\underbrace{0}}+\underset{\in U_2}{\underbrace{0}}$. From the uniqueness of the representation of the zero, we conclude $u=0$. The intersection is therefore trivial, i.e. $U_1\cap U_2=\{0\}$.

Inner direct sum and disjoint union of sets

We can imagine the sum of two subspaces as a structure-preserving union: Forming the sum is "structure-preserving" because the result is again a subspace. This means that the vector space structure is preserved when forming the sum. We can also think of this construction as a union because the sum contains both subspaces. The subspaces $U$ and $W$ are subsets of the sum $U+W$. The sum $U+W$ is the smallest subspace that contains the two subspaces $U$ and $W$. Just as you can form unions with sets, the sums of subspaces also work in the same way.

The direct sum is a special case of the sum of subspaces. This means that every direct sum is also a structure-preserving union. "Being direct" is a property of a sum of subspaces. We now want to see whether there is a property of the union of sets that corresponds to the directness of a sum.

Direct sums are characterized by the fact that the decomposition of the vectors in the sum is unique. If we have a vector $v\in U\oplus W$ with $v=u+w$, where $u\in U$ and $w\in W$, then the vectors $u$ and $w$ are unique. For a union $X\cup Y$ of sets $X$ and $Y$, each element $a\in X\cup Y$ lies in $X$ or in $Y$. The element can also lie in both, which means that we generally do not clearly know where they lie. We cannot assign $a$ unambiguously if $a\in X\cap Y$, i.e. in the intersection. This means that the assignment of elements $a\in X\cup Y$ is unique if $X\cap Y$ is empty. In fact, this criterion corresponds exactly to the criterion for a sum to be direct: We want $U\cap W=\{0\}$, which is the smallest possible vector space, so the intersection contains nothing more from $U$ and $W$ (except the zero, which it must contain anyway as a vector space). This is exactly the definition of a disjoint union. In other words, the direct sum of subspaces intuitively corresponds to the disjoint union of sets.

Basis and dimension

We have seen that the direct sum is a special case of a sum of subspaces. So we can transfer everything we know about the vector space sum to the direct sum. We have already seen that the union of bases of $U$ and $W$ is a generating system of $U+W$. This means if $B_U$ is a basis of $U$ and if $B_W$ is a basis of $W$, then $B_U\cup B_W$ is a generating system of $U+W$. If $U$ and $W$ are finite dimensional, we can use the dimension formula. $\dim (U+W)=\dim (U)+\dim (W)-\dim (U\cap W).$

If the sum $U+W$ is direct, i.e. if $U+W=U\oplus W$, then we even have $U\cap W=\{0\}$. Since $\dim (\{0\})=0$, the following sum formula applies in the finite-dimensional case: $\dim (U+W)=\dim (U)+\dim (W)-\underset{=0}{\underbrace{\dim (U\cap W)}}=\dim (U)+\dim (W).$ So the dimension of the sum space $U+W$ is exactly the sum of the dimensions $\dim (U)$ and $\dim (W)$. If $B_U$ is a basis of $U$ and if $B_W$ is a basis of $W$, then we can conclude $\dim (U+W)=\dim (U)+\dim (W)=|B_U|+|B_W|.$ Since $U\cap W=\{0\}$, the union of the bases of $U$ and $W$ is disjoint, i.e. $B_U\uplus B_W$. Therefore, we get $|B_U|+|B_W|=|B_U\cup B_W|$. Because $B_U\cup B_W$ is a generating system of $U+W$ and because $\dim (U+W)=|B_U\cup B_W|$, we conclude that $B_U\cup B_W$ is a basis of $U+W$.

We have thus seen that in finite dimensions, the union of the bases of $U$ and $W$ is a basis of $U\oplus W$. This also applies in general:

Theorem (Basis of the direct sum)

Let $U$ and $W$ be two subspaces of a $K$-vector space $V$. Assume that the sum of $U$ and $W$ is direct, that is, we can write $U\oplus W$. Let $B_U$ be a basis of $U$ and $B_W$ a basis of $W$. Then the union of $B_U$ and $B_W$ is disjoint and $B_U\cup B_W$ is a basis of $U\oplus W$.

