How to Find the Null Space of a Matrix

Step-by-Step Guide

1. 1

   Step 1: 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}}}
2. 2

   Step 2: Row-reduce to reduced row-echelon form (RREF).

   For large matrices, you can usually use a calculator.
   
   Recognize that row-reduction 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. , 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}}} , 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+s-3t} and x3=−2s+2t.{\displaystyle x_{3}=-2s+2t.} x=(2r+s−3tr−2s+2tst){\displaystyle {\mathbf {x} }={\begin{pmatrix}2r+s-3t\\r\\-2s+2t\\s\\t\end{pmatrix}}} , The weights will be the free variables.
   
   Because they can be anything, you can write the solution as a span.
   
   Nul⁡A=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 Nul⁡A{\displaystyle \operatorname {Nul} A} confirms that they are orthogonal.
3. 3

   Step 3: Write out the RREF matrix in equation form.

4. 4

   Step 4: Reparameterize the free variables and solve.

5. 5

   Step 5: Rewrite the solution as a linear combination of vectors.

Detailed Guide

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