5.1 Invertible matrices

Invertibility is one of the first places where linear algebra becomes more than a procedure for solving one system. A square matrix is invertible exactly when it can be undone by another square matrix, and that idea turns out to be equivalent to many other statements: row reduction to the identity, consistency of every system $Ax = b$ , linear independence of the columns, and the ability to write every vector as a linear combination of those columns.

This note develops those equivalences carefully. The goal is not only to know what the symbol $A^{-1}$ means, but also to recognize when it exists and how to use it without guessing.

Before you start

Invertibility uses several earlier ideas at the same time. Before reading this section, you should be comfortable with the following habits.

Matrix multiplication is order-sensitive: in general AB and BA are different expressions.
The identity matrix $I_p$ is the matrix that leaves every compatible vector unchanged.
Row reduction is a controlled sequence of reversible row operations when each elementary operation is legal.
The homogeneous system $Ax = 0$ detects hidden freedom in the columns of $A$ .
A system $Ax = b$ has a unique solution exactly when there is no free variable after reduction.

The main new idea is that these earlier facts are not isolated. For a square matrix, they collapse into one theorem package: being invertible, row-reducing to the identity, having no nonzero null-space vector, and solving every system uniquely all say the same thing in different languages.

Left and right inverses

Before the square case, it is useful to separate two one-sided notions.

Definition

Left inverse and right inverse

Let $A$ be a $p \times q$ matrix.

A $q \times p$ matrix $H$ is a left inverse of $A$ if $HA = I_q$ .
A $q \times p$ matrix $G$ is a right inverse of $A$ if $AG = I_p$ .

These definitions matter because matrix multiplication is not commutative. For rectangular matrices, a left inverse and a right inverse need not both exist. The square case is special.

The dimensions also explain the names. If $H$ is a left inverse, then $H$ sits on the left of $A$ in $HA = I_q$ . If $G$ is a right inverse, then $G$ sits on the right of $A$ in $AG = I_p$ . These are different requirements because the two products have different sizes and different meanings.

Common mistake

Do not treat one-sided inverse statements as symmetric

From $HA = I_q$ you cannot simply reverse the order and conclude $AH = I_p$ . Matrix multiplication has direction. The square invertible case is powerful precisely because both directions hold at once.

Definition

Invertible matrix

Let $A$ be a $p \times p$ square matrix. We say that $A$ is invertible if there exists a $p \times p$ matrix $B$ such that

BA = AB = I_p.

The matrix $B$ is called the inverse of $A$ , and we write $B = A^{-1}$ .

Theorem

The inverse is unique

If $B$ is a left inverse of $A$ and $C$ is a right inverse of $A$ , then $B = C$ . So an invertible matrix has exactly one inverse.

Proof

Why the inverse is unique

What invertibility means

Invertibility is a reversibility statement. Applying $A$ changes a vector, but if $A$ is invertible then $A^{-1}$ undoes that change exactly.

That is why the identity matrix appears in the definition. The identity matrix does nothing:

I_p x = x

for every compatible vector x. An inverse is precisely a matrix that brings you back to that unchanged state.

Worked example

A diagonal matrix is easy to invert

Let

D = \operatorname{diag}(2, -1, 3).

Then

D^{-1} = \operatorname{diag}\left(\tfrac{1}{2}, -1, \tfrac{1}{3}\right).

This works because each diagonal entry is replaced by its reciprocal, and the off-diagonal zeros stay zero. Multiplying $D$ by $D^{-1}$ gives $I_3$ .

Row reduction and the inverse

The most practical way to test invertibility is to row-reduce. The key point has two parts:

row-operation matrices are invertible, with inverse given by the reverse row operation;
a square matrix is invertible exactly when it can be row-reduced to $I_p$ .

Theorem

Row-operation matrices are invertible

If $\rho$ is a row operation on matrices with p rows, and $\bar{\rho}$ is the reverse row operation, then the corresponding row-operation matrices $M[\rho]$ and $M[\bar{\rho}]$ satisfy

M[\rho]^{-1} = M[\bar{\rho}], \qquad M[\bar{\rho}]^{-1} = M[\rho].

