Linear subspaces/Sum/Section

Definition

For a ${}K$-vector space ${}V$ and a family of linear subspaces ${}U_{1},\ldots ,U_{n}\subseteq V$, we define the sum of these linear subspaces by

${}U_{1}+\cdots +U_{n}={\left\{u_{1}+\cdots +u_{n}\mid u_{i}\in U_{i}\right\}}\,.$

For this, we also write ${}\sum _{i=1}^{n}U_{i}$. The sum is again a linear subspace. In case

${}V=U_{1}+\cdots +U_{n}\,,$

we say that ${}V$ is the sum of the linear subspaces ${}U_{1},\ldots ,U_{n}$. The following theorem describes an important relation between the dimension of the sum of two linear subspaces and the dimension of their intersection.

Theorem

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let ${}U_{1},U_{2}\subseteq V$ denote linear subspaces. Then

$\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}=\dim _{K}{\left(U_{1}\cap U_{2}\right)}+\dim _{K}{\left(U_{1}+U_{2}\right)}\,.$

Proof

Let ${}w_{1},\ldots ,w_{k}$ be a basis of ${}U_{1}\cap U_{2}$. On one hand, we can extend this basis, according to fact, to a basis ${}w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n}$ of ${}U_{1}$, on the other hand, we can extend it to a basis ${}w_{1},\ldots ,w_{k},v_{1},\ldots ,v_{m}$ of ${}U_{2}$. Then

$w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n},v_{1},\ldots ,v_{m}$

is a generating system of ${}U_{1}+U_{2}$. We claim that it is even a basis. To see this, let

${}a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}+c_{1}v_{1}+\cdots +c_{m}v_{m}=0\,.$

This implies that the element

${}a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}=-c_{1}v_{1}-\cdots -c_{m}v_{m}\,$

belongs to ${}U_{1}\cap U_{2}$. From this, we get directly ${}b_{i}=0$ for ${}i=1,\ldots ,n$, and ${}c_{j}=0$ for ${}j=1,\ldots ,m$. From the equation before, we can then infer that also ${}a_{\ell }=0$ holds for all ${}\ell$. Hence, we have linear independence. This gives altogether

${}{\begin{aligned}\dim _{K}{\left(U_{1}\cap U_{2}\right)}+\dim _{K}{\left(U_{1}+U_{2}\right)}&=k+k+n+m\\&=k+n+k+m\\&=\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}.\end{aligned}}$

$\Box$

The intersection of two planes (through the origin) in ${}\mathbb {R} ^{3}$ is "usually“ a line; it is the plane itself if the same plane is taken twice, but it is never just a point. This observation is generalized in the following statement.

Corollary

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space of dimension ${}n$. Let ${}U_{1},U_{2}\subseteq V$ denote linear subspaces of dimensions ${}\dim _{K}{\left(U_{1}\right)}=n-k_{1}$ and ${}\dim _{K}{\left(U_{2}\right)}=n-k_{2}$. Then

${}\dim _{K}{\left(U_{1}\cap U_{2}\right)}\geq n-k_{1}-k_{2}\,.$

Proof

Due to fact, we have

${}{\begin{aligned}\dim _{K}{\left(U_{1}\cap U_{2}\right)}&=\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}-\dim _{K}{\left(U_{1}+U_{2}\right)}\\&=n-k_{1}+n-k_{2}-\dim _{K}{\left(U_{1}+U_{2}\right)}\\&\geq n-k_{1}+n-k_{2}-n\\&=n-k_{1}-k_{2}.\end{aligned}}$

$\Box$

Recall that, for a linear subspace ${}U\subseteq V$, the difference ${}\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}$ is called the codimension of ${}U$ in ${}V$. With this concept, we can paraphrase the statement above by saying that the codimension of an intersection of linear subspaces equals at most the sum of their codimensions.

Corollary

Let a homogeneous system of linear equations with ${}k$ equations in ${}n$ variables be given. Then the dimension

of the solution space of the system is at least ${}n-k$.

Proof

The solution space of one linear equation in ${}n$ variables has dimension ${}n-1$ or ${}n$. The solution space of the system is the intersection of the solution spaces of the individual equations. Therefore, the statement follows by applying fact to the individual solution spaces.

$\Box$