Rank of a matrix — definition, row rank = column rank theorem: Question bank

Before you start, recall the three faces of rank you need here: row rank = dimension of the row space, column rank = dimension of the column space, and pivots = the leading nonzero entries after elimination. The central theorem says all three are one number.

Figure — Rank of a matrix — definition, row rank = column rank theorem

Where a symbol appears repeatedly in the "Spot the error" and "Why" sections, we use the rank factorization $A = C R$ . Here is the one definition you need for all of it:

True or false — justify

Every

m \times n

matrix has

rank (A) = rank (A^{T})

TRUE. Row rank of

A

equals column rank of

A

, and transposing swaps rows with columns, so

rank (A^{T})

counts the same number from the other side.

If a matrix has a row of all zeros, its rank must be less than the number of rows.

TRUE. A zero row contributes nothing to the row space, so at most

m - 1

rows can be independent, forcing

rank < m

5 \times 3

matrix can have rank

4

FALSE. Rank

\leq min (m, n) = min (5, 3) = 3

, because only

3

columns live in the picture and they can span at most a

3

-dimensional space.

If every column of

A

is nonzero, then

rank (A)

equals the number of columns.

FALSE. Nonzero columns can still be dependent; e.g.

(1, 1)

and

(2, 2)

are both nonzero but span only a line, giving rank

1

Row operations can change the actual column space of a matrix.

TRUE — and this is the subtle part. Row ops change which vectors the columns are, so the column space (the set of reachable outputs) can genuinely move; only its dimension is preserved.

rank (2 A) = 2 rank (A)

for a nonzero matrix

A

FALSE. Scaling by

2

scales every column by

2

but never changes independence, so

rank (2 A) = rank (A)

; rank is not a magnitude.

A

n \times n

and

rank (A) = n

, then

A

is invertible.

TRUE. Full column rank means the columns are independent and span

R^{n}

, which is exactly the invertible condition.

rank (A + B) = rank (A) + rank (B)

whenever

A, B

have no zero rows.

FALSE. Only

rank (A + B) \leq rank (A) + rank (B)

holds in general; overlapping column spaces destroy additivity regardless of zero rows.

A matrix and its reduced echelon form always have the same rank.

TRUE. Elimination replaces rows by linear combinations of rows, which never changes the row space or its dimension, so the pivot count is preserved.

The rank of a matrix can be zero.

TRUE — but only for the zero matrix. If any entry is nonzero, that row (and column) is independent, giving rank

\geq 1

Spot the error

"I reduced

A

to echelon form; the pivots are in columns 1 and 3, so columns 1 and 3 of the reduced matrix form a basis of the column space."

The positions are right but the vectors are wrong: you must take columns 1 and 3 of the original

A

, because row ops changed the column vectors themselves.

A

has two identical rows, so I delete one; that changes the rank because I removed a row."

Deleting a duplicate row removes nothing from the row space (it was already in the span), so the rank is unchanged — the row was never contributing.

A

has two identical columns, so I delete one; that shrinks the rank because the matrix now has fewer columns."

Wrong for the same reason as rows: a duplicate column is already in the span of the others, so the column space and its dimension are untouched — deleting it changes the size of the matrix, not its rank.

"Since

rank (A B) \leq rank (A)

, we always have

rank (A B) = rank (A)

when

B

is nonzero."

The bound is only

\leq

. If

B

collapses directions that

A

needed, the product can have strictly smaller rank; equality is not guaranteed by

B

being nonzero.

"The matrix has 4 nonzero rows, therefore its rank is 4."

Nonzero rows can be dependent; you must reduce to echelon form and count pivots, not visible nonzero rows.

"Column rank counts columns, row rank counts rows, so for a

3 \times 5

matrix they can't be equal — different totals."

They count dimensions of spans, not raw totals; the central theorem forces both spans to share the same dimension even though there are

5

columns and

3

rows.

"In

A = C R

the inner dimension

r

proves the rank is

r

only if

r

is the smaller of

m, n

No — recall

C

m \times r

with independent columns and

R

r \times n

with independent rows, so

r

is forced to be the shared rank; it can be any value up to

min (m, n)

, not necessarily the smaller dimension itself.

Why questions

Why does the factorization

A = C R

prove row rank

\leq

column rank?

Recall

C

m \times r

and

R

r \times n

. Row

i

A

equals (row

i

C

) times

R

, so every row of

A

is a combination of the

r

rows of

R

, capping the row rank at

r

= column rank.

Why do we take pivot columns from the original matrix for a column-space basis, but read the count from the reduced one?

The count (dimension) is invariant under row ops, but the actual column vectors change; the reduced pivots only reveal which positions are independent, and those positions point back to the untouched original columns.

Why is

rank (A) \leq min (m, n)

and never more?

The columns are

n

vectors living in

R^{m}

, so their span uses at most

n

vectors and fits inside at most

m

dimensions — whichever is smaller is the ceiling.

Why does rank measure "output dimension" of the map

x \mapsto A x

Every output

A x

is a combination of the columns, so the reachable set is exactly the column space, whose dimension is the rank.

Why can't we prove row rank = column rank just by "counting pivots once"?

Counting pivots gives one number, but it doesn't explain why the two spaces agree; the

A = C R

argument is what proves both spans genuinely have that dimension.

Why does adding a linearly dependent row leave the rank unchanged?

A dependent row already lies in the span of the others, so it adds no new direction to the row space and the dimension stays put.

Edge cases

What is the rank of the

3 \times 3

zero matrix?

Zero. There are no nonzero rows or columns, so both spaces are just the origin, a

0

-dimensional space.

What is

rank (0 \cdot A)

for any matrix

A

, and how does it contrast with

rank (2 A)

Zero — multiplying by the scalar

0

turns

A

into the zero matrix, collapsing every direction, whereas multiplying by

2

(or any nonzero scalar) preserves independence and keeps the rank equal to

rank (A)

What is the rank of an

n \times n

identity matrix?

n

. Its columns are the standard basis vectors — fully independent — so it has full rank and is invertible.

What is the rank of a single nonzero column vector (an

m \times 1

matrix)?

One. A single nonzero vector spans a line, so both column rank and row rank equal

1

What is the rank of a matrix all of whose entries equal the same nonzero constant

c

One. Every row is

c \cdot (1, 1, \dots, 1)

, so all rows are scalar multiples of one vector, giving a

1

-dimensional row space.

For an

m \times n

matrix with

rank (A) = n

(full column rank), what is

dim ker (A)

Zero, by the Rank–Nullity Theorem:

rank (A) + dim ker (A) = n

forces the nullity to vanish, so

A x = 0

has only the trivial solution.

A

m \times n

with

m < n

, can

A

have a trivial kernel?

No. Rank

\leq m < n

, so by rank–nullity

dim ker (A) = n - rank (A) \geq n - m > 0

; there is always a nonzero solution to

A x = 0

Can a nonsquare matrix be "full rank"?

Yes — full rank just means

rank = min (m, n)

; a

2 \times 5

matrix has full rank

2

if its two rows are independent, even though it is far from square.

Connections

Row Space and Column Space
Gaussian Elimination & Echelon Form
Rank–Nullity Theorem
Linear Independence and Basis
Matrix Multiplication as Linear Combination
Invertible Matrix Theorem