Nullspaces

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Nullspaces provide an important way of constructing subspaces of $\mathbb R^n$ .

If $A$ is an $m\times n$ matrix with entries in $\mathbb R$ , the nullspace of A is the set $N(A) := \{{\bf x}\in \mathbb R^n\ |\ A*{\bf x} = {\bf 0}\}\subset \mathbb R^n$ In other words, $N(A)$ is the set of all solutions to the homogeneous matrix equation $A*{\bf x} = {\bf 0}$ .

For any $m\times n$ matrix $A$ with entries in $\mathbb R$ , $N(A)$ is a subspace of $\mathbb R^n$ .

Proof

Note that ${\bf 0}\in N(A)$ . So we need to show that $N(A)$ satisfies the two closure axioms.

Suppose ${\bf v}, {\bf w}\in N(A)$ . Then $A*({\bf v} + {\bf w}) = A*{\bf v} + A*{\bf w} = {\bf 0} + {\bf 0} = {\bf 0}$ Hence $N(A)$ is closed under addition.

Suppose $\alpha \in \mathbb R$ and ${\bf v}\in N(A)$ . Then $A*(\alpha {\bf v}) = \alpha (A*{\bf v}) = \alpha {\bf 0} = {\bf 0}$ Hence $N(A)$ is closed under scalar multiplication. This completes the proof. $\blacksquare$

Nullspaces can be computed using methods that have already been discussed. If $A$ is an $m\times n$ matrix the equation $A*{\bf x} = {\bf 0}$ is consistent. Either the equation only admits the trivial solution ${\bf x} = {\bf 0}$ , or we have a parametrized family of solutions. In either case we want to find a spanning set for $N(A)$ .

Suppose $A$ is a $5\times 5$ matrix for which $rref(A) = \begin {bmatrix} 1 & 0 & 0 & -1 & 2\\0 & 1 & 0 & 2 & 3\\0 & 0 & 1 & -4 & -3\\ 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end {bmatrix}$ . The columns which do not contain leading ones are the fourth and fifth. Thus $x_4$ and $x_5$ are the natural parameters for the parametrized solution set. We can derive a spanning set for $N(A)$ as follows:

Write the all of the original set of variables $x_1, x_2, x_3, x_4, x_5$ in terms of the free variables $x_4, x_5$ :
Use this to write the general solution vector $\bf x$ in terms of $x_4$ and $x_5$ : ${\bf x} = \begin {bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end {bmatrix} = \begin {bmatrix}x_4 - 2x_5\\-2x_4 -3x_5\\4x_4 + 3x_5\\x_4\\x_5\end {bmatrix}$
Separate the general solution into its homogeneous components: ${\bf x} = \begin {bmatrix}x_4 - 2x_5\\-2x_4 -3x_5\\4x_4 + 3x_5\\x_4\\x_5\end {bmatrix} = \begin {bmatrix}x_4\\-2x_4\\4x_4\\x_4\\0\end {bmatrix} + \begin {bmatrix}-2x_5\\-3x_5\\3x_5\\0\\x_5\end {bmatrix} = x_4\begin {bmatrix}1\\-2\\4\\1\\0\end {bmatrix} + x_5\begin {bmatrix}-2\\-3\\3\\0\\1\end {bmatrix}$
Use this last expression for the general solution finally express $N(A)$ as the span of a set of vectors: $N(A) = Span\left \{\begin {bmatrix}1\\-2\\4\\1\\0\end {bmatrix},\begin {bmatrix}-2\\-3\\3\\0\\1\end {bmatrix}\right \}$

Suppose $A$ is the $4\times 6$ matrix $\begin {bmatrix} 3 & 4 & 7 & -1 & 1 & 5\\1 & -5 & 11 & 9 & -3 & 12\\8 & 13 & 15 & 11 & 6 & 0\\ 10 & 3 & 1 & 2 & -9 & -2\\4 & 0 & 3 & 0 & 2 & 1\end {bmatrix}$ Using MATLAB or Octave, compute $rref(A)$ (hint: do this in rational format by entering the command ”format rat” before executing the rref command). Then use the resulting output to construct a spanning set for $N(A)$ . See if you can find a spanning set consisting of vectors with integer entries.

For the same matrix $A$ as in Exercise 1, in either MATLAB or Octave use the “null” command applied to $A$ and enter “null(A)” on the command line. Compare your answer to that in Exercise 1. Is it the same? Next, enter “help null” in the command line to to read about what that command does, and how to interpret the output.