Rank and nullity from RREF

A worked example

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Question

Let \[ A= \begin{pmatrix} 1&0&2&1\\ 0&1&-1&3\\ 1&1&1&4\\ 0&1&2&7\\ 2&1&3&5 \end{pmatrix}. \]

Its RREF is \[ \mathrm{RREF}(A)= \begin{pmatrix} 1&0&2&0\\ 0&1&-1&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}. \]

1. Find \(\mathrm{rank}(A)\).
2. Find \(\mathrm{nullity}(A)\).
3. Find the left nullity of \(A\).
4. Identify the pivot columns of \(A\).
5. Find a basis for the column space \(C(A)\).
6. Find a basis for the null space \(N(A)\).
7. (*) Find a basis for the row space \(C(A^T)\).
8. Let \(\vec{u}_i\) be the \(i\)th column of \(A\). Are \(\{\vec{u}_1,\vec{u}_2,\vec{u}_3\}\) linearly independent? Are \(\{\vec{u}_1,\vec{u}_3,\vec{u}_4\}\) linearly independent?

Solutions

1. \(\mathrm{rank}(A)=3\).
   There are 3 pivots (leading 1s) in \(\mathrm{RREF}(A)\).

2. \(\mathrm{nullity}(A)=1\).
   \(A\) has \(n=4\) columns, so by rank–nullity: \(\mathrm{nullity}(A)=4-3=1\).

3. The left nullity of \(A\) is \(2\).
   \(A\) has \(m=5\) rows, and \(\mathrm{rank}(A)=\mathrm{rank}(A^T)=3\), so \(\mathrm{nullity}(A^T)=5-3=2\).

4. The pivot columns of \(A\) are columns \(1,2,4\) (pivots in \(\mathrm{RREF}(A)\) occur in those columns).

5. A basis for the column space \(C(A)\) is given by the corresponding pivot columns of the original matrix \(A\): \[ \left\{ \begin{pmatrix}1\\0\\1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\1\\1\\1\end{pmatrix}, \begin{pmatrix}1\\3\\4\\7\\5\end{pmatrix} \right\}. \]

6. A basis for the null space \(N(A)\): solve \(\mathrm{RREF}(A)\vec{x}=\vec{0}\) with \(\vec{x}=(x_1,x_2,x_3,x_4)^T\). From the RREF, \(x_4=0\), \(x_1+2x_3=0\), \(x_2-x_3=0\).
   Let \(x_3=t\). Then \(x_1=-2t\), \(x_2=t\), \(x_4=0\), so \[ \vec{x}=t\begin{pmatrix}-2\\1\\1\\0\end{pmatrix}. \] A basis is \[ \left\{ \begin{pmatrix}-2\\1\\1\\0\end{pmatrix} \right\}. \]

7. One approach is to find the pivot columns of \(A^T\) from the RREF of \(A^T\) (that you have to compute), similar to how we found a basis for the column space in the previous problem. Then the first, second, and the last columns of \(A^T\) are pivot columns, that is the first, second, and the last rows of \(A\) are pivot rows.
   Can we find the solution without computing the RREF of \(A^T\) again? Unlike pivot columns, the “pivot rows” of \(A\) may change position under row operations (for example, by row swaps). Therefore, we cannot simply read off corresponding rows from the original matrix \(A\) using the RREF of \(A\) the way we did for the column space.
   (*) Instead, we use the fact that row operations preserve the row space. Hence, a basis for the row space \(C(A^T)\) can be read directly from the nonzero rows of \(\mathrm{RREF}(A)\).
   One possible basis is \[ \left\{ \begin{pmatrix}1&0&2&0\end{pmatrix}, \begin{pmatrix}0&1&-1&0\end{pmatrix}, \begin{pmatrix}0&0&0&1\end{pmatrix} \right\}. \]

8. Let \(\vec{u}_i\) be the \(i\)th column of \(A\).

• \(\{\vec{u}_1,\vec{u}_2,\vec{u}_3\}\) is not linearly independent.
  In the RREF, column 3 is not a pivot column (when the first three columns are present), so it is a linear combination of the pivot columns 1 and 2.

• \(\{\vec{u}_1,\vec{u}_3,\vec{u}_4\}\) is linearly independent.
  Look at the RREF and consider only columns 1, 3, and 4: \[ \begin{pmatrix} 1&2&0\\ 0&-1&0\\ 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix}. \] In this reduced matrix, columns 1, 3, and 4 (of the original matrix) each contain a pivot.
  Therefore, \(\{\vec{u}_1,\vec{u}_3,\vec{u}_4\}\) is linearly independent.