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Rank as the number of pivots, the two comparisons that settle a solution set, and why a homogeneous system with more unknowns than equations always has a non-zero solution.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to count the rank of a matrix from its reduced form, decide consistency by comparing the rank of the coefficient matrix with the rank of the augmented matrix, count the free variables as the number of unknowns less the rank, bound the rank by the number of rows and of columns and draw the consequences for over- and underdetermined systems, explain why a homogeneous system is always consistent and when it has a non-zero solution, and build a matrix to a prescribed rank rather than only reading one off.
The last lesson read a whole solution set off a reduced matrix. Most of what it read depended on one thing only: where the pivots were, and how many of them there were. This lesson keeps the count and throws the rest away, which turns out to be enough to answer every question about how many solutions a system has.
The rank of a matrix is the number of pivots in its reduced row echelon form, written $\operatorname{rank} A$. A system is consistent if it has at least one solution. A system is homogeneous if every right-hand side is zero, so that it reads $Ax = 0$. A system with more unknowns than equations is underdetermined; one with more equations than unknowns is overdetermined. The trivial solution of a homogeneous system is $x = 0$.
Everything about the size of a solution set follows from the rank, asked twice.
First: is it consistent? Compare $\operatorname{rank} A$ with $\operatorname{rank} [A \mid b]$. Attaching the column $b$ to $A$ cannot lower the rank and cannot raise it by more than one, so there are exactly two cases. If the ranks are equal, the system is consistent. If $\operatorname{rank} [A \mid b] = \operatorname{rank} A + 1$, the extra pivot is in the augmented column, which is a row reading $0 = c$ with $c$ non-zero: no solutions.
Second, only if the first said yes: how many? Compare $\operatorname{rank} A$ with $n$, the number of unknowns. Every column is a pivot column or a free one, so
$$\text{(number of free variables)} = n - \operatorname{rank} A.$$
If the rank is $n$ there are no free variables and the solution is unique. If the rank is less than $n$ there is at least one, and the solutions form a set of dimension $n - \operatorname{rank} A$.
The order matters. A free column says nothing at all until consistency is settled, because an inconsistent system has free columns and no solutions.
Homogeneous systems skip the first question. $x = 0$ always satisfies $Ax = 0$, so a homogeneous system is consistent, always, and only the second comparison is left. It has a non-zero solution exactly when $\operatorname{rank} A < n$.
Another way: picture
Picture the pivots as claims staked on the columns. Each pivot claims one column and one row, and no two pivots share either — that is the staircase. The claimed columns are the basic variables, whose values are dictated; the unclaimed ones are free, and each unclaimed column is one direction you may travel without leaving the solution set. A pivot staked on the augmented column is a claim against nothing, because there is no variable there, and that is exactly the picture of an impossible equation.
Another way: steps
Each pivot needs a row to itself and a column to itself, so for an $m \times n$ matrix
$$\operatorname{rank} A \le \min(m, n).$$
That single inequality carries most of the folklore about over- and underdetermined systems, and corrects the rest of it.
More unknowns than equations ($n > m$). Then $\operatorname{rank} A \le m < n$, so there is always at least one free column. A consistent such system therefore has infinitely many solutions — and a homogeneous one is automatically consistent, which gives the most useful corollary in the unit: a homogeneous system with more unknowns than equations always has a non-zero solution. It is used later to prove that $n+1$ vectors in $n$ dimensions are never independent, and it is proved here, by counting.
More equations than unknowns ($m > n$). Here the folklore says "no solution", and the folklore is wrong. It says only that consistency is not guaranteed: $\operatorname{rank}[A \mid b]$ may exceed $\operatorname{rank} A$. An overdetermined system with repeated or dependent equations can be perfectly consistent, and one with rank $n$ that happens to be consistent has exactly one solution. What is true is that a typical overdetermined system is inconsistent, and that is why least squares exists.
Square ($m = n$). Rank $n$ gives exactly one solution for every $b$; rank less than $n$ gives no solutions for some $b$ and infinitely many for others. There is no middle case, and this is the first appearance of the list of equivalent conditions that this course keeps returning to.
The solutions of $Ax = 0$ are worth knowing even when the problem in front of you has a non-zero right-hand side, because of the decomposition from the last lesson: every solution of $Ax = b$ is one particular solution plus a solution of $Ax = 0$.
So the homogeneous system controls uniqueness and the right-hand side controls existence, and the two are independent questions:
| $Ax = 0$ has | $Ax = b$ has |
|---|---|
| only $x = 0$ | no solution or exactly one, depending on $b$ |
| a non-zero solution | no solution or infinitely many, depending on $b$ |
Notice what is missing from the table: no row offers "exactly one" alongside "infinitely many". Whether a consistent system has one solution or a continuum is a fact about $A$ alone, settled before $b$ is even looked at.