This gives a clean interpretation of row reduction: every row operation is actually multiplication on the left by an invertible matrix.

Theorem

Invertibility and row reduction

For a square matrix $A$ , the following are equivalent:

$A$ is invertible.
$A$ is row-equivalent to $I_p$ .
$A$ is a product of $p \times p$ row-operation matrices.
$A$ is nonsingular.

The practical consequence is very concrete: if row operations transform $A$ to $I_p$ , then those same operations, applied to $[A | I_p]$ , transform it to $[I_p | A^{-1}]$ .

Read and try

Follow one inverse-by-row-reduction example

The live demo lets you step through [A | I] until the left block becomes I.

Start from [A | I]. If A is invertible, row reduction will turn the left block into I.

1	2	1	1	0	0
0	1	1	0	1	0
2	3	4	0	0	1

The live demo above is the shortest way to see the logic. It is not the definition. It is the computational method that matches the definition.

Read and try

Trace one full row-reduction path

The live stepper walks through one complete elimination path, showing the row operation, the pivot you are focusing on, and the matrix produced at each step.

1	2	2	4
1	3	3	5
2	6	5	6

Row operation

Choose the first pivot in column 1.

What to notice

Column 1 already has a convenient pivot 1 in the first row, so we do not need a row swap.

Start with the augmented matrix. The first pivot should help us clear the entries underneath it.

The second widget shows the shape of a full elimination path. In an invertible case, the left block eventually becomes $I_p$ , and that is the moment when the right block becomes the inverse.

Equivalent formulations

Invertibility is useful because it has a dictionary of equivalent conditions. This is the main bridge between algebra, row reduction, and systems of linear equations.

Theorem

Equivalent ways to recognize invertibility

Let $A$ be a $p \times p$ matrix. Then the following statements are equivalent:

$A$ is invertible.
$A$ is row-equivalent to $I_p$ .
$A$ is a product of row-operation matrices.
$A$ has a left inverse.
$A$ has a right inverse.
$A$ is nonsingular.
For every column vector b with p entries, the system $Ax = b$ is consistent.
For every column vector b with p entries, the system $Ax = b$ has the unique solution $x = A^{-1}b$ .

Two of these statements are especially important in practice.

Statement 7 says that the columns of $A$ span $R^p$ .
Statement 8 says that invertibility gives you a complete solution formula, not just existence.

That is why invertibility is the exact algebraic condition behind solving a linear system by a matrix inverse.

How to use the invertibility dictionary

The equivalent statements above are not meant to be memorized as a long list. They are a toolkit. In a problem, choose the statement that is closest to the information you already have.

If you are given a row-reduction computation, read pivot positions: a square matrix is invertible exactly when every column is a pivot column.
If you are given $A^{-1}$ or a one-sided inverse for a square matrix, use algebraic multiplication with the identity matrix.
If you are given a nonzero vector v with $Av = 0$ , use the null-space condition to prove that $A$ is not invertible.
If you are asked about all systems $Ax = b$ , translate the question into "does every b have exactly one preimage under $A$ ?"

Worked example

A nonzero null-space vector proves non-invertibility

Suppose $A$ is a square matrix and

v = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \ne 0, \qquad Av = 0.

Then $A$ cannot be invertible. Indeed, if $A^{-1}$ existed, multiplying $Av = 0$ on the left by $A^{-1}$ would give

A^{-1}Av = A^{-1}0.

The left side is v, while the right side is 0. This would force $v = 0$ , contradicting the given vector. Therefore the existence of one nonzero solution of $Ax = 0$ already proves that $A$ is singular.

Worked example

Use an inverse to solve a system

Suppose

A^{-1} = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix}, \qquad b = \begin{bmatrix} 5 \\ -1 \end{bmatrix}.

If $Ax = b$ , then multiply both sides by $A^{-1}$ :

x = A^{-1}b = \begin{bmatrix} 2 & -1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ -1 \end{bmatrix} = \begin{bmatrix} 11 \\ -3 \end{bmatrix}.

The inverse is not just a symbol for undoing $A$ ; it is an explicit formula for the unique solution.