Proof (Basis of the direct sum)

We have already seen that $B_U\cup B_W$ is a generating system of $U+W$. Therefore, we only have to show that the union $B_U\uplus B_W$ is disjoint and linearly independent.

Proof step: $B_U\uplus B_W$

Suppose we have $v\in B_U\cap B_W$. Then $v\in U\cap W=\{0\}$, so $v=0$. However, this is a contradiction to $v\in B_U$ and $v\in B_W$, as a basis cannot contain the zero vector. Therefore, there can be no $v\in B_U\cap B_W$, i.e. $B_U\cap B_W=\mathrm{\varnothing }$.

Proof step: $B_U\cup B_W$ is linearly independent

Let $0=\sum _{i=1}^n{\alpha }_i{u}_i+\sum _{j=1}^m{\beta }_j{w}_j$ for any $n,m\in \mathbb{N}$, $u_i\in B_U$ and $w_j\in B_W$ pairwise different, as well as ${\alpha }_i,{\beta }_j\in K$. We must show that all ${\alpha }_i$ and ${\beta }_j$ are equal to $0$. This corresponds exactly to the definition of the linear independence of $B_U\cup B_W$.

From $0=\sum _{i=1}^n{\alpha }_i{u}_i+\sum _{j=1}^m{\beta }_i{w}_i$ we conclude $\sum _{i=1}^n{\alpha }_i{u}_i=\sum _{j=1}^m-{\beta }_j{w}_j.$ This term is in $U$ (as a linear combination of elements in $B_U$) as well as in $W$ (as a linear combination of elements in $B_W$). Since $U\oplus W$ is a direct sum, we obtain $\sum _{i=1}^n{\alpha }_i{u}_i=0=\sum _{j=1}^m-{\beta }_j{w}_j.$ From the linear independence of $B_U$ we conclude ${\alpha }_i=0$ for all $i$ and from the linear independence of $B_W$ we conclude ${\beta }_j=0$ for all $j$.

From this theorem, we may immediately conclude that $\dim (U\oplus W)=\dim (U)+\dim (W).$

Exercises

Exercise

Let $K=\mathbb{Z}/3\mathbb{Z}$ and let $V=K^3$. Consider the two subspaces $U=\mathrm{span}\{(1,1,0)\}$ and $W=\mathrm{span}\{(0,2,2),(0,0,1)\}$. Show that $U\oplus W=V$ and determine $u\in U$ and $w\in W$ so that $(2,0,2)=u+w$ holds.

Solution

To show $U\oplus W=V$, we need to prove two things: First, that the sum of $U$ and $W$ is direct, i.e. $U\cap W=\{0\}$. Secondly, we must show that the sum of $U$ and $W$ equals $V$, i.e. $U+W=V$.

Proof step: $U\cap W=\{0\}$

Because $U$ and $W$ contain the zero vector as subspaces, $\{0\}\subseteq U\cap W$ is obvious. For the proof of the reverse inclusion, let $v\in U\cap W$ be arbitrary. Then

$v=\lambda \left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)={\mu }_1\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)+{\mu }_2\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$

for certain $\lambda ,{\mu }_1,{\mu }_2\in K$. From the first line of the vectors we get $\lambda \cdot 1={\mu }_1\cdot 0+{\mu }_2\cdot 0=0$. So $\lambda =0$ and therefore $v=0$.

Proof step: $U+W=V$

By definition, $U+W\subseteq V$. The two vectors that span $W$ are obviously linearly independent, so $\dim (W)=2$. Furthermore, $\dim (U)=1$ and $\dim (V)=\dim (K^3)=3$. The dimension formula for subspaces then renders

$\dim (U+W)=\dim (U)+\dim (W)-\dim (U\cap W)=1+2-0=3=\dim (V).$

The dimensions of the subspaces are therefore equal and $U+W\subseteq V$ follows from $U+W=V$.

Alternatively, you could prove the equality by showing that every $v\in V$ can be written as the sum of a $u\in U$ and a $w\in W$.

