How to Find the Null Space of a Matrix
The null space of a matrix A{\displaystyle A} is the set of vectors that satisfy the homogeneous equation Ax=0.{\displaystyle A\mathbf {x} =0.} Unlike the column space ColA,{\displaystyle \operatorname {Col} A,} it is not immediately obvious what the relationship is between the columns of A{\displaystyle A} and NulA.{\displaystyle \operatorname {Nul} A.}
Every matrix has a trivial null space  the zero vector. This article will demonstrate how to find nontrivial null spaces.
Steps

Consider a matrix A{\displaystyle A} with dimensions of m×n{\displaystyle m\times n}. Below, your matrix is 3×5.{\displaystyle 3\times 5.}
 A=(−36−11−71−223−12−458−4){\displaystyle A={\begin{pmatrix}3&6&1&1&7\\1&2&2&3&1\\2&4&5&8&4\end{pmatrix}}}

Rowreduce to reduced rowechelon form (RREF). For large matrices, you can usually use a calculator. Recognize that rowreduction here does not change the augment of the matrix because the augment is 0.
 (1−20−130012−200000){\displaystyle {\begin{pmatrix}1&2&0&1&3\\0&0&1&2&2\\0&0&0&0&0\end{pmatrix}}}
 We can clearly see that the pivots  the leading coefficients  rest in columns 1 and 3. That means that x1{\displaystyle x_{1}} and x3{\displaystyle x_{3}} have their identifying equations. The result is that x2,x4,x5{\displaystyle x_{2},x_{4},x_{5}} are all free variables.

Write out the RREF matrix in equation form.
 x1−2x2−x4+3x5=0x3+2x4−2x5=0{\displaystyle {\begin{aligned}x_{1}2x_{2}x_{4}+3x_{5}&=0\\x_{3}+2x_{4}2x_{5}&=0\end{aligned}}}

Reparameterize the free variables and solve.
 Let x2=r,x4=s,x5=t.{\displaystyle x_{2}=r,x_{4}=s,x_{5}=t.} Then x1=2r+s−3t{\displaystyle x_{1}=2r+s3t} and x3=−2s+2t.{\displaystyle x_{3}=2s+2t.}
 x=(2r+s−3tr−2s+2tst){\displaystyle \mathbf {x} ={\begin{pmatrix}2r+s3t\\r\\2s+2t\\s\\t\end{pmatrix}}}

Rewrite the solution as a linear combination of vectors. The weights will be the free variables. Because they can be anything, you can write the solution as a span.
 NulA=Span{(21000),(10−210),(−30201)}{\displaystyle \operatorname {Nul} A=\operatorname {Span} \left\{{\begin{pmatrix}2\\1\\0\\0\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\2\\1\\0\end{pmatrix}},{\begin{pmatrix}3\\0\\2\\0\\1\end{pmatrix}}\right\}}
 This null space is said to have dimension 3, for there are three basis vectors in this set, and is a subset of R5,{\displaystyle \mathbb {R} ^{5},} for the number of entries in each vector.
 Notice that the basis vectors do not have much in common with the rows of A{\displaystyle A} at first, but a quick check by taking the inner product of any of the rows of A{\displaystyle A} with any of the basis vectors of NulA{\displaystyle \operatorname {Nul} A} confirms that they are orthogonal.
Tips
 For an m×n{\displaystyle m\times n} matrix A,{\displaystyle A,} if n>m,{\displaystyle n>m,} then there will always be a nontrivial null space of A,{\displaystyle A,} because there will not be a pivot in every column (and therefore will have free variables).
 Remember that for a m×n{\displaystyle m\times n} matrix A,{\displaystyle A,} NulA{\displaystyle \operatorname {Nul} A} is always a subset of Rn.{\displaystyle \mathbb {R} ^{n}.}
 The dimension of the null space comes up in the rank theorem, which posits that the rank of a matrix is the difference between the dimension of the null space and the number of columns.
 RankA=dimColA−dimNulA{\displaystyle \operatorname {Rank} A=\operatorname {dim} \operatorname {Col} A\operatorname {dim} \operatorname {Nul} A}
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