Reduced Echelon Form

06 Reduced Echelon Form and Row Equivalence

We have used Gauss's Method to solve linear systems of equations. This method uses row operations to put a linear system or matrix into echelon form. [From now on, I will just talk in terms of matrices.] Chapter 1, Section III covers reduced echelon form. In an echelon form matrix, any all-zero row is at the bottom, and for rows that are not all-zero (after the first row), the leading entry in that row lies to the right of the leading entry in the previous row. A reduced echelon form matrix has the additional properties that (1) every leading entry is a 1 and (2) in any column that contains a leading entry, that leading entry is the only non-zero entry in that column.

After a matrix has been put into echelon form, it is easy to apply additional row operations to put the matrix into reduced echelon form. Just divide each non-zero row by the leading entry in that row, making the leading entry equal to one. Then, for any column that contains a leading entry for row $i$ , convert all the other entries in that column to zero by subtracting appropriate multiples of row $i$ from the other rows. (For the second stage, work from right to left.) For example: $\begin{aligned} (\begin{array}{c} 2 & - 4 & 1 & 6 & 3 \\ 0 & 0 & - 4 & 2 & 0 \\ 0 & 0 & 0 & 3 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}) & \overset{\begin{array}{c} \frac{1}{2} ρ_{1} \\ - \frac{1}{4} ρ_{2} \\ \frac{1}{3} ρ_{3} \end{array}}{\to} (\begin{array}{c} 1 & - 2 & \frac{1}{2} & 3 & \frac{3}{2} \\ 0 & 0 & 1 & - \frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}) \\ \overset{\begin{array}{c} - 3 ρ_{3} + ρ 1 \\ \frac{1}{2} ρ_{3} + ρ_{2} \end{array}}{\to} (\begin{array}{c} 1 & - 2 & \frac{1}{2} & 0 & - \frac{1}{2} \\ 0 & 0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}) \\ \overset{\begin{array}{c} - \frac{1}{2} ρ_{2} + ρ_{1} \end{array}}{\to} (\begin{array}{c} 1 & - 2 & 0 & 0 & - \frac{2}{3} \\ 0 & 0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 & 0 & 0 \end{array}) \end{aligned}$ Once a system is in reduced row-echelon form, it is very easy to solve.

But the more interesting aspect of Section 1.III is the definition of row equivalence of matrices, the proof that row equivalence is an equivalence relation on the set of all $m \times n$ matrices, and the proof that given any $m \times n$ matrix, $A$ , there is one and only one reduced echelon form matrix that is row equivalent to $A .$

The really important consequence of all this is that no matter what specific sequence of row operations that you apply to a matrix to put it into reduced echelon form, you will always end up with the same reduced echelon form matrix.

We defined free variables and leading variables for echelon form matrices. But starting from a matrix, $A$ , you can get many different echelon form matrices by applying different sequences of row operations. Do those various echelon form matrices that are row-equivalent to $A$ have the same free variables? Do they even have the same number of free variables? We now know that the answer to these questions is "yes." The reason is that all of those echelon form matrices must reduce to the same reduced echelon form matrix, and going from echelon form to reduced echelon form does not change the set of free variables (since a free variable just corresponds to a column that does not contain a leading entry).

Finally, there is result stated as a Lemma in Section 1.III.2 that will be important for us later: In an echelon form matrix, a non-zero row cannot be a linear combination of the other non-zero rows. Another way of saying this is that the non-zero rows in an echelon form matrix are linearly independent.

There is some fairly sophisticated mathematics in Section 1.III that will require some work to understand, especially for people who have not taken Math 135.

There is the idea of "equivalence relation" on a set. A relation on a set is an operation that takes any two elements of the set and tells you whether or not they are related in the sense defined by that relation. If I say that $\sim$ is a relation on a set $X$ , then for any $a, b \in X$ , we would write $a \sim b$ if $a$ and $b$ are related according to that relation. For example, $\leq$ is a relation on the set $R$ , and we write $x \leq y$ to mean that $x$ and $y$ are related according to that relation.

An equivalence relation is a type of relation that breaks up a set into non-empty subsets, where the intersection of any two subsets is empty. This is called a partition of the set. All of the elements a given subset are related to each other, and two elements from different subsets are not related. Given a relation, you can show that it is an equivalence relation by showing that it has certain properties—reflexive, symmetric, and transitive—but usually we are most interested in the way the relation partitions the set. (For the record, a relation $\sim$ on a set $X$ is reflexive if $x \sim x$ for all $x \in X$ . It is symmetric if whenever $x \sim y$ , it is also true that $y \sim x$ . And it is transitive if whenever $x \sim y$ and $y \sim z$ , it is also true that $x \sim z$ .)