Row-equivalence through invertible matrices

We can push the row-operation viewpoint one step further. Instead of thinking about row-equivalence as a long list of elementary moves, package the whole list into one invertible matrix on the left.

Theorem

Row-equivalence is left multiplication by an invertible matrix

Suppose $A$ and $B$ are matrices with p rows. Then the following are equivalent:

$A$ and $B$ are row-equivalent.
There exists an invertible $p \times p$ matrix $G$ such that

B = GA.

Moreover, once $B = GA$ , we also have

A = G^{-1}B.

This theorem is not a new computational trick. It is a cleaner language for the same phenomenon. A sequence of row operations can always be compressed into one invertible matrix $G$ , and the reverse row operations are encoded by $G^{-1}$ .

Worked example

Reading a row-equivalence as one matrix equality

Let

A = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}, \qquad G = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.

The matrix $G$ is the row-operation matrix for the move

R_2 \leftarrow R_2 - R_1.

GA = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 2 \\ 0 & 1 & 1 \end{bmatrix}.

If we call this new matrix $B$ , then $B = GA$ . That single equation records the entire row operation. Since $G$ is invertible, $A$ and $B$ are row-equivalent.

The gain is conceptual. Once you know that row-equivalence means multiplication by an invertible matrix on the left, you can explain many invariants in one line instead of by repeating row-operation arguments.

Theorem

Row operations preserve linear relations among corresponding columns

Let $A$ and $B$ be row-equivalent $p \times q$ matrices, and write their columns as

A = [a_1 \; a_2 \; \cdots \; a_q], \qquad B = [b_1 \; b_2 \; \cdots \; b_q].

a_j = \alpha_1 a_{k_1} + \alpha_2 a_{k_2} + \cdots + \alpha_n a_{k_n},

then

b_j = \alpha_1 b_{k_1} + \alpha_2 b_{k_2} + \cdots + \alpha_n b_{k_n}.

In particular, linear dependence and linear independence among corresponding columns are preserved by row-equivalence.

The proof is short once $B = GA$ is known. Multiply the relation for the columns of $A$ by $G$ . Because matrix multiplication is linear,

Ga_j = \alpha_1 G a_{k_1} + \alpha_2 G a_{k_2} + \cdots + \alpha_n G a_{k_n},

which is exactly the corresponding relation amongst the columns of $B$ .

This is the bridge from row reduction to column language. Row operations change the actual columns, but they do not change which columns are redundant or which column relations are forced by the others.

Why the reduced row-echelon form is unique

The reduced row-echelon form in a row-equivalence class is unique. That fact is easy to overlook, but it is what makes later definitions mathematically legitimate.

Theorem

A row-equivalence class has exactly one reduced row-echelon form

Suppose $A$ is a matrix, and suppose $B$ and $C$ are both reduced row-echelon forms. If $B$ is row-equivalent to $A$ and $C$ is row-equivalent to $A$ , then

B = C.

A standard proof uses induction on the rank. The basic strategy is:

compare the pivot columns from left to right,
use preserved linear relations to force the same pivot positions, and then
show that every free column must have the same coefficients in terms of the pivot columns.

So reduced row-echelon form is not merely a convenient final answer. It is the final answer inside a row-equivalence class.

Definition

Rank

The rank of a matrix is the number of pivots in its reduced row-echelon form.

This definition works only because the reduced row-echelon form is unique. If different reduction paths could produce different reduced forms with different numbers of pivots, then rank would depend on the calculation. The uniqueness theorem rules that out.

Column independence and linear combinations

Invertibility can also be read directly from the columns.

Theorem

Invertibility and the columns of a square matrix

For a $p \times p$ matrix $A$ , the following are equivalent:

$A$ is invertible.
The columns of $A$ are linearly independent.
Every column vector in $R^p$ is a linear combination of the columns of $A$ .

These are not separate facts. They are three ways of reading the same structural statement.

If the columns are linearly independent, then no column is redundant. If they span $R^p$ , then every target vector can be built from them. For a square matrix, those two conditions coincide exactly when the matrix is invertible.

Why the transpose also matters

Invertibility behaves well under transpose.

