Expressing Nonpivot Columns Through Pivot Columns

From a Reduced Matrix to the Original Columns

Row reduction changes the entries of a matrix, so a column in its reduced form is generally not the corresponding original column. Yet the reduced row-echelon form contains useful information about how the original columns relate to one another. In particular, it tells us how to express each original nonpivot column as a combination of the original pivot columns. The key is to use the entries of the reduced matrix as coefficients, while using the columns of the original matrix in the resulting combination.

Let $\mathbf{A}\in\mathbb{R}^{m\times n}$, and write $\mathbf{R}=\operatorname{rref}(\mathbf{A})$. A pivot column is a column containing a leading 1 in $\mathbf{R}$; all other columns are nonpivot columns. Suppose the pivot columns are numbered $p_1,\dots,p_r$. In $\mathbf{R}$, these pivot columns are the first $r$ standard coordinate vectors in $\mathbb{R}^m$: the pivot in row $i$ is in column $p_i$. Any rows below the last pivot are zero rows.

Theorem: Let $\mathbf{A}\in\mathbb{R}^{m\times n}$, and let $\mathbf{R}=\operatorname{rref}(\mathbf{A})$ have pivot columns $p_1,\dots,p_r$. For every nonpivot column $j$, the original columns satisfy $$ \mathbf{a}_j=r_{1j}\mathbf{a}_{p_1}+\cdots+r_{rj}\mathbf{a}_{p_r}, $$ where $r_{ij}$ is the entry in row $i$, column $j$ of $\mathbf{R}$. If there are no pivot columns, the sum is the zero vector.

The formula uses two different matrices in two different ways: the coefficients $r_{ij}$ come from $\mathbf{R}$, but the vectors in the combination are columns of $\mathbf{A}$. This distinction matters because row operations change the columns themselves.

Why the Coefficients Work

Proof. We prove the formula for any chosen nonpivot column $j$. Each elementary row operation is multiplication on the left by an invertible matrix: swapping two rows reverses itself, multiplying a row by a nonzero scalar is reversed by multiplying by its reciprocal, and adding a multiple of one row to another is reversed by subtracting that multiple. Therefore, if row operations reduce $\mathbf{A}$ to $\mathbf{R}$, there is an invertible matrix $\mathbf{E}$ such that $\mathbf{R}=\mathbf{E}\mathbf{A}$. In reduced row-echelon form, the pivot column $p_i$ has a 1 in row $i$ and zeros elsewhere. Thus the column $j$ of $\mathbf{R}$ is the combination of its pivot columns with coefficients given by its entries in the pivot rows: $$ \mathbf{R}_{j}=\sum_{i=1}^{r}r_{ij}\mathbf{R}_{p_i}. $$ Using $\mathbf{R}=\mathbf{E}\mathbf{A}$, this becomes $$ \mathbf{E}\mathbf{a}_j=\sum_{i=1}^{r}r_{ij}\mathbf{E}\mathbf{a}_{p_i} =\mathbf{E}\left(\sum_{i=1}^{r}r_{ij}\mathbf{a}_{p_i}\right). $$ Since $\mathbf{E}$ is invertible, multiplying both sides by $\mathbf{E}^{-1}$ gives the required identity for the original columns. If $r=0$, then $\mathbf{R}$ is the zero matrix, so $\mathbf{E}\mathbf{a}_j=\mathbf{0}$; invertibility gives $\mathbf{a}_j=\mathbf{0}$, as the empty-sum case requires. $\square$

Corollary: Each nonpivot column gives a dependence relation among the original columns: for a nonpivot column $j$, $$ \mathbf{a}_j-\sum_{i=1}^{r}r_{ij}\mathbf{a}_{p_i}=\mathbf{0}. $$

Proof. The theorem gives $\mathbf{a}_j=\sum_{i=1}^{r}r_{ij}\mathbf{a}_{p_i}$; moving the right side to the left gives the displayed equation. It is a nontrivial relation because the coefficient on $\mathbf{a}_j$ is $1$, and $j$ is not one of the pivot-column indices. $\square$

This sign convention is worth noting. The column expression has coefficients $r_{ij}$ on the pivot columns. When written as a homogeneous dependence relation with everything on the left, those coefficients appear with minus signs, while the nonpivot column has coefficient $1$.

