In mathematics, the Zassenhaus algorithm^[1] is a method to calculate a basis for the intersection and sum of two subspaces of a vector space. It is named after Hans Zassenhaus, but no publication of this algorithm by him is known.^[2] It is used in computer algebra systems.^[3]

Algorithm

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 ${\displaystyle U\cap W}$.

Example

Consider the two subspaces ${\displaystyle U=\left\langle \left({\begin{array}{r}1\\-1\\0\\1\end{array))\right),\left({\begin{array}{r}0\\0\\1\\-1\end{array))\right)\right\rangle }$ and ${\displaystyle W=\left\langle \left({\begin{array}{r}5\\0\\-3\\3\end{array))\right),\left({\begin{array}{r}0\\5\\-3\\-2\end{array))\right)\right\rangle }$ of the vector space ${\displaystyle \mathbb {R} ^{4))$.

Using the standard basis, we create the following matrix of dimension ${\displaystyle (2+2)\times (2\cdot 4)}$:

${\displaystyle \left({\begin{array}{rrrrrrrr}1&-1&0&1&&1&-1&0&1\\0&0&1&-1&&0&0&1&-1\\\\5&0&-3&3&&0&0&0&0\\0&5&-3&-2&&0&0&0&0\end{array))\right).}$

Using elementary row operations, we transform this matrix into the following matrix:

${\displaystyle \left({\begin{array}{rrrrrrrrr}1&0&0&0&&\bullet &\bullet &\bullet &\bullet \\0&1&0&-1&&\bullet &\bullet &\bullet &\bullet \\0&0&1&-1&&\bullet &\bullet &\bullet &\bullet \\\\0&0&0&0&&1&-1&0&1\end{array))\right)}$ (Some entries have been replaced by "${\displaystyle \bullet }$" because they are irrelevant to the result.)

Therefore ${\displaystyle \left(\left({\begin{array}{r}1\\0\\0\\0\end{array))\right),\left({\begin{array}{r}0\\1\\0\\-1\end{array))\right),\left({\begin{array}{r}0\\0\\1\\-1\end{array))\right)\right)}$ is a basis of ${\displaystyle U+W}$, and ${\displaystyle \left(\left({\begin{array}{r}1\\-1\\0\\1\end{array))\right)\right)}$ is a basis of ${\displaystyle U\cap W}$.