Theorem

Transpose and powers

If $A$ is invertible, then:

$A^t$ is invertible, and $(A^t)^{-1} = (A^{-1})^t$ .
$A^n$ is invertible for every integer n, and $(A^n)^{-1} = (A^{-1})^n$ .

The transpose result is useful when you want to turn a statement about columns into a statement about rows. The power rule is useful when a repeated transformation appears in a calculation.

Products that equal the identity

Some practice problems do not hand you a matrix and ask for an inverse directly. Instead, they give a product such as

ABCD=I

and ask which shorter products or cyclic reorderings are forced to be invertible. The correct method is not to commute matrices. It is to group the product carefully and use one-sided inverse information for square matrices.

Theorem

A one-sided identity is enough in the square case

Let $P$ and $Q$ be square matrices of the same size. If

PQ=I,

then both $P$ and $Q$ are invertible, and

Q=P^{-1},\qquad QP=I.

This theorem is one reason the square case is special. For rectangular matrices, a one-sided inverse may fail to be a two-sided inverse. For square matrices in this course, the invertibility dictionary lets us promote a one-sided identity to a genuine inverse relation.

Worked example

Cyclic identities from $ABCD=I$

Suppose A,B,C,D are $5 \times 5$ matrices and

ABCD=I_5.

Group the product as

(ABC)D=I_5.

By the one-sided identity theorem, ABC and $D$ are inverses of each other. Therefore

DABC=I_5.

Next group the same original product cyclically:

(DAB)C=I_5 \qquad\text{and}\qquad (CDA)B=I_5,

after using the already justified cyclic identities. This gives

CDAB=I_5, \qquad BCDA=I_5.

So BCDA, CDAB, and DABC are forced to equal $I_5$ . Arbitrary reorderings such as DCBA or DBAC are not forced by $ABCD=I_5$ ; matrix multiplication still does not commute.

The safe workflow is:

keep the given order unless a theorem justifies changing it;
group adjacent factors to form a product $PQ=I$ ;
use square invertibility to reverse that grouped product;
repeat only with identities already proved.

How to approach invertibility problems

When a question asks whether a matrix is invertible, do not start by trying random inverse entries. First decide which evidence is cheapest.

Check that the matrix is square. Non-square matrices do not have a two-sided inverse in the sense of this section.
If entries are given, row-reduce and look for a pivot in every column.
If a determinant is already available later in the course, use $det(A) \ne 0$ ; before determinants, use pivots, rank, or null space.
If a candidate inverse is given, multiply in the required order and verify the identity matrix.
If a nonzero null-space vector or a column relation is given, conclude non-invertibility immediately.

This approach keeps the work tied to the theorem package. The point is not to perform the longest possible computation; it is to choose a valid equivalent condition and apply it cleanly.

Worked example

Find an inverse by row reduction

Let

A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}.

Start from $[A | I_2]$ :

\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 5 & 0 & 1 \end{array}\right].

Eliminate the entry below the first pivot:

\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -1 & -3 & 1 \end{array}\right].

Now scale the second row and clear the entry above the second pivot:

\left[\begin{array}{cc|cc} 1 & 0 & -5 & 2 \\ 0 & 1 & 3 & -1 \end{array}\right].

A^{-1} = \begin{bmatrix} -5 & 2 \\ 3 & -1 \end{bmatrix}.

The computation is not the point by itself. The point is that the right block of the augmented matrix records the inverse because the left block has been driven to the identity.

Practice-style inverse computations

In extended practice, inverse questions are often less tidy than the first $2 \times 2$ example. A matrix may contain a parameter, or the row reduction may be shown only in compressed stages. The underlying rule is unchanged:

[A \mid I_p] \sim [I_p \mid B] \quad\Longrightarrow\quad B = A^{-1}.

If the left block cannot be reduced to $I_p$ , then the matrix is not invertible. If the left block becomes $I_p$ , the right block is not an auxiliary calculation; it is the inverse.

Worked example

A parameterized inverse from row reduction

For a real number $\alpha$ , let

A_\alpha = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 1 & 3 & 1 & 2 \\ 0 & 0 & 1 & \alpha \\ 1 & 2 & 1 & 3 \end{bmatrix}.