Reading and Checking Column Relations

Worked Example: Two Nonpivot Columns

Find expressions for the nonpivot columns of $\mathbf{A}=\begin{bmatrix}1&2&1&3\\0&1&1&1\\1&3&2&4\end{bmatrix}$ using the original pivot columns, and check each expression.

First reduce $\mathbf{A}$. Replace $R_3$ by $R_3-R_1$, then replace the new $R_3$ by $R_3-R_2$. Finally replace $R_1$ by $R_1-2R_2$:

\begin{bmatrix}1&2&1&3\\0&1&1&1\\1&3&2&4\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3-R_1}} \begin{bmatrix}1&2&1&3\\0&1&1&1\\0&1&1&1\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3-R_2}} \begin{bmatrix}1&2&1&3\\0&1&1&1\\0&0&0&0\end{bmatrix} \mathrel{\mathop{\sim}^{R_1\to R_1-2R_2}} \begin{bmatrix}1&0&-1&1\\0&1&1&1\\0&0&0&0\end{bmatrix}.

The pivot columns are 1 and 2, so columns 3 and 4 are nonpivot columns. In column 3 of the RREF, the entries in the pivot rows are $-1$ and $1$. In column 4 they are $1$ and $1$. The theorem therefore gives $\mathbf{a}_3=-\mathbf{a}_1+\mathbf{a}_2$ and $\mathbf{a}_4=\mathbf{a}_1+\mathbf{a}_2$.

Check: The original columns are $\mathbf{a}_1=(1,0,1)^{\top}$, $\mathbf{a}_2=(2,1,3)^{\top}$, $\mathbf{a}_3=(1,1,2)^{\top}$, and $\mathbf{a}_4=(3,1,4)^{\top}$. Then $-\mathbf{a}_1+\mathbf{a}_2=(-1,0,-1)^{\top}+(2,1,3)^{\top}=(1,1,2)^{\top}=\mathbf{a}_3$, and $\mathbf{a}_1+\mathbf{a}_2=(1,0,1)^{\top}+(2,1,3)^{\top}=(3,1,4)^{\top}=\mathbf{a}_4$. Both identities hold in the original matrix.

The pivot columns need not be the first columns. A zero column, for example, cannot contain a pivot. The same coefficient rule applies whatever the pivot positions are.

Worked Example: Pivot Columns 2 and 3

For the matrix below, identify the pivot columns and express every nonpivot column using the original pivot columns:

\mathbf{A}= \begin{bmatrix}0&1&0&2\\0&0&1&3\\0&1&1&5\end{bmatrix}.

Replace $R_3$ by $R_3-R_1$, then replace the new $R_3$ by $R_3-R_2$:

\begin{bmatrix}0&1&0&2\\0&0&1&3\\0&1&1&5\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3-R_1}} \begin{bmatrix}0&1&0&2\\0&0&1&3\\0&0&1&3\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3-R_2}} \begin{bmatrix}0&1&0&2\\0&0&1&3\\0&0&0&0\end{bmatrix}.

This is in RREF, with pivot columns 2 and 3. Columns 1 and 4 are nonpivot columns. Column 1 of the RREF has zeros in both pivot rows, so $\mathbf{a}_1=0\mathbf{a}_2+0\mathbf{a}_3=\mathbf{0}$. Column 4 has entries $2$ and $3$ in those rows, so $\mathbf{a}_4=2\mathbf{a}_2+3\mathbf{a}_3$.

Check: The original columns are $\mathbf{a}_1=(0,0,0)^{\top}$, $\mathbf{a}_2=(1,0,1)^{\top}$, $\mathbf{a}_3=(0,1,1)^{\top}$, and $\mathbf{a}_4=(2,3,5)^{\top}$. We have $2\mathbf{a}_2+3\mathbf{a}_3=(2,0,2)^{\top}+(0,3,3)^{\top}=(2,3,5)^{\top}=\mathbf{a}_4$; the zero-column identity also holds.