The solution set of $Ax = 0$ is also closed under addition and scaling — if $Au = 0$ and $Av = 0$ then $A(su + tv) = 0$ — which the solution set of $Ax = b$ is not, unless $b = 0$. That closure is what makes it a subspace, and the vocabulary for saying so is the next unit.
Reading free columns before checking consistency. "There is a free variable, so there are infinitely many solutions" is false for an inconsistent system, which can have as many free columns as you like and no solutions at all. Consistency first, always.
Counting a pivot in the augmented column. It is not a pivot for any variable. Its only job is to announce that the system is inconsistent, and including it in $\operatorname{rank} A$ makes the arithmetic of $n - \operatorname{rank} A$ meaningless.
Believing more equations means no solution. It means consistency is not guaranteed. Repeated equations, or equations that are combinations of the others, cost nothing.
Believing more unknowns means a solution exists. It does not. $x + y = 1$ together with $x + y = 2$ is underdetermined and inconsistent. What extra unknowns guarantee is a free variable, hence never exactly one solution — not that there is one.
Thinking a homogeneous system can fail. It cannot. $x = 0$ works. The only question about a homogeneous system is whether anything else does.
Confusing the number of equations with the rank. Equations that repeat each other, or that are sums of others, contribute nothing. The rank counts independent constraints, which is why it has to be computed rather than read off the page.
$\left(\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \end{array}\right)$ reduces to $\left(\begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right)$: $\operatorname{rank} A = \operatorname{rank}[A \mid b] = 1$.
Consistent: the ranks agree.
Three unknowns, rank $1$, so $3 - 1 = 2$ free variables and a plane of solutions.
Second comparison.
Change the last entry to $9$ and it reduces to a bottom row $(0\ 0\ 0 \mid 1)$: $\operatorname{rank} A = 1$ but $\operatorname{rank}[A \mid b] = 2$. No solutions — with the same two free columns as before.
Which is why the order of the two questions matters.
$x + y = 3$, $x - y = 1$, $2x = 4$: three equations, two unknowns.
Overdetermined, and not yet condemned.
The third equation is the sum of the first two, so reduction leaves a zero row: $\operatorname{rank} A = \operatorname{rank}[A \mid b] = 2$.
The extra equation was not independent.
Rank $2$ equals the number of unknowns, so there is exactly one solution, $x = 2$, $y = 1$. Change the third equation to $2x = 5$ and the ranks disagree: no solution.
The count of equations never decided anything; the rank did.
Homogeneous, so $x = 0$ is a solution and the system is consistent. The first comparison is settled without any work.
Existence is free here.
The rank is at most the number of rows, so $\operatorname{rank} A \le 3 < 5$.
The shape bounds the rank.
So there are at least $5 - 3 = 2$ free variables, and the system has infinitely many solutions besides the zero one.
A consistent system has $7$ unknowns, and the rank of its coefficient matrix is $3$. Fill in the three counts.
| How many | |
|---|---|
| Pivot columns | |
| Free variables | |
| Dimension of the solution set |
Row reduction of an augmented matrix ends at $\left(\begin{array}{rr|r} 1 & 4 & 3 \\ 0 & 0 & 5 \end{array}\right)$. How many solutions has the system?
What is the largest the rank of an $3 \times 6$ matrix can be?
Answer:
A system $Ax = b$ has $5$ unknowns. Match each condition on the ranks to what it settles.
| Exactly one solution | No solution | Infinitely many solutions | Exactly $5$ solutions | |
|---|---|---|---|---|
| The rank of $A$ is smaller than the rank of $[A \mid b]$ | ||||
| $A$ and $[A \mid b]$ both have rank $5$ | ||||
| $A$ and $[A \mid b]$ both have rank $1$ |
Select every statement that is true.
This task has no paper form; do it on a device.
Write a $2 \times 2$ matrix of rank exactly $1$ in which no entry is zero.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A consistent system has $7$ unknowns, and the rank of its coefficient matrix is $3$. Fill in the three counts.
| How many | |
|---|---|
| Pivot columns | |
| Free variables | |
| Dimension of the solution set |
You can compute a rank, decide consistency from the two ranks, and count the free variables. Say in your own words why a homogeneous system with more unknowns than equations must have a solution other than zero. Next: matrix arithmetic, and the algebra that these systems have been written in all along.
10. Your turn: a homogeneous system has $5$ unknowns and $3$ equations. What can you say?, step 3
More unknowns than equations, homogeneous: always.