Row-reducing the augmented matrix $[A_\alpha \mid I_4]$ gives

\left[\begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 5-\alpha & -2 & -1 & \alpha-2 \\ 0 & 1 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & \alpha & 0 & 1 & -\alpha \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right].

Since the left block is $I_4$ for every real $\alpha$ , the matrix $A_\alpha$ is invertible for every real $\alpha$ , and

A_\alpha^{-1} = \begin{bmatrix} 5-\alpha & -2 & -1 & \alpha-2 \\ -1 & 1 & 0 & 0 \\ \alpha & 0 & 1 & -\alpha \\ -1 & 0 & 0 & 1 \end{bmatrix}.

Notice what this conclusion does and does not say. The parameter remains in the inverse; it is not a value to be solved for. The row reduction has shown that no value of $\alpha$ creates a missing pivot in the left block.

Common mistake

Do not treat every parameter as a restriction

Sometimes a parameter appears because the answer is a family of inverses, not because the matrix fails for special values. The restriction comes from a missing pivot or an illegal division by zero, not from the mere presence of a symbol.

Reading supplied row-reduction tables

Longer inverse questions often give only parts of a row-reduction chain. That is still enough information if you read the chain structurally. The important question is always:

What is the left block, and what does that force the right block to mean?

If a sequence of row operations sends

[A \mid I_p] \quad\text{to}\quad [I_p \mid D],

then $A$ is invertible and $D=A^{-1}$ . If the final left block is merely an echelon form with a missing pivot column, then $A$ is not invertible, no matter how complicated the right block looks.

Worked example

Recovering $A$ and solving a transposed system from a reduction table

Suppose a reduction table shows that $[A \mid I_4]$ is row-equivalent to $[I_4 \mid D]$ , where

D= \begin{bmatrix} 3 & -2 & 3 & -2 \\ 0 & 1 & -2 & 1 \\ -3 & 1 & -2 & 2 \\ 1 & 0 & 1 & -1 \end{bmatrix}.

Then the row-reduction result itself already says

A^{-1}=D.

If the earlier part of the table lets us reverse the first row operations, the original left block can be recovered as

A= \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 3 \\ 2 & 1 & 3 & 3 \\ 3 & 2 & 4 & 3 \end{bmatrix}.

Now let

g= \begin{bmatrix} 1 \\ 0 \\ 3 \\ 0 \end{bmatrix}.

To solve $A^t x=g$ , do not row-reduce again from the beginning. Since $A$ is invertible, $A^t$ is also invertible and

(A^t)^{-1}=(A^{-1})^t=D^t.

Therefore

x=(A^t)^{-1}g=D^t g = \begin{bmatrix} -6 \\ 1 \\ -3 \\ 4 \end{bmatrix}.

The computation is short because the reduction table has already done the hard work.

Worked example

Four inverse computations read by pivot columns

For each matrix below, the method is the same: row-reduce $[A \mid I_4]$ , look only at the pivot pattern of the left block first, and read the inverse from the right block only if the left block becomes $I_4$ .

Case (a). For

A_1= \begin{bmatrix} 1 & 0 & 3 & 3 \\ 2 & 1 & 2 & 3 \\ -1 & 3 & 0 & 3 \\ 4 & 2 & 2 & 4 \end{bmatrix},

one echelon stage has left block

A_1^\sharp= \begin{bmatrix} 1 & 0 & 3 & 3 \\ 0 & 1 & -4 & -3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

There is no pivot in the fourth column. Hence $A_1$ is not invertible.

Case (b). For

A_2= \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 3 \\ 2 & 1 & 3 & 3 \\ 3 & 2 & 4 & 3 \end{bmatrix},

the reduction reaches $[I_4 \mid A_2^{-1}]$ , with

A_2^{-1}= \begin{bmatrix} 3 & -2 & 3 & -2 \\ 0 & 1 & -2 & 1 \\ -3 & 1 & -2 & 2 \\ 1 & 0 & 1 & -1 \end{bmatrix}.