We want to write $v=(2,0,2)$ as the sum of a vector in $U$ and a vector in $W$. That means, we are looking for ${\lambda }_1,{\lambda }_2,{\lambda }_3\in K=\mathbb{Z}/3\mathbb{Z}$ with

$\left(\begin{array}{c}2\\ 0\\ 2\end{array}\right)={\lambda }_1\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)+{\lambda }_2\left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)+{\lambda }_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$

We may write this as a linear system:

$\begin{array}{cc}2& =\lambda 1\\ 0& ={\lambda }_1+2{\lambda }_2\\ 2& =2{\lambda }_2+{\lambda }_3\end{array}$

From the first line we conclude ${\lambda }_1=2\in \mathbb{Z}/3\mathbb{Z}$. Plugging this into the second line gives ${\lambda }_2=-2=2\in \mathbb{Z}/3\mathbb{Z}$. Again plugging this into the third line finally yields ${\lambda }_3=1\in \mathbb{Z}/3\mathbb{Z}$. Therefore, $v=u+w$ holds with

$u=2\cdot \left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c}2\\ 2\\ 0\end{array}\right)\in U\text{and}w=2\cdot \left(\begin{array}{c}0\\ 2\\ 2\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\in W.$

For the following two exercises, you should know what a linear map is.

Exercise (Self-inverse linear maps and subspaces)

Let $V$ be a $ℝ$-vector space and $f:V\to V$ a linear map.

1. Show that the subsets $U=\{v\in V\mid v=f(v)\}$ and $W=\{v\in V\mid f(v)=-v\}$ are subspaces of $V$.
2. Let additionally $f\circ f={\mathrm{id}}_V$, where ${\mathrm{id}}_V$ denotes the identity map on $V$. (A linear mapping with this property is called self-inverse.) Show that then $V=U\oplus W$ holds for the two subspaces from the first exercise part.

Solution (Self-inverse linear maps and subspaces)

Solution sub-exercise 1:

We use the subspace criterion and show that $U$ and $W$ are non-empty subsets of $V$ that are closed under linear combinations. We only provide the proof for $W$. The proof for $U$ works in the same way, you just have to replace all equations of the form "$f(v)=-v$" with "$f(v)=v$".

Proof step: $W\subseteq V$

This holds by definition of $W$.

Proof step: $W$ is nonempty.

Since $f(0_V)=0_V=-0_V$ we have $0_V\in W$. So $W$ is nonempty.

Proof step: $W$ is closed under linear combinations.

Let $u,v\in W$ and $\lambda ,mu\in ℝ$ be arbitrary. Then

$\begin{array}{cc}& f(\lambda \cdot u+\mu \cdot v)\\ & \downarrow \text{linearity of }f\\ & =\lambda \cdot f(u)+\mu \cdot f(v)\\ & \downarrow u,v\in W\text{and definition of }W\\ & =\lambda \cdot (-u)+\mu \cdot (-v)\\ & \downarrow \text{distributive law}\\ & =-(\lambda \cdot u+\mu \cdot v),\\ \end{array}$

So the linear combination $\lambda \cdot u+\mu \cdot v$ also lies in $W$.

Solution sub-exercise 2:

In order to show $U\oplus W=V$, we need to prove two things: First, that the sum of $U$ and $W$ is direct, i.e. $U\cap W=\{0\}$. Secondly, we must show that the sum of $U$ and $W$ equals $V$, i.e. that every vector $v\in V$ can be written as the sum of a vector $u\in U$ and a vector $w\in W$.

Proof step: $U\cap W=\{0\}$

Because $U$ and $W$ contain the zero vector as subspaces, $\{0\}\subseteq U\cap W$ is obvious. For the proof of the reverse inclusion, let $v\in U\cap W$ be arbitrary. Then

$v\stackrel{v\in U}{=}f(v)\stackrel{v\in W}{=}-v,$

i.e., $2\cdot v=0$, so $v=0$. Because $v\in U\cap W$ was arbitrary, we have thus shown that $U\cap W=\{0\}$.

