Direct sum of two linear subspaces

Vector spaces: More about subspaces

Direct sum of two linear subspaces

The Dimension theorem for linear subspaces gives a relation between the dimensions of the sum and the intersection of two linear subspaces of a given vector space. We now focus on a special situation in which two subspaces span the entire vector space and their intersection is trivial (that is, $\{\vec{0}\}$ ).

The sum $U+W$ of the linear subspaces $U$ and $W$ of a vector space $V$ is called direct if

$U+W =V$ and
$U\cap W = \{\vec{0}\}$ .

In this case, $W$ is called a complement of $U$ in $V$ .

We write $V = U\oplus W$ to indicate $V$ is a direct sum of $U$ and $W$ is.

Of course, $U$ is also a complement of $W$ in $V$ .

If $V = \mathbb{R}^2$ , $U =\linspan{\rv{1,0}}$ (the $x$ -axis), and $W =\linspan{\rv{0,1}}$ (the $y$ -axis), then $V$ is the direct sum of $U$ and $W$ .

In general, the complement of a subspace is far from unique. Consider $V = \mathbb{R}^2$ and $U=\linspan{\rv{1,0}}$ the $x$ -axis. For any real number $a$ the linear subspace $W_a=\linspan{\rv{a,1}}$ is a complement of $U$ . The $y$ -axis, that is, the subspace $W_0$ , is just one of many complements of $U$ in $V$ .

The vector space $V$ is always a direct sum of $V$ and $\{\vec{0}\}$ . This trivial direct sum is not interesting. Often, the direct sum is reduced to reduce all kinds of questions about $V$ to questions about the two subspaces in the direct sum. For example, we can construct a basis of $V$ by finding a basis of $U$ and basis of $W$ ; see below.

For more than two linear subspaces, the direct sum is also defined. For example, if $T$ , $U$ , and $W$ are linear subspaces of $V$ , then we call $V$ the direct sum of $T$ , $U$ , and $W$ if $V = T\oplus (U+W) = U\oplus (T+W) = W\oplus(T+U)$ In particular, it is not sufficient to require that every pair of $T$ , $U$ , and $W$ yields a direct sum of their span.

If $V$ has basis $\basis{\vec{a}_1,\ldots,\vec{a}_n}$ , then $V = \linspan{\vec{a}_1}\oplus \cdots\oplus\linspan{\vec{a}_n}$

The following statements for linear subspaces $U$ and $W$ of a vector space $V$ are equivalent.

$V = U\oplus W$
For each vector $\vec{v}$ in $V$ there are unique vectors $\vec{u}$ in $U$ and $\vec{w}$ in $W$ such that $\vec{v}=\vec{u}+\vec{w}$ .

If $\dim{V}$ is finite, then each of these statements is also equivalent to each of the following two statements:

$\dim{U}+\dim{W} = \dim{U+W} = \dim{V}$ .
A basis of $U$ together with a basis of $W$ is a basis of $V$ .

Set $V= U\oplus W$ . By the definition of the sum of linear subspaces, there are vectors $\vec{u}$ in $U$ and $\vec{w}\in W$ such that $\vec{v}=\vec{u}+\vec{w}$ . If $\vec{v} = \vec{u_1}+\vec{w_1}$ for certain vectors $\vec{u_1}$ in $U$ and $\vec{w_1}\in W$ , then rewriting the equality gives

$\vec{u}-\vec{u_1} = \vec{w_1}-\vec{w}\in U\cap W$

Because $U\cap W = \{\vec{0}\}$ , this implies $\vec{u}-\vec{u_1} = \vec{w_1}-\vec{w}= \vec{0}$ , so $\vec{u_1}=\vec{u}$ and $\vec{w_1}=\vec{w}$ . Thus, the uniqueness of $\vec{u}$ and $\vec{w}$ have been established.

Conversely, suppose that, for each vector $\vec{v}$ in $V$ there are unique vectors $\vec{u}$ in $U$ and $\vec{w}$ in $W$ such that $\vec{v}=\vec{u}+\vec{w}$ . Then each vector $\vec{v}$ in $V$ belongs to $U+W$ so $U+W = V$ . we will show that $U\cap W = \{\vec{0}\}$ . Suppose, for this purpose, that $\vec{u}\in U\cap W$ . Then $\vec{u}$ can be written in two ways as the sum of a vector in $U$ and a vector in $W$ , namely

$\begin{array}{rclcrclcr}\vec{u}&=&\vec{0}&+&\vec{u} &=& \vec{u}&+&\vec{0} \\ &&\text{in}&&\text{in}&&\text{in}&&\text{in}\\ &&U&&W&&U&&W\end{array}$

The uniqueness condition in this case means that $\vec{u}=\vec{0}$ . This proves that $U\cap W$ equals $\{\vec{0}\}$ .

