Simpler system of linear equations
What is the simpliest system of linear equation you can think of?
How about solving \(x = 1 \\ y = 2 \\ z = 3 \ ?\)
Pretty simple, right? If we represented this system as augmented matrix then it would look like \(\begin{pmatrix} 1 & 0 & 0 &\bigg| & 1 \\ 0 & 1 & 0 &\bigg| & 2 \\ 0 & 0 & 1 &\bigg| & 3 \end{pmatrix}\) The key observation from inspecting this is that we have a lot of $0$’s in the coefficient matrix. In fact, we only have a single entry in each column that is nonzero.
Of course, not all systems are so simple.
Question: Can we operate, in some way, on a matrix $[A|b]$ that
- does not change the set of solutions to the system but
- simplifies the coefficient matrix by zeroing out more entries?
The answer is yes, as we will see shortly.
In fact, we can bring the augmented matrix $[A|b]$ to a canonical form that, while not as simple as the example above, is incredibly useful for solving the system.
Row Echelon Form
Definition: We say a matrix $A$ is in row echelon form if
- all zero rows are at the bottom of $A$ and
- reading left to right, the first nonzero entry of a row is always strictly to the right first nonzero entry of the row above it.
It helps our communication and understanding to be more precise and to do so we need to introduce some notation. For an matrix $A = (a_{ij})$, let’s write $A_{\bullet s}$ as shorthand for the vector that is the $s$-th row of $A$. So in the example \(A = \begin{pmatrix} 7 & -1 & 0 & 2 \\ 3 & 3 & 3 & -1 \\ -1 & 2 & -4 & 0 \end{pmatrix}\) we have \(\mathbf{R}_1^A = \begin{pmatrix} 7 & -1 & 0 & 2 \end{pmatrix} \\ \mathbf{R}_2^A = \begin{pmatrix} 3 & 3 & 3 & -1 \end{pmatrix} \\ \mathbf{R}_3^A = \begin{pmatrix} -1 & 2 & -4 & 0 \end{pmatrix}.\)
Similarly we introduce the notation $A_{t \bullet}$ for the $t$-th column vector of $A$. Continuing the example we have \(\mathbf{C}^A_{1} = \begin{pmatrix} 7 \\ 3 \\ -1 \end{pmatrix} \ , \ \mathbf{C}^A_{2} = \begin{pmatrix} -1 \\ 3 \\ 2 \end{pmatrix} \\ \mathbf{C}^A_{3} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} \ , \ \mathbf{C}^A_{4} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}\) We will often omit the $A$ superscript if the context is clear.
We say a vector $\mathbf{v}$ is non-zero if at least one entry is not $0$.
Definition: For a non-zero vector $\mathbf{v}$, the pivot of $\mathbf{v}$ is the first non-zero entry of $\mathbf{v}$.
For example, the pivot of \(\begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}\) is the second entry in list, $3$. For \(\begin{pmatrix} -1 & 2 & -4 & 0 \end{pmatrix}\) the pivot is the first entry in the list, $-1$.
Note that we can only talk about pivots for non-zero vectors.
Given these notions, we can rephrase the definition row echelon form as
Definition: A matrix $A$ is in row echelon form if
- all zero rows are at the bottom of $A$ and
- for all nonzero rows $A_{\bullet s}$ the index of the pivot of $A_{\bullet s}$ is strictly greater that the index of the pivot of $A_{\bullet (s-1)}$. In less words, \(\operatorname{index} (\operatorname{Pivot}(A_{\bullet (s-1)})) \lneq \operatorname{index} (\operatorname{Pivot}(A_{\bullet s}))\)
We say an augmented matrix $[A|b]$ is in row echelon form if $A$ is in row echelon form.
For example, \(A = \begin{pmatrix} 7 & -1 & 0 & 2 \\ 3 & 3 & 3 & -1 \\ -1 & 2 & -4 & 0 \end{pmatrix}\) is not in row echelon form since
- all rows are non-zero so none need to be at the bottom but
- the indices of the pivots are all three rows are $1$; they do not strictly decrease as we move down the matrix.
The pivots of the nonzero rows should look like a staircase. Here are two examples of matrices in row echelon form. \(B = \begin{pmatrix} 7 & -1 & 0 & 2 \\ 0 & 3 & 3 & -1 \\ 0 & 0 & -4 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 7 & -1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}\)
There is a more refined form row echelon row.
Reduced row echelon form
Definition: A matrix $A$ is in reduced row echelon form if
- $A$ is row echelon form,
- all the pivots of the rows of $A$ have value $1$, and
- each pivot is the only non-zero entry in its column.
We say an augmented matrix $[A|b]$ is in reduced row echelon form if $A$ is in reduced row echelon form.
All the examples we have seen so far are not in reduced row echelon form, even if they were in row echelon form.
An example in reduced row echelon is \(\begin{pmatrix} 1 & 0 & 5 \\ 0 & 1 & 3 \end{pmatrix}\)
We will introduce operations which will allow us to transform any matrix into reduced row echelon. For an augmented matrix $[A|b]$, the operations will preserve the set of solutions to the associated linear system.