Two matrices $A$ and $B$ are said to be row equivalent if there is a sequence of row operations that transforms $A$ to $B$ . Of course, this can't happen unless $A$ and $B$ have the same numbers of rows and columns. For any given positive integers $n$ and $m$ , row equivalence is a relation on the set of $m \times n$ matrices. It is easy to show that this relation is reflexive, symmetric, and transitive. That means that it is an equivalence relation, and it partitions the set of $m \times n$ matrices into non-empty, non-intersecting subsets.

The most difficult theorem we have encountered so far (Theorem 2.6 in Section 1.III.2) says that each of those subsets contains exactly one reduced echelon form matrix. Since two matrices are row equivalent if and only if they are in the same subset, this means that every matrix is row equivalent to one and only one reduced echelon form matrix. Furthermore, two matrices are row equivalent if and only if they reduce to the same reduced echelon form matrix. The proof of the theorem uses proof by mathematical induction.

Proof by induction is actually very natural. It amounts to showing that something is true for $n = 1$ , then that it follows from that that it is also true for $n = 2$ , then that it follows from that that it is also true for $n = 3$ , then that it follows from that that it is also true for $n = 4$ , and so on. That is, each step follows logically from the previous step. Obviously, phrased this way it becomes ridiculously tedious. Imagine trying to prove that it's true for $n = 10000$ . Proof by induction is a standard way of presenting this type of proof that doesn't require proving each step individually. Proof by induction is used to show that some statement is true for all positive integers $n$ . The proof is done in just two parts:

Base case: Prove that the statement is true for $n = 1$ .
Inductive case: Prove that whenever the statement is true for some positive integer $k$ , then it is also true for $k + 1$ .

The inductive case takes care of proving all of the infinitely many individual steps—from 1 to 2, from 2 to 3, from 3 to 4, from 4 to 5, and so on—all at once. (Sometimes, the induction starts with a base case other than $n = 1$ . For example, if the base case is $n = 5$ , then you prove that the statement is true for all integers greater than or equal to 5.)

A typical first example of induction is to prove that $1 + 2 + 3 + \dots + n = \frac{n (1 + n)}{2}$ for any positive integer $n$ . To convince you of this without using induction, I might say the following and ask you to convince yourself that there is a pattern that allows you to derive each step from the previous step, and that that pattern will keep working forever:

For $n = 1$ , $1 = \frac{1 (1 + 1)}{2}$ . ✓
For $n = 2$ , $(1) + 2 = \frac{1 (1 + 1)}{2} + 2 = \frac{1 (1 + 1) + 2 \cdot 2}{2} = \frac{1 \cdot 2 + 2 \cdot 2}{2} = \frac{3 \cdot 2}{2} = \frac{2 (1 + 2)}{2}$ . ✓
For $n = 3$ , $(1 + 2) + 3 = \frac{2 (1 + 2)}{2} + 3 = \frac{2 (1 + 2) + 2 \cdot 3}{2} = \frac{2 \cdot 3 + 2 \cdot 3}{2} = \frac{4 \cdot 3}{2} = \frac{3 (1 + 3)}{2}$ . ✓
For $n = 4$ , $(1 + 2 + 3) + 4 = \frac{3 (1 + 3)}{2} + 4 = \frac{3 (1 + 3) + 2 \cdot 4}{2} = \frac{3 \cdot 4 + 2 \cdot 4}{2} = \frac{5 \cdot 4}{2} = \frac{4 (1 + 4)}{2}$ . ✓
For $n = 5$ , $(1 + 2 + 3 + 4) + 5 = \frac{4 (1 + 4)}{2} + 5 = \frac{4 (1 + 4) + 2 \cdot 5}{2} = \frac{4 \cdot 5 + 2 \cdot 5}{2} = \frac{6 \cdot 5}{2} = \frac{5 (1 + 5)}{2}$ . ✓

Or I might give a formal proof by induction, which should be easier to understand and absolutely convincing:

Base Case: For $n = 1$ , we have $1 = \frac{1 (1 + 1)}{2}$ , so the theorem is true in this case.

Inductive Case: Suppose that $k$ is any positive integer and that we already know that $1 + 2 + \dots + k = \frac{k (1 + k)}{2}$ . We must show $1 + 2 + \dots + k + (k + 1) = \frac{(k + 1) (1 + (k + 1))}{2}$ . But $\begin{aligned} 1 + 2 + \dots + k + (k + 1) & = (1 + 2 + \dots + k) + (k + 2) \\ = \frac{k (1 + k)}{2} + (k + 1) \\ = \frac{k (1 + k) + 2 (k + 1)}{2} \\ = \frac{(k + 2) (k + 1)}{2} \\ = \frac{(k + 1) (2 + k)}{2} \\ = \frac{(k + 1) (1 + (k + 1))}{2} \end{aligned}$

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