How to Find the Null Space of a Matrix

Last Updated: March 5, 2024

The null space of a matrix $A$ is the set of vectors that satisfy the homogeneous equation $A{\mathbf {x}}=0.$ Unlike the column space $\operatorname {Col}A,$ it is not immediately obvious what the relationship is between the columns of $A$ and $\operatorname {Nul}A.$

Every matrix has a trivial null space - the zero vector. This article will demonstrate how to find non-trivial null spaces.

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1
Consider a matrix $A$ with dimensions of $m\times n$ . ^{[1]
X
Research source} Below, your matrix is $3\times 5.$
- $A={\begin{pmatrix}-3&6&-1&1&-7\\1&-2&2&3&-1\\2&-4&5&8&-4\end{pmatrix}}$
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2
Row-reduce to reduced row-echelon form (RREF). ^{[2]
X
Research source} 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.
- ${\begin{pmatrix}1&-2&0&-1&3\\0&0&1&2&-2\\0&0&0&0&0\end{pmatrix}}$
- We can clearly see that the pivots - the leading coefficients - rest in columns 1 and 3. That means that $x_{{1}}$ and $x_{{3}}$ have their identifying equations. The result is that $x_{{2}},x_{{4}},x_{{5}}$ are all free variables.
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3
Write out the RREF matrix in equation form. ^{[3]
X
Research source}
- ${\begin{aligned}x_{{1}}-2x_{{2}}-x_{{4}}+3x_{{5}}&=0\\x_{{3}}+2x_{{4}}-2x_{{5}}&=0\end{aligned}}$
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4
Reparameterize the free variables and solve. ^{[4]
X
Research source}
- Let $x_{{2}}=r,x_{{4}}=s,x_{{5}}=t.$ Then $x_{{1}}=2r+s-3t$ and $x_{{3}}=-2s+2t.$
- ${\mathbf {x}}={\begin{pmatrix}2r+s-3t\\r\\-2s+2t\\s\\t\end{pmatrix}}$
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5
Rewrite the solution as a linear combination of vectors. ^{[5]
X
Research source} The weights will be the free variables. Because they can be anything, you can write the solution as a span.
- $\operatorname {Nul}A=\operatorname {Span}\left\{{\begin{pmatrix}2\\1\\0\\0\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\-2\\1\\0\end{pmatrix}},{\begin{pmatrix}-3\\0\\2\\0\\1\end{pmatrix}}\right\}$
- This null space is said to have dimension 3, for there are three basis vectors in this set, and is a subset of ${\mathbb {R}}^{{5}},$ for the number of entries in each vector.
- Notice that the basis vectors do not have much in common with the rows of $A$ at first, but a quick check by taking the inner product of any of the rows of $A$ with any of the basis vectors of $\operatorname {Nul}A$ confirms that they are orthogonal.

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How do you find the basis for the column space of a matrix?

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Row reduce the matrix and pick the columns that have pivot points. Let us use the matrix A: 1 3 5, 5 6 7, 10 12 14. If we now reduce it, we get the following matrix: 1 0 -1, 0 1 2, 0 0 0. The first two columns have pivot positions, but the last column does not. So we go back to the original matrix A and the first two columns of the original matrix A form the basis of the column space.

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For an $m\times n$ matrix $A,$ if $n>m,$ then there will always be a non-trivial null space of $A,$ because there will not be a pivot in every column (and therefore will have free variables).

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Remember that for a $m\times n$ matrix $A,$ $\operatorname {Nul}A$ is always a subset of ${\mathbb {R}}^{{n}}.$

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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.
- $\operatorname {Rank}A=\operatorname {dim}\operatorname {Col}A-\operatorname {dim}\operatorname {Nul}A$
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