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Null Space of a Matrix

  • Let \(A\) be a \(m \times n\) matrix
  • The subspace \(W = \{x \in \mathbb{R}^n | Ax = 0\}\) of \(\mathbb{R}^n\) is called the Solution Space of the homogenous system \(Ax = 0\) or the Null Space of \(A\).
  • Note that the null space is the subspace of \(\mathbb{R}^n\). The dimension of the null space is the nullity of \(A\).
\[x,y \in W \implies Ax = Ay = 0 \implies A(x-y) = 0 \implies Ax + Ay = 0 + 0 = 0\]
\[ \lambda \in \mathbb{R} \implies A(\lambda x) = \lambda (Ax) = 0 \implies \lambda 0 = 0 \]
\[\therefore \ \lambda \in W\]

Finding the nullity and a basis for the null space

  • We have seen how to find the dimension and a basis for the row space of A using row reduction.
  • We will use row reduction to also find the nullity and a basis for the null space of A.

  • First how to find the solution space for a system Ax b i.e. Gaussian elimination.

    • Form the augmented matrix \([A |b]\)
    • Applying the elementary row operations to the augmented matrix we reduce the matrix \(A\) to its reduced row echelon form and obtain \([R | c]\) where \(R\) is the reduced row echelon form of \(A\) and \(c\) is the reduced row echelon form of \(b\).
    • If the \(i^{th}\) column has the leading entry of some row, we call \(x_i\) a dependent variable.
    • If the \(i^{th}\) column has no leading entry, we call \(x_i\) an independent variable.
    \[ nullity(A) = \text{number of independent variables}\]

  • Assign arbitrary values \(t_i\) to the \(i^{th}\) independent variable.

  • Compute the value of each dependent variable in terms of \(t_is\) from the unique row it occurs in.
  • Every solution is obtained by letting \(t_is\) be any real number.

Example

  • Consider the homogeneous system \(Ax = 0\) where \(A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{bmatrix}\)
  • The augmented matrix is \([A | 0] = \begin{bmatrix} 1 & 1 & 1 |& 0 \\ 2 & 2 & 2 |& 0 \\ 3 & 3 & 3 |& 0 \end{bmatrix}\)
  • Row reduce the augmented matrix to obtain \([R | c] = \begin{bmatrix} 1 & 1 & 1 |& 0 \\ 0 & 0 & 0 |& 0 \\ 0 & 0 & 0 |& 0 \end{bmatrix}\)
  • Here \(x_1\) is a dependent variable and \(x_2 , x_3\) are independent variables.
    • The first column has the leading entry of the first row hence \(x_1\) is a dependent variable.
    • The second and the third column dont have the leading entry of any row , hence \(x_2 , x_3\) are independent variables.

  • Hence the nullity of \(A\) is 2 because there are 2 independent variables.

  • Put \(x_2 = t_1\) and \(x_3 = t_2\) that yeilds \(x_1 = -t_1 - t_2\).

    • The null space of \(A\) is \(\{ (-t_1-t_2, t_1,t_2)\}\)
    • Now put \(t_1 = 0 , t_2 = 1\) and you will get \((-1 , 0 ,1)\).
    • Now put \(t_1 = 1 , t_2 = 0\) and you will get \((-1 , 1 , 0)\).
    • Hence the basis vector is \(\{(-1,0,1), (-1,1,0)\}\)

  • Consider the matrix \(A = \begin{bmatrix} 1 & 2 & 0 & 3 \\ 2 & 3 & 0 & 3 \\ 1 & 1 & 1 & 2 \end{bmatrix}\)

    • Applying row reductions on the matrix \(A\) we obtain \([R | c] = \begin{bmatrix} 1 & 2 & 0 & 3 |& 0 \\ 0 & 1 & 0 & 3 |& 0 \\ 0 & 0 & 1 & 2 |& 0 \end{bmatrix}\)
    • \(x_1,x_2 \text{ and } x_3\) are dependent variables and \(x_4\) are independent variables.
    • Hence the nullity of \(A\) is 1. The null space of \(A\) is \(\{ (3t,-3t,-2t,t)\}\)

The Rank-Nullity Theorem

  • Let \(A\) be a \(m \times n\) matrix.
  • The rank is number of linearly dependent variables in \(A\).
  • The nullity is the number of linearly independent variables in \(A\).
\[rank(A) + nullity(A) = n , \text{where n is the number of columns in A}\]