Null space (kernel) and column space (image) — basis, dimension: Question bank

True or false — justify

For a matrix $A \in R^{m \times n}$ (that is: $m$ rows, $n$ columns — $n$ inputs, $m$ outputs), decide and say why.

The column space of

A

and the column space of its RREF are always equal.

False. Row operations scramble entries within each column, so they move the actual output vectors; only the pattern of dependence (which columns are pivots) is preserved, not the vectors themselves.

A

3 \times 5

then its null space contains vectors with

5

components.

True. Null space vectors are inputs

x

, and

A

has

5

columns, so

x

needs

5

entries — one to multiply each column.

dim Null (A) = 0

then

A

must be a square matrix.

False. It only forces full column rank (

r = n

), which can happen with

m > n

(a tall matrix), e.g. Example 2 in the parent note is

3 \times 2

with trivial kernel.

Adding a column to

A

can never decrease its rank.

True. The old columns are still present, so their span is still reachable; a new column can only add a direction or be redundant — never remove a dimension already there.

The nullity equals the number of zero rows in the RREF.

False. Zero rows count dependent rows (this is

m - r

, not

n - r

); nullity counts free columns

= n - r

. These coincide only by accident when

m = n

If every column of

A

is a pivot column, the map

x \mapsto A x

is injective.

True. Full column rank means no free variables, so

A x = 0

forces

x = 0

; distinct inputs give distinct outputs.

Col (A)

and

Null (A)

are subspaces of the same space.

False in general.

Col (A) \subseteq R^{m}

(outputs) and

Null (A) \subseteq R^{n}

(inputs); they only live in the same space when

m = n

, and even then they need not overlap.

rank (A) = m

(full row rank), then

A x = b

is solvable for every

b

True. Full row rank means

Col (A)

fills all of

R^{m}

, so every output

b

is reachable — the map is surjective.

Two different matrices can have exactly the same null space and the same column space.

True. E.g.

A

and

2 A

have identical column and null spaces; scaling never changes which directions are reachable or crushed.

Spot the error

Each item states a piece of reasoning. Find the flaw.

A

4 \times 6

with rank

4

, so nullity

= 4 - 6 = - 2

Wrong sign and wrong subtraction: nullity

= n - r = 6 - 4 = 2

, always

n

(columns) minus rank, never

m - r

. Nullity can never be negative.

"To get a Col(A) basis I row-reduced and took the pivot columns of the RREF."

The positions are right but the vectors are wrong. Take those pivot columns from the original

A

; the RREF columns no longer point in the correct output directions.

A x = 0

has only the trivial solution, so

A

is invertible."

Only valid if

A

is square. Trivial kernel gives injectivity (

r = n

); invertibility additionally needs

r = m = n

, i.e. a square matrix of full rank.

"There are

3

free variables, so I write

3

special solutions, then a

4

th for the pivot variable."

Pivot variables never get their own special solution — they are determined by the free ones. The number of basis vectors equals the number of free variables exactly, here

3

"The columns of

A

are dependent, so the null space is all of

R^{n}

Dependence only guarantees at least one free variable, so

dim Null \geq 1

; it equals

n

only if

A

is the zero matrix (every column free).

b

is not a column of

A

, so

A x = b

has no solution."

b

need not be a literal column — it must lie in the span of the columns, i.e. in

Col (A)

. A combination of columns can reach

b

even if

b

isn't one of them. See Solving Ax=b.

"Row rank and column rank might differ, so I'll compute both to be safe."

They are always equal — that single number is the rank. This is a theorem (Rank of a matrix), not a coincidence, so one computation suffices.

Why questions

Why do we read Col(A) from the original matrix but Null(A) from the reduced one?

Row operations preserve the solution set of

A x = 0

(so the null space is unchanged and safe to read from RREF), but they distort column entries (so Col must come from the untouched original). Different spaces, different rules.

Why must

dim Col (A) + dim Null (A)

equal

n

and not

m

It's a count of columns: each of the

n

columns is either a pivot (counted in rank) or free (counted in nullity), so the two dimensions partition the

n

inputs. See Rank–Nullity Theorem.

Why does a free variable produce exactly one basis vector for the null space?

Setting that free variable to

1

and all other free variables to

0

pins down every pivot variable uniquely, giving one specific solution; doing this per free variable yields independent vectors that span all solutions.

Why can the column space never be "curved" or shifted off the origin?

Because

A \cdot 0 = 0

(so it contains the origin) and

A (α x + β y) = α A x + β A y

(closed under combinations); linearity forces a flat sheet through

0

— a subspace, never an affine or bent surface.

Why does full column rank (

r = n

) make the map injective but not necessarily surjective?

r = n

kills the null space (injective), but if

m > n

the column space is an

n

-dimensional slab inside a bigger

R^{m}

, so most outputs are unreachable (not surjective).

Why is knowing pivot positions enough to name the independent columns, even though RREF changed the values?

The dependence relations among columns are exactly the vectors in the null space, and the null space is unchanged by row reduction — so RREF reports the same independence pattern the original columns had.

Edge cases

The zero matrix

0_{m \times n}

: what are its column space, null space, rank, nullity?

Col = {0} \subseteq R^{m}

with

dim 0

, so rank

0

;

Null = R^{n}

(everything is crushed) with nullity

n

. Check:

0 + n = n

. ✓

A single column matrix

A \in R^{m \times 1}

that is nonzero: describe both spaces.

Col (A)

is the line spanned by that one column (dim

1

, rank

1

);

Null (A) = {0} \subseteq R^{1}

since

x

is a single scalar and

A x = 0

forces

x = 0

(nullity

0

A wide matrix with more columns than rows (

n > m

): can its null space ever be trivial?

No. Rank is at most

min (m, n) = m < n

, so nullity

= n - r \geq n - m > 0

; there is always at least one free variable, hence a nonzero vector in the kernel.

A matrix whose only nonzero entries are on the diagonal, one of them zero (e.g.

diag (3, 0, 5)

): what's the nullity?

Nullity

1

— the second column is a zero column (free variable

x_{2}

), giving the special solution

(0, 1, 0)^{⊤}

; the other two diagonal entries are pivots, so rank

2

A

is square with

det A \neq = 0

: state Col, Null, rank, nullity in one breath.

Col (A) = R^{n}

(surjective),

Null (A) = {0}

(injective), rank

= n

, nullity

= 0

— the map is a bijection, i.e. invertible.

b = 0

, how many solutions does

A x = b

have, and how does that relate to the null space?

Exactly the null space itself is the full solution set — the particular solution can be taken as

0

, so all solutions are

0 + Null (A)

. If nullity

> 0

there are infinitely many; if nullity

= 0

, only

x = 0

Connections

Rank of a matrix — the single number both bases hinge on.
Rank–Nullity Theorem — the counting identity every trap here tests.
Row reduction & RREF — why positions transfer but values don't.
Linear independence and basis — pivots $\Leftrightarrow$ independent columns.
Injective and surjective linear maps — trivial kernel vs full row rank.
Four fundamental subspaces — where each space lives.
Solving Ax=b — solvability lives in the column space.

#flashcards/maths