Proof step: $U+W=V$

Because $U$ and $W$ are subsets of $V$, $U+W\subseteq V$ is obvious. For the reverse inclusion, let $v\in V$ be arbitrary. Then the following then applies:

$v=\frac{1}{2}v+\frac{1}{2}v+\frac{1}{2}f(v)-\frac{1}{2}f(v)=\underset{=:u}{\underbrace{\frac{1}{2}(v+f(v))}}+\underset{=:w}{\underbrace{\frac{1}{2}(-f(v)+v)}}=u+w.$

Because $f\circ f={\mathrm{id}}_V$, it follows from the linearity of $f$ that

$f(u)=f(\frac{1}{2}(v+f(v)))=\frac{1}{2}(f(v)+f(f(v)))=\frac{1}{2}(f(v)+{\mathrm{id}}_V(v))=\frac{1}{2}(f(v)+v)=u.$

So $u\in U$. Analogously, one shows $\frac{1}{2}w\in W$:

$f(w)=f(\frac{1}{2}(v-f(v)))=\frac{1}{2}(f(v)-f(f(v)))=\frac{1}{2}(f(v)-v)=-w.$

So $v$ is a sum of a vector from $U$ and a vector from $W$. Since $v\in V$ was arbitrary, we conclude $V\subseteq U+W$ .

For this exercise, you need to know what the kernel and the image of a linear map are.

Exercise (Idempotent mappings)

Let $f:V\to V$ be a linear map with $f\circ f=f$. (A linear map with this property is called idempotent or a projection). Show: $V=\mathrm{im}(f)\oplus \ker (f)$.

Solution (Idempotent mappings)

We show that $V=\mathrm{im}(f)+\ker (f)$ und $\{0\}=\mathrm{im}(f)\cap \ker (f)$. By definition of the direct sum , the sum of $\ker (f)+\mathrm{im}(f)$ is therefore indeed direct.

Proof step: $V=\mathrm{im}(f)+\ker (f)$

Since both the kernel and the image of $f$ are subspaces of $V$, we immediately get $\ker (f)+\mathrm{im}(f)\subseteq V$. Let us now show the reverse inclusion $V\subseteq \ker (f)+\mathrm{im}(f)$.

Let $v\in V$ be arbitrary. From the condition $f=f\circ f$, we get $f(v)=f(f(v))$, or in other words $f(v)-f(f(v))=0$. Due to the linearity of $f$, we conclude $f(v-f(v))=0$. Therefore, the element $v-f(v)$ lies in the kernel of $f$. Furthermore, $f(v)$ lies in the image of $f$ by definition. Thus

$v=v-f(v)+f(v)=\underset{\in \ker (f)}{\underbrace{(v-f(v))}}+\underset{\in \mathrm{im}(f)}{\underbrace{f(v)}}$

is the sum of an element from $\ker (f)$ and an element from $\mathrm{im}(f)$. So $v$ is in $\mathrm{im}(f)+\ker (f)$. Because $v\in V$ was arbitrary, we have shown $V\subseteq \ker (f)+\mathrm{im}(f)$.

Proof step: $\ker (f)\cap \mathrm{im}(f)=\{0\}$

Because $U$ and $W$ contain the zero vector as subspaces, $\{0\}\subseteq U\cap W$ is obvious. For the proof of the reverse inclusion, let $v\in \ker (f)\cap \mathrm{im}(f)$ be arbitrary. Then $v$ is an element of the kernel of $f$ and $f(v)=0$ applies. Because $v$ is also in the image of $f$, there is a $w\in V$ so that $v=f(w)$. Because $f=f\circ f$, we have

$0=f(v)=f(f(w))=f(w)=v.$

Since $v\in \ker (f)\cap \mathrm{im}(f)$ was arbitrary, we have thus shown $\ker (f)\cap \mathrm{im}(f)=\{0\}$.

In ${ℝ}^2$ we can illustrate the statement from the previous exercise:

Example (Projektions in ${ℝ}^2$)

Let $f:{ℝ}^2\to {ℝ}^2$ be $(x,y)\mapsto (x,x)$. Then $f$ is linear. In addition, $f\circ f=f$ applies: For each vector $(x,y)\in {ℝ}^2$, we have

$f(f(x,y))=f(x,x)=(x,x)=f(x,y).$

The map $f$ is therefore a projection. Clearly, $f$ projects vectors in ${ℝ}^2$ along the $y$-axis onto the first angle bisector $\mathrm{span}\{(1,1)\}$. In particular, $\mathrm{im}(f)=\mathrm{span}\{(1,1)\}$. Further, $f$ maps the $y$-axis to the zero vector, i.e. $\ker (f)=\mathrm{span}\{(0,1)\}$. Therefore, we indeed have ${ℝ}^2=\mathrm{im}(f)\oplus \ker (f)$ as proven in the exercise.