How to find the null space of a matrix

To find a null space, we want every vector that a matrix sends to zero. Row reduction makes those inputs easier to see. Once we identify the free variables, we can write the whole solution set in vector form.

Learn Linear Algebra with meStep-by-step videos, lesson notes, and practice for matrices, vectors, and transformations.

Explore Linear Algebra →

Find the inputs that give the zero vector

The null space of $A$, written $N\left(A\right)$, is the set of vectors $x$ satisfying $Ax=0$. If $A$ has $n$ columns, these input vectors have $n$ components.

$A=\left[\begin{array}{rr}1& 2\\ 2& 4\end{array}\right]$

The equation $Ax=0$ gives $x_1+2x_2=0$ and $2x_1+4x_2=0$. The second equation is twice the first. Let $x_2=t$, and then $x_1=-2t$.

$x=t(-2,1)$

Every real value of t gives an input in the null space. The resulting vectors form a line through the origin.

Check the three subspace properties

The zero vector belongs to the null space because $A0=0$. If $Au=0$ and $Av=0$, distributing gives $A(u+v)=Au+Av=0$. Also, $A\left(cu\right)=cAu=0$ for any real scalar $c$.

So the null space contains zero and is closed under addition and scalar multiplication. These are the three properties of a subspace.

Set up the homogeneous system

$A=\left[\begin{array}{rrr}2& 1& -3\\ 4& 2& -6\\ 1& -1& -6\end{array}\right]$

Use $Ax=0$ and attach a column of zeros. Swap the first and third rows, then use the first row to clear the entries below it.

$\left[\begin{array}{rrrrr}1& -1& -6& |& 0\\ 0& 6& 18& |& 0\\ 0& 3& 9& |& 0\end{array}\right]$

Find the free variable

Divide the second row by 6, clear the entry below it, and then clear the second entry in the first row.

$\left[\begin{array}{rrrrr}1& 0& -3& |& 0\\ 0& 1& 3& |& 0\\ 0& 0& 0& |& 0\end{array}\right]$

The first two columns have pivots. The third variable is free, so let $x_3=t$. Then $x_1=3t$ and $x_2=-3t$.

Write the full null space as a span

$x=t(3,-3,1)$

The null space is the span of $(3,-3,1)$. One nonzero vector forms a basis, so the null space has dimension $1$. Multiplying $A$ by $(3,-3,1)$ gives zero in every row.

Full Linear Algebra course

Learn Linear Algebra with me

We’ll work through matrices, vectors, and transformations, one step at a time.

Explore my Linear Algebra course →

• Video lessons
• Lesson notes
• Practice problems

Included with membership, along with every other Krista King Math course.