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LinearAlgebra

 NullSpace
 compute a basis for the nullspace (kernel) of a Matrix

 Calling Sequence NullSpace(A, options)

Parameters

 A - Matrix options - (optional); constructor options for the result object

Description

 • The NullSpace(A) function computes a basis for the nullspace (kernel) of the linear transformation defined by Matrix A. The result is a (possibly empty) set of Vectors.
 • The constructor options provide additional information (readonly, shape, storage, order, datatype, and attributes) to the Vector constructor that builds the result. These options may also be provided in the form outputoptions=[...], where [...] represents a Maple list.  If a constructor option is provided in both the calling sequence directly and in an outputoptions option, the latter takes precedence (regardless of the order).  If constructor options are specified in the calling sequence, each resulting Vector has the same specified options.
 • This function is part of the LinearAlgebra package, and so it can be used in the form NullSpace(..) only after executing the command with(LinearAlgebra). However, it can always be accessed through the long form of the command by using LinearAlgebra[NullSpace](..).

Examples

 > $\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$
 > $A≔⟨⟨6,3,0⟩|⟨4,2,0⟩|⟨2,1,0⟩⟩$
 ${A}{≔}\left[\begin{array}{ccc}{6}& {4}& {2}\\ {3}& {2}& {1}\\ {0}& {0}& {0}\end{array}\right]$ (1)
 > $\mathrm{kern}≔\mathrm{NullSpace}\left(A\right)$
 ${\mathrm{kern}}{≔}\left\{\left[\begin{array}{c}{-}\frac{{1}}{{3}}\\ {0}\\ {1}\end{array}\right]{,}\left[\begin{array}{c}{-}\frac{{2}}{{3}}\\ {1}\\ {0}\end{array}\right]\right\}$ (2)
 > $\mathrm{.}\left(A,{\mathrm{kern}}_{1}\right)$
 $\left[\begin{array}{c}{0}\\ {0}\\ {0}\end{array}\right]$ (3)
 > $\mathrm{.}\left(A,{\mathrm{kern}}_{2}\right)$
 $\left[\begin{array}{c}{0}\\ {0}\\ {0}\end{array}\right]$ (4)
 > $\mathrm{NullSpace}\left(\mathrm{IdentityMatrix}\left(3\right)\right)$
 ${\varnothing }$ (5)
 > $B≔\mathrm{Matrix}\left(\left[\left[\frac{1}{3},\frac{1}{2}\right],\left[\frac{1}{2},\frac{3}{4}\right],\left[1,\frac{3}{2}\right]\right],\mathrm{datatype}=\mathrm{float}\right)$
 ${B}{≔}\left[\begin{array}{cc}{0.333333333333333}& {0.500000000000000}\\ {0.500000000000000}& {0.750000000000000}\\ {1.}& {1.50000000000000}\end{array}\right]$ (6)
 > $\mathrm{NullSpace}\left(B\right)$
 $\left\{\left[\begin{array}{c}{0.832050294337844}\\ {-0.554700196225229}\end{array}\right]\right\}$ (7)