Important: The linalg package has been deprecated. Use the superseding packages LinearAlgebra[IntersectionBasis], instead.

- For information on migrating linalg code to the new packages, see examples/LinearAlgebraMigration.

A basis for the intersection of the vector spaces spanned by the vectors in Si is returned.

A basis for the zero-dimensional space is an empty set or list.

The command with(linalg,intbasis) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{linalg}\right)\:$

$\mathrm{v1}≔\mathrm{vector}\left(\left[1\,0\,1\,0\right]\right)$

$\mathrm{v1}≔\left[\begin{array}{cccc}1& 0& 1& 0\end{array}\right]$

$\mathrm{v2}≔\mathrm{vector}\left(\left[0\,1\,0\,1\right]\right)$

$\mathrm{v2}≔\left[\begin{array}{cccc}0& 1& 0& 1\end{array}\right]$

$\mathrm{v3}≔\mathrm{vector}\left(\left[1\,2\,1\,1\right]\right)$

$\mathrm{v3}≔\left[\begin{array}{cccc}1& 2& 1& 1\end{array}\right]$

$\mathrm{v4}≔\mathrm{vector}\left(\left[-1\,-2\,1\,0\right]\right)$

$\mathrm{v4}≔\left[\begin{array}{cccc}\mathrm{-1}& \mathrm{-2}& 1& 0\end{array}\right]$

$\mathrm{intbasis}\left(\left\{\mathrm{v1}\,\mathrm{v2}\right\}\,\left\{\mathrm{v3}\,\mathrm{v4}\right\}\right)$

$\varnothing$

$\mathrm{intbasis}\left(\left[\mathrm{v1}\,\mathrm{v2}\,\mathrm{v3}\right]\,\left[\mathrm{v3}\,\mathrm{v4}\right]\,\left[\mathrm{v2}\,\mathrm{v3}\right]\right)$

$\left\{\left[\begin{array}{cccc}1& 2& 1& 1\end{array}\right]\right\}$

$u≔\mathrm{evalm}\left(\mathrm{v1}-\mathrm{v2}\right)$

$u≔\left[\begin{array}{cccc}1& \mathrm{-1}& 1& \mathrm{-1}\end{array}\right]$

$v≔\mathrm{evalm}\left(\mathrm{v1}+\mathrm{v2}\right)$

$v≔\left[\begin{array}{cccc}1& 1& 1& 1\end{array}\right]$

$w≔\mathrm{evalm}\left(\mathrm{v1}-\mathrm{v3}\right)$

$w≔\left[\begin{array}{cccc}0& \mathrm{-2}& 0& \mathrm{-1}\end{array}\right]$

$\mathrm{intbasis}\left(\left\{u\,v\,w\right\}\,\left\{\mathrm{v1}\,\mathrm{v2}\,\mathrm{v3}\right\}\right)$

$\left\{\left[\begin{array}{cccc}0& 2& 0& 1\end{array}\right]\,\left[\begin{array}{cccc}\mathrm{-1}& 1& \mathrm{-1}& 1\end{array}\right]\,\left[\begin{array}{cccc}\mathrm{-2}& 0& \mathrm{-2}& 0\end{array}\right]\right\}$