Invertible matrix theorem — 12+ equivalent conditions: Question bank

Definition Vocabulary this page assumes (built here so the bank is self-contained)

Square matrix: same number of rows as columns, written $n \times n$ . Only square matrices can be "invertible."
Invertible: there is a matrix $A^{- 1}$ with $A A^{- 1} = A^{- 1} A = I$ , the identity. Picture a machine you can always run backwards.
Null space $Nul A$ : all inputs $x$ with $A x = 0$ — the vectors the machine crushes to the origin. If only $x = 0$ is crushed, the null space is "trivial" ${0}$ .
Column space $Col A$ : every output you can build as a combination of $A$ 's columns — the set of $b$ that $A x = b$ can actually reach.
Pivot: after row reduction, a leading nonzero entry marking a "real" direction; a free column has no pivot and creates a free variable.
Rank: the number of pivots = dimension of the column space. Nullity = $dim Nul A$ . They obey rank–nullity: $rank A + dim Nul A = n$ .
Injective (one-to-one): different inputs never share an output. Surjective (onto): every possible output is reached.

Recall The IMT conditions this page cites (restated so you needn't leave the page)

For a square $n \times n$ matrix, these are all equivalent — the numbers match the parent IMT list:

(1) $A$ is invertible. (2) $A$ is row-equivalent to $I_{n}$ (RREF $= I$ ). (3) $A$ has $n$ pivots. (4) $A x = 0$ has only $x = 0$ (trivial null space). (5) columns are linearly independent. (6) the map $x \mapsto A x$ is injective. (7) $A x = b$ solvable for every $b$ . (8) columns span $R^{n}$ . (9) the map is surjective. (10) left inverse exists ( $C A = I$ ). (11) right inverse exists ( $A D = I$ ). (12) $A^{T}$ is invertible. (13) $det A \neq = 0$ . (14) $0$ is not an eigenvalue. (15) $rank A = n$ . (16) columns form a basis. (17) $dim Nul A = 0$ .

The two pictures below are the mental model behind almost every trap — keep glancing back at them.

Figure — Invertible matrix theorem — 12+ equivalent conditions

True or false — justify

Are the columns of a

5 \times 3

matrix being linearly independent enough to conclude it is invertible?

False — invertibility (and the whole IMT) is defined only for square matrices; a

5 \times 3

matrix isn't square, so "invertible" doesn't even apply to it. See Linear Independence and Span.

A

4 \times 4

and its columns are linearly independent, is

A

invertible?

True — for a square matrix, independent columns force a pivot in every column (condition 3), hence full rank

n

(15), hence RREF

= I

(2), hence invertibility (1); the chain 5 ⟹ 3 ⟹ 2 ⟹ 1 does the work.

Is it true that a matrix with a very small determinant like

1 0^{- 9}

is "almost not invertible"?

False as a theorem statement — invertibility is the strict condition

det A \neq = 0

(13), and

1 0^{- 9} \neq = 0

, so

A

is fully invertible; smallness is a conditioning issue (see the "condition number" note below), not an invertibility one.

For a square matrix, does

A D = I

(a right inverse only) already prove

A

is invertible?

True — for square

A

, a one-sided inverse is automatically two-sided (conditions 10 "

C A = I

" and 11 "

A D = I

" are both in the IMT), so

A D = I

forces

D A = I

Is it true that

A x = b

having a solution for some particular

b

tells you

A

is invertible?

False — condition 7 requires a solution for every

b

R^{n}

; one lucky

b

that happens to lie in the column space proves nothing about the others.

0

is an eigenvalue of

A

, is

A

non-invertible?

True — a zero eigenvalue means some

x \neq = 0

satisfies

A x = 0 x = 0

, a nonzero null vector, which breaks the trivial-null-space condition (4) and condition 14. See Eigenvalues and Eigenvectors.

Does

A^{T}

being invertible guarantee

A

is invertible?

True — condition 12; since

det A^{T} = det A

, one determinant is nonzero exactly when the other is.

Is a matrix whose columns span

R^{n}

but are linearly dependent possible for a square matrix?

False — for a square matrix, spanning

R^{n}

needs

n

pivots (condition 8 forces 3), and

n

pivots also force independence (5); you cannot separate the two properties. See Basis and Dimension.

Spot the error

A

3 \times 5

with rank

3

, so it spans

R^{3}

, so it is invertible." — where's the flaw?

Spanning

R^{3}

is fine, but "invertible" is undefined for a

3 \times 5

non-square matrix; the map is onto but not one-to-one (it has

5 - 3 = 2

free variables), so no two-sided inverse exists.

"Since

A x = 0

has the trivial solution

x = 0

, the columns are independent." — what's wrong?

Every homogeneous system has the trivial solution; independence (condition 5, equivalently 4) requires that the trivial solution be the only one. The speaker mislabels the always-true case.

