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 — and 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.

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. Writing r for the rank of A and assuming 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.

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, meaning 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