Rank, Nullity & Solution Sets

The last lesson asked how many independent directions a clutch of arrows holds. Stack those arrows as the columns of a matrix, and that same count earns a name. It turns out to answer, at a single glance, the question we ground through elimination two lessons ago: does Ax = b have one solution, infinitely many, or none? The shortcut is to count.

Two counts: rank and nullity

The rank of a matrix is the number of genuinely independent rows it holds — equally, the number of pivots elimination eventually produces. A row that is a combination of the others is redundant and adds nothing to the count.

A fact is being smuggled into that sentence, and it deserves a stop because nothing about it is obvious: the count of independent rows and the count of independent columns are always the same number. A 3 × 5 matrix has five columns living in 3-D space and three rows living in 5-D space — different worlds entirely. Yet the two tallies never disagree.

That shared value is the rank, which is why you may compute it whichever way is easier, and why it can never exceed either dimension: rank ≤ min(m, n).

Its partner is the nullity — the number of free directions, the dimension of the null space (the set of x solving Ax = 0). If only x = 0 works, the nullity is 0; each independent direction you can slide along and stay at zero adds one. Put plainly, nullity is just how many free variables the system has.

The rank-nullity theorem

For any m × n matrix A — that is, n columns — the two counts always add up to the number of columns:

The reason it is exact: each of the n columns is either a pivot column (counted by rank) or a free column (counted by nullity) — there is no third kind. And that one equation classifies Ax = b. Write r for the rank of A and assume the system is consistent (b is actually reachable, so [A | b] has the same rank as A — no 0 = nonzero row):

• One solution when r = n (full column rank): no free variables, so a single point.
• Infinitely many when consistent with r < n: there are n − r free variables, a whole family.
• No solution when inconsistent: rank[A | b] > rank A, a contradiction row.

The third branch is the one that stays abstract until you meet it, so meet it now.

Take x + y = 2 alongside 2x + 2y = 5. Here rank A = 1 (the second row is twice the first). Subtracting 2 × R1 from R2 in the augmented matrix leaves the row 0 = 1, so rank[A | b] = 2 > 1 and there is no solution.

Change that 5 to a 4 and the same subtraction leaves 0 = 0. Now the system is consistent with r = 1 < n = 2, one free variable, and infinitely many solutions. Same matrix A, different b, two of the three branches.

A consequence drops straight out. A wide matrix (more columns than rows, m < n) can have rank at most m, so r ≤ m < n, forcing nullity = n − r > 0. More unknowns than equations can never pin down a single point.

A worked example, traced

Watch dependent rows collapse. In A, every row is a multiple of [1 2 3]:

A = [ 1  2  3 ]   R2 → R2 − 2R1,  R3 → R3 − 3R1   [ 1  2  3 ]
    [ 2  4  6 ]  ──────────────────────────────►  [ 0  0  0 ]
    [ 3  6  9 ]                                     [ 0  0  0 ]

One pivot survives, so rank A = 1 — the three rows held only one independent direction.

Now a matrix with a genuine second direction:

B = [ 1  2  3 ]              echelon              [ 1  2  3 ]
    [ 0  1  4 ]   ──────────────────────────────► [ 0  1  4 ]
    [ 2  5  10]                                     [ 0  0  0 ]

Two pivots, so rank B = 2. With n = 3 columns the theorem hands you the nullity with no extra work:

nullity = n − rank = 3 − 2 = 1

So Bx = 0 has a one-parameter family of solutions — set the single free variable to t and every solution is a multiple of one direction.

And a 2-equation, 3-unknown system has a 2 × 3 matrix, so rank ≤ 2 < 3: its nullity is at least 1. It can never have a unique answer — only infinitely many (if consistent) or none.

A question to carry forward

For a square matrix, “full rank” is the line between a unique solution and trouble.

It would be handy to test that with a single number, computed once, rather than reducing the whole matrix every time. Here is the thread onward: is there one number attached to a square matrix that is non-zero exactly when its rank is full — when its columns are independent and Ax = b has a unique answer?

In one breath

• Rank = independent rows/columns = number of pivots; nullity = dimension of the null space (Ax=0) = number of free variables.
• Rank-nullity theorem: rank + nullity = n, where n is the number of COLUMNS (every column is a pivot or a free column).
• Classify Ax=b (consistent): r = n → unique; r < n → infinitely many (n−r free vars); inconsistent (rank[A|b] > rank A) → none.
• A wide matrix (m < n) has nullity > 0, so Ax = 0 always has non-zero solutions — more unknowns than equations never gives a unique point.
• Nullity is almost always n − rank straight from the theorem.

Practice