For the proof of equivalence of these statements with the third statement, first assume that $\dim{U}+\dim{W} = \dim{U+W}$ . Then, thanks to the Dimension theorem for linear subspaces, $\dim{U\cap W}=0$ , which is equivalent to $U\cap W=\{\vec{0}\}$ . If $\dim{U+W} = \dim{V}$ , we must have $U+W= V$ in view of Property 1 of linear subspaces. From this we conclude that the equalities $\dim{U}+\dim{W} = \dim{U+W} = \dim{V}$ imply that $V$ is the direct sum of $U$ and $W$ .

Conversely, if $V=U\oplus W$ , then $\dim{U+W} = \dim{V}$ because $V = U+W$ , and, because of the Dimension Theorem for linear subspaces, $0=\dim{U\cap W} = \dim{U}+\dim{W}-\dim{U+W} = \dim{U}+\dim{W}-\dim{V}$ so $\dim{V} = \dim{U}+\dim{W}$ . Thus we have derived both equalities $\dim{U+W} = \dim{V} = \dim{U}+\dim{W}$ .

Finally, we deduce that the third statement is equivalent to the last statement. If a basis of $U$ and a basis of $W$ together form a basis of $V$ , then $V$ has a basis of $\dim{U}+\dim{W}$ vectors, so $\dim{V } = \dim{U}+\dim{W}$ . Since each vector of $V$ can be written as a linear combination of such a basis, it is also a sum of a vector from $U$ , and a vector from $W$ , so $V = U+W$ , and therefore also $\dim{V} = \dim{U+W}$ .

Conversely: Set $\dim{U}+\dim{W} = \dim{U+W} = \dim{V}$ and let $\basis{\vec{u_1},\ldots,\vec{u_m}}$ be a basis of $U$ , and $\basis{\vec{w_1},\ldots,\vec{w_n}}$ a basis $W$ . Because $U+W = V$ , their union $\basis{ \vec{u_1},\ldots,\vec{u_m},\vec{w_1},\ldots,\vec{w_n}}$ spans the entire vector space $V$ .

Suppose now that a linear combination of these vectors is equal to zero:

$\lambda_1\vec{u_1}+\cdots+\lambda_m\vec{u_m}+\mu_1\vec{w_1}+\cdots+\mu_n\vec{w_n} = \vec{0}$

where $\lambda_1,\ldots,\lambda_m,\mu_1,\ldots,\mu_n$ are scalars. Then $\lambda_1\vec{u_1}+\cdots+\lambda_m\vec{u_m}=-\mu_1\vec{w_1}-\cdots-\mu_n\vec{w_n}$ is a vector of $U\cap W$ , which is equal to $\{\vec{0}\}$ , so the vector is equal to $\vec{0}$ . The left-hand side is a linear combination of vectors from a basis of $U$ , so each $\lambda_i$ is equal to $0$ , and the right-hand side is a linear combination of vectors from a basis of $W$ , so each $\mu_j$ is equal to $0$ . This shows that all coefficients of the linear combination are equal to $0$ , with implies that the union is linearly independent. The conclusion is that $\basis{ \vec{u_1},\ldots,\vec{u_m},\vec{w_1},\ldots,\vec{w_n}}$ is a basis of $V$ .

Let $P$ be the vector space of all polynomials in $x$ . Show that this vector space is the direct sum of linear subspaces of even and odd polynomials:
$\begin{array}{rcl} U &=& \{f(x)\in P\mid f(-x) = f(x)\}\\ W &=& \{f(x)\in P \mid f(-x) = -f(x)\}\end{array}$

We first show that every polynomial $f(x)$ in $P$ is the sum of an even and an odd polynomial. For this purpose we consider the polynomials $f_+(x) = \frac{1}{2}(f(x)+f(-x))\phantom{xx}\text{ and }\phantom{xx} f_-(x) = \frac{1}{2}(f(x)-f(-x))$
From $f_+(-x) = \frac{1}{2}(f(-x)+f(x)) = f_+(x)$ it follows $f_+(x)$ is even, and from $f_-(-x) = \frac{1}{2}(f(-x)-f(x)) = -f_-(x)$ that $f_-(x)$ is odd. Thus, we have shown that $P = U+W$ .

To complete the proof that $P$ is the direct sum of $U$ and $W$ , we need to show that $U\cap W$ consists only of the zero vector. If $f(x)\in U\cap W$ , then $\begin{array}{rcl}f(x)& =& f(-x)\\ &&\phantom{xxx}\color{blue}{f(x)\in U}\\&=&-f(x)\\&&\phantom{xxx}\color{blue}{f(x)\in W}\end{array}$ so $2f(x) = 0$ , which implies that $f(x) = 0$ . The proof is complete.

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