Reading Coefficients When Pivots Are Separated

Worked Example: A Nonpivot Column Between Pivot Columns

Reduce the matrix, then write the relations for its nonpivot columns using its original pivot columns:

\mathbf{A}= \begin{bmatrix}1&2&1&2\\0&0&1&3\\1&2&0&-1\end{bmatrix}.

Replace $R_3$ by $R_3-R_1$, then add $R_2$ to the new $R_3$, and finally replace $R_1$ by $R_1-R_2$:

\begin{bmatrix}1&2&1&2\\0&0&1&3\\1&2&0&-1\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3-R_1}} \begin{bmatrix}1&2&1&2\\0&0&1&3\\0&0&-1&-3\end{bmatrix} \mathrel{\mathop{\sim}^{R_3\to R_3+R_2}} \begin{bmatrix}1&2&1&2\\0&0&1&3\\0&0&0&0\end{bmatrix} \mathrel{\mathop{\sim}^{R_1\to R_1-R_2}} \begin{bmatrix}1&2&0&-1\\0&0&1&3\\0&0&0&0\end{bmatrix}.

The pivots are in columns 1 and 3. Thus columns 2 and 4 are nonpivot columns. Reading the entries in the pivot rows gives $\mathbf{a}_2=2\mathbf{a}_1+0\mathbf{a}_3$ and $\mathbf{a}_4=-\mathbf{a}_1+3\mathbf{a}_3$.

Check: The original columns are $\mathbf{a}_1=(1,0,1)^{\top}$, $\mathbf{a}_2=(2,0,2)^{\top}$, $\mathbf{a}_3=(1,1,0)^{\top}$, and $\mathbf{a}_4=(2,3,-1)^{\top}$. We obtain $2\mathbf{a}_1=(2,0,2)^{\top}=\mathbf{a}_2$, and $-\mathbf{a}_1+3\mathbf{a}_3=(-1,0,-1)^{\top}+(3,3,0)^{\top}=(2,3,-1)^{\top}=\mathbf{a}_4$.

A Common Mistake: Using Reduced Columns as Original Columns

In the first worked example, the third column of the RREF is $(-1,1,0)^{\top}$. A tempting but wrong calculation is to say that the original third column is $(-1,1,0)^{\top}$. It is not: the original third column is $(1,1,2)^{\top}$. The reduced column provides coefficients; it does not replace the original column.

The correction is to read $(-1,1)$ from the pivot rows of that reduced column, then use those numbers on the original pivot columns: $\mathbf{a}_3=-\mathbf{a}_1+\mathbf{a}_2$. Indeed, $-(1,0,1)^{\top}+(2,1,3)^{\top}=(1,1,2)^{\top}$. The coefficient rule and the use of original columns are both essential.

The same distinction explains how a column expression becomes a dependence relation. From $\mathbf{a}_4=-\mathbf{a}_1+3\mathbf{a}_3$, for example, move all terms to one side to obtain $\mathbf{a}_1+\mathbf{a}_4-3\mathbf{a}_3=\mathbf{0}$. The coefficient on the nonpivot column is $1$; the signs on the pivot-column terms change when they are moved across the equality.

These relations give more than a dependence test. They describe exactly how each nonpivot column contributes no new direction beyond the pivot columns. In practical calculations, retain the original matrix, record the pivot-column indices from its RREF, and read each nonpivot column's coefficients from the pivot rows. Then check each proposed identity by combining the corresponding original columns.

Check Your Understanding

For each question, distinguish the entries in the RREF from the columns of the original matrix.

If the pivot columns of an RREF are columns 1 and 3, what original columns should be used to express a nonpivot column?
For a nonpivot column whose pivot-row entries are $4$ and $-2$, write the column expression in terms of the original pivot columns.
Turn the expression in question 2 into a homogeneous dependence relation.
Why is a column of the RREF generally not the corresponding original column?
If a matrix has no pivot columns, what does the theorem say about each of its original columns?

Expressing Nonpivot Columns Through Pivot Columns

What You'll Learn

From a Reduced Matrix to the Original Columns

Why the Coefficients Work