Input

Let V be a vector space and U, W two finite-dimensional subspaces of V with the following spanning sets:

${\displaystyle U=\langle u_{1},\ldots ,u_{n}\rangle }$

and

${\displaystyle W=\langle w_{1},\ldots ,w_{k}\rangle .}$

Finally, let ${\displaystyle B_{1},\ldots ,B_{m}}$ be linearly independent vectors so that ${\displaystyle u_{i}}$ and ${\displaystyle w_{i}}$ can be written as

${\displaystyle u_{i}=\sum _{j=1}^{m}a_{i,j}B_{j}}$

and

${\displaystyle w_{i}=\sum _{j=1}^{m}b_{i,j}B_{j}.}$

Output

The algorithm computes the base of the sum ${\displaystyle U+W}$ and a base of the intersection ${\displaystyle U\cap W}$.

Algorithm

The algorithm creates the following block matrix of size ${\displaystyle ((n+k)\times (2m))}$:

${\displaystyle {\begin{pmatrix}a_{1,1}&a_{1,2}&\cdots &a_{1,m}&a_{1,1}&a_{1,2}&\cdots &a_{1,m}\\\vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\a_{n,1}&a_{n,2}&\cdots &a_{n,m}&a_{n,1}&a_{n,2}&\cdots &a_{n,m}\\b_{1,1}&b_{1,2}&\cdots &b_{1,m}&0&0&\cdots &0\\\vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\b_{k,1}&b_{k,2}&\cdots &b_{k,m}&0&0&\cdots &0\end{pmatrix}}}$

Using elementary row operations, this matrix is transformed to the row echelon form. Then, it has the following shape:

${\displaystyle {\begin{pmatrix}c_{1,1}&c_{1,2}&\cdots &c_{1,m}&\bullet &\bullet &\cdots &\bullet \\\vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\c_{q,1}&c_{q,2}&\cdots &c_{q,m}&\bullet &\bullet &\cdots &\bullet \\0&0&\cdots &0&d_{1,1}&d_{1,2}&\cdots &d_{1,m}\\\vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\0&0&\cdots &0&d_{\ell ,1}&d_{\ell ,2}&\cdots &d_{\ell ,m}\\0&0&\cdots &0&0&0&\cdots &0\\\vdots &\vdots &&\vdots &\vdots &\vdots &&\vdots \\0&0&\cdots &0&0&0&\cdots &0\end{pmatrix}}}$

Here, ${\displaystyle \bullet }$ stands for arbitrary numbers, and the vectors ${\displaystyle (c_{p,1},c_{p,2},\ldots ,c_{p,m})}$ for every ${\displaystyle p\in \{1,\ldots ,q\}}$ and ${\displaystyle (d_{p,1},\ldots ,d_{p,m})}$ for every ${\displaystyle p\in \{1,\ldots ,\ell \}}$ are nonzero.

Then ${\displaystyle (y_{1},\ldots ,y_{q})}$ with

${\displaystyle y_{i}:=\sum _{j=1}^{m}c_{i,j}B_{j}}$

is a basis of ${\displaystyle U+W}$ and ${\displaystyle (z_{1},\ldots ,z_{\ell })}$ with

${\displaystyle z_{i}:=\sum _{j=1}^{m}d_{i,j}B_{j}}$

is a basis of ${\displaystyle U\cap W}$.

Proof of correctness

First, we define ${\displaystyle \pi _{1}:V\times V\to V,(a,b)\mapsto a}$ to be the projection to the first component.

Let ${\displaystyle H:=\{(u,u)\mid u\in U\}+\{(w,0)\mid w\in W\}\subseteq V\times V.}$ Then ${\displaystyle \pi _{1}(H)=U+W}$ and ${\displaystyle H\cap (0\times V)=0\times (U\cap W)}$.

Also, ${\displaystyle H\cap (0\times V)}$ is the kernel of ${\displaystyle {\pi _{1}|}_{H}}$, the projection restricted to H. Therefore, ${\displaystyle \dim(H)=\dim(U+W)+\dim(U\cap W)}$.

The Zassenhaus algorithm calculates a basis of H. In the first m columns of this matrix, there is a basis ${\displaystyle y_{i}}$ of ${\displaystyle U+W}$.

The rows of the form ${\displaystyle (0,z_{i})}$ (with ${\displaystyle z_{i}\neq 0}$) are obviously in ${\displaystyle H\cap (0\times V)}$. Because the matrix is in row echelon form, they are also linearly independent. All rows which are different from zero (${\displaystyle (y_{i},\bullet )}$ and ${\displaystyle (0,z_{i})}$) are a basis of H, so there are ${\displaystyle \dim(U\cap W)}$ such ${\displaystyle z_{i}}$s. Therefore, the ${\displaystyle z_{i}}$s form a basis of ${\displaystyle U\cap W}$.