"Row 2 of

A

equals row 1, so the columns are dependent." — is the reasoning sound?

The conclusion (not invertible) is correct, but skipping a step. Two equal rows make the rows dependent, so the row rank is below

n

. Because row rank always equals column rank (

rank A = rank A^{T}

), the column rank is also below

n

, so a column is free and the columns are dependent — that link, not "rows look like columns," is why it holds. See Rank and Nullity.

det A = 0

, so

A

has no eigenvalues." — fix it.

det A = 0

means

0

is an eigenvalue (condition 14 failing), not that eigenvalues are absent;

A

still has

n

eigenvalues (counting multiplicity) whose product is

det A = 0

A

is invertible, so its RREF has some free columns." — where's the mistake?

Invertible ⟹ RREF is exactly

I_{n}

(condition 2), which has a pivot in every column and therefore no free columns. See Elementary Matrices and Row Reduction.

C A = I

shows

C

is a left inverse, so

A

might still not be invertible." — correct for square

A

No — for square

A

, a left inverse (condition 10) is already in the IMT list, so

C A = I

proves

A

is fully invertible and

C = A^{- 1}

"The columns are dependent, hence

det A

is a small nonzero number." — fix it.

Dependent columns make

det A

exactly zero, not merely small; dependence and

det A \neq = 0

(condition 13) are mutually exclusive. See Determinant.

Why questions

Why does "one-to-one" (injective) coincide with "onto" (surjective) for square matrices but not in general?

Because for a square

A

n

pivots simultaneously deliver independence — injective (condition 6) — and spanning — onto (9); a non-square matrix can have at most one of these, so the two properties split apart. See Linear Transformations — Injective and Surjective.

Why is checking

det A \neq = 0

often preferred over computing

A^{- 1}

directly?

det A

(condition 13) is a single scalar test for the same yes/no question, while computing

A^{- 1}

is far more expensive; the IMT lets you pick the cheapest equivalent condition and know all the others follow.

Why does a free (non-pivot) column immediately kill invertibility?

A free column corresponds to a free variable, giving a nonzero solution to

A x = 0

; that nonzero null vector breaks condition 4 and hence every IMT condition. See Rank and Nullity.

Why does

A \sim I

(row equivalent to the identity) imply

A

is invertible?

Every row operation is the same as multiplying on the left by an elementary matrix

E_{i}

— the identity with that one operation applied, and each

E_{i}

is invertible because every row operation can be undone (swaps swap back, scalings divide, additions subtract). If

E_{k} \dots E_{1} A = I

then

B = E_{k} \dots E_{1}

is a product of invertibles (hence invertible) and

B A = I

, so

B = A^{- 1}

(condition 2 ⟹ 1). See Elementary Matrices and Row Reduction.

Why is

dim Nul A = 0

equivalent to full rank for a square matrix?

By rank–nullity,

rank A + dim Nul A = n

; so nullity

0

(condition 17) forces rank

n

(15), and vice versa. See Rank and Nullity.

What does "condition number" mean, and why is a tiny determinant not the same as being non-invertible?

The condition number measures how much a small wiggle in

b

(or in

A

) can amplify into a large wiggle in the solution

x

A x = b

— a "sensitivity to error" gauge, not a yes/no property. A matrix can be perfectly invertible (

det \neq = 0

) yet ill-conditioned (huge condition number), so solving it numerically is shaky even though the theorem happily calls it invertible.

Edge cases

Is the

1 \times 1

matrix

[0]

invertible?

No — its determinant is

0

, its single column is the zero vector (dependent), and it maps everything to

0

; every IMT condition fails.

Is the zero matrix

0_{n \times n}

ever invertible?

Never — its columns are all

0

, so

det = 0

, rank

= 0

, and its null space is all of

R^{n}

; it fails the IMT for every

n \geq 1

Does the IMT apply to a

0 \times 0

(empty) matrix?

By convention the empty matrix is invertible with

det = 1

, since it vacuously maps the trivial space

{0}

bijectively to itself — a harmless boundary case rather than a real trap.

If a square

A

has two identical columns, what does the IMT say instantly?

Two equal columns are dependent, so condition 5 fails; therefore

det A = 0

(13), rank

< n

(15),

Nul A \neq = {0}

(4/17),

0

is an eigenvalue (14), and

A

is not invertible — all at once.

Can a square matrix be onto (

Col A = R^{n}

) yet fail to be one-to-one?

No — for a square matrix onto forces

n

pivots, which also forces trivial null space; the two cannot be separated, unlike in the non-square world.

Recall The single question behind every item

Almost every trap above is really one of two errors: (1) forgetting the matrix must be square before invoking the IMT, or (2) confusing "there exists" with "for all" — one solution versus every solution, one $b$ versus every $b$ . Ask "is it square?" and "am I quantifying over one case or all cases?" and most traps disarm themselves.