Download 22 Echelon Forms

Echelon Forms and the General Solution to Ax = b
Learning Goals: to see that elimination is still the technique that gets our answers, but that going
farther than needed in practice gets us more in theory.
The goal of elimination is to reduce A, by eliminations and row swaps to a matrix U
which is upper triangular. Let’s stop worrying about what kind of trouble we get into if there is a
missing pivot, and allow general m × n matrices. This guarantees that there will be some row or
column (or both!) that do not hold a pivot, just by the pigeonhole principle. What we will do is
elimination and row swaps to clear out columns as usual. If a column is missing a pivot—there
is no row swap that will fix elimination—just move on to the next column. Yes, a column is
broken. Life goes on.
⎡1 2 3 4 ⎤
⎢2 4 6 8 ⎥
⎢
⎥
Example: apply elimination to ⎢ 1 1 1 1 ⎥ . The first column needs us to subtract two times
⎢
⎥
⎢1 3 5 7 ⎥
⎢⎣ 2 3 4 6 ⎥⎦
row one from row two, row one from rows three and four, and two times row one from row 5:
⎡1 2 3 4 ⎤
⎢0 0 0 0 ⎥
⎢
⎥
⎢ 0 −1 −2 −3 ⎥ . Now swap rows two and three and continue elimination in the second
⎢
⎥
3⎥
⎢0 1 2
⎢⎣ 0 −1 −2 −2 ⎥⎦
⎡1 2 3 4 ⎤
⎢ 0 −1 −2 −3⎥
⎢
⎥
column to get: ⎢ 0 0 0 0 ⎥ . The third column is hopeless. We swap rows to get a pivot
⎢
⎥
⎢0 0 0 0 ⎥
⎢⎣ 0 0 0 1 ⎥⎦
3 4⎤
⎡1 2
⎢ 0 −1 −2 −3⎥
⎢
⎥
in the fourth column and end up with ⎢ 0 0 0 1 ⎥ . We’d do further elimination in the
⎢
⎥
⎢0 0 0 0 ⎥
⎢⎣ 0 0 0 0 ⎥⎦
fourth column, but it is unneeded.
A matrix such as this result, which has each row starting with a pivot and zeroes below
the pivot is said to be in row-echelon form (REF).
Each step of the elimination, as we know, can be carried out by a row operation matrix,
which is invertible. Let B = EA where E is any matrix that is a product of elementary row
operations—eliminations above or below the diagonal, row swaps, and constant multiples of
rows. Then A and B are said to be row-equivalent. We have really just shown:
Theorem: every matrix is row-equivalent to one in row-echelon form.
The form is not unique, since we could do various eliminations upward or multiply rows by nonzero constants to obtain another row-equivalent echelon matrix. Note that row-equivalence is an
equivalence relation on the set of m × n matrices.
Example: each invertible matrix is row-equivalent to I. Just do Gauss-Jordan.
The really nice thing about row-equivalence is that it doesn’t alter the nullspace. But the
nullspace is much easier to deal with if the matrix is in row-echelon, or even better reduced rowechelon form. The nullspace is unchanged because elimination doesn’t affect the right-hand side
of Ax = 0. And now the matrix is ready for back substitution.
If every column has a pivot, we have only the trivial solution. So the nullspace is Z. This
is by far the most common and important case. We have a unique solution to our system. But
we have to be prepared for all eventualities.
So any variable that corresponds to a column missing its pivot will be called a free
variable. The free variables can be set to whatever we want, and we can still solve the system. If
looked at as equations, we could put all the free variables on the right-hand side, and give them
any values whatsoever. What remains on the left is an n × n invertible system (pivot in every
remaining column!) so we can solve for any right-hand side.
For instance, in the matrix above, there is no choice but x4 = 0. But x3 is free. Let’s set
its value to t. Then there is no choice for x2 but –2t. And then there is no choice for x1 but t. So
we get any vector of the form (t, –2t, t, 0) = t(1, –2, 1, 0).
In solving systems in the practical world, we would stop here. In theory, though, it’s
more convenient to continue with Gauss-Jordan elimination, eliminating above the pivots and
dividing through by them. This produces the reduced row-echelon form. For the example
⎡ 1 0 −1 0 ⎤
⎢0 1 2 0 ⎥
⎢
⎥
matrix above we would get ⎢ 0 0 0 1 ⎥ .
⎢
⎥
⎢0 0 0 0 ⎥
⎢⎣ 0 0 0 0 ⎥⎦
Theorem: every matrix is row-equivalent to a unique reduced row-echelon form matrix.
Proof: the equivalence to some such matrix is easy, since we’re still doing row
operations. Why is it unique? Let R1 and R2 be to RREF matrices, and let ER1 = R2.
Notice that ER1 combines rows of R1 to obtain rows of R2. But each row has a pivot in a
different column. Adding in multiples of earlier rows puts nonzero numbers in earlier
positions in the row. Adding in later rows puts nonzero numbers in the places where
there are supposed to be zeroes above and below the pivots. Row swapping ruins the
echelon form, and multiplying rows by constants ruins the ones in the pivot position. So
no combination of these operations will turn one RREF into a different one.
Again, from the reduced row-echelon form, it is easy to read off the “special” solutions to
the nullspace. From this matrix, the free variable is x3. So we know the other non-zero entries in
the “special” vector are the opposites of the entries in the third column: (1, –2, 1, 0) as seen
before.
In more complicated matrices, the reduced form allows us to pick out the special vectors
more easily, since it is really doing back substitution in advance. And the reduced form will help
us identify some other subspaces associated with a matrix.
Finally, notice that if a matrix has more columns than rows, it is guaranteed to have free
variables. There just aren’t enough rows to put a pivot in each column! Such matrices are
guaranteed to have nontrivial nullspaces.
Reading: 3.2
Problems: 3.2: 1 – 4, 9, 11, 12, 13, 14, 15 (do this!), 19, 20, 21 – 25, 26, 27, 28, 35 – 37