Case (c). For

A_3= \begin{bmatrix} 1 & 2 & -1 & 30 \\ 3 & 2 & 0 & 15 \\ 0 & 1 & 3 & 15 \\ 1 & 1 & 1 & 10 \end{bmatrix},

one echelon stage has left block

A_3^\sharp= \begin{bmatrix} 1 & 2 & -1 & 30 \\ 0 & 1 & 3 & 15 \\ 0 & 0 & 5 & -5 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

Again a pivot column is missing, so $A_3$ is not invertible.

Case (d). For

A_4= \begin{bmatrix} 1 & 1 & 2 & 1 \\ 2 & 3 & 4 & 1 \\ 3 & 3 & 3 & 1 \\ 1 & 2 & 3 & 1 \end{bmatrix},

the reduction reaches $[I_4 \mid A_4^{-1}]$ , with

A_4^{-1}= \begin{bmatrix} 1 & 1 & 0 & -2 \\ -2 & -2 & 1 & 3 \\ 1 & 2 & -1 & -2 \\ 0 & -3 & 1 & 3 \end{bmatrix}.

This example is deliberately repetitive. The repetition is the point: in every case, the left block decides invertibility before the right block is interpreted as an inverse.

Worked example

A final identity block controls the whole chain

Suppose a row-operation chain carries $[A \mid I_5]$ through an intermediate matrix $[B \mid C]$ and finally to $[D \mid E]$ . After the last few operations, the final augmented matrix is

[D \mid E] = \left[ \begin{array}{ccccc|ccccc} 1 & 0 & 0 & 0 & 0 & 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 9 & 1 & -18 & -3 & -2 \\ 0 & 0 & 1 & 0 & 0 & -6 & 2 & 11 & 2 & 0 \\ 0 & 0 & 0 & 1 & 0 & -3 & 0 & 6 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -6 & 0 & 12 & 2 & 1 \end{array} \right].

The left block is $I_5$ , so $D=I_5$ and

A^{-1}=E= \begin{bmatrix} 1 & 0 & -2 & 0 & 0 \\ 9 & 1 & -18 & -3 & -2 \\ -6 & 2 & 11 & 2 & 0 \\ -3 & 0 & 6 & 1 & 0 \\ -6 & 0 & 12 & 2 & 1 \end{bmatrix}.

It follows immediately that $A^t$ is invertible and

(A^t)^{-1}=E^t = \begin{bmatrix} 1 & 9 & -6 & -3 & -6 \\ 0 & 1 & 2 & 0 & 0 \\ -2 & -18 & 11 & 6 & 12 \\ 0 & -3 & 2 & 1 & 2 \\ 0 & -2 & 0 & 0 & 1 \end{bmatrix}.

The same chain can also be read as one left multiplication. Let $H$ be the product of all row-operation matrices in the chain. Since $HA=I_5$ and $HI_5=E$ , we have $H=A^{-1}$ . Therefore, if the same row operations carry $F+I_5$ to $A^3+3A+I_5$ , then

A^3+3A+I_5=A^{-1}(F+I_5).

Multiplying by $A$ on the left gives

F+I_5=A(A^3+3A+I_5)=A^4+3A^2+A,

F=A^4+3A^2+A-I_5.

The lesson is that a row-reduction table is not just numerical work. It also records an invertible matrix multiplying on the left.

Proof practice without determinants

Some proof exercises deliberately ask for invertibility arguments that do not use determinants. That is good practice: many inverse facts are really statements about products, identities, and null spaces.

Worked example

A vector identity that proves invertibility

Suppose $A$ is a $7 \times 7$ matrix and

A^2x = Ax + x

for every vector $x \in R^7$ . Move the terms involving $A$ to the left:

(A^2-A)x = x \qquad\text{for every }x.

Since $A^2-A=(A-I_7)A$ , this says

((A-I_7)A)x = I_7x \qquad\text{for every }x.

Two matrices that agree on every vector are the same matrix, so

(A-I_7)A = I_7.

Thus $A-I_7$ is a left inverse of $A$ . Because $A$ is square, a one-sided identity is enough: $A$ is invertible, and its inverse is $A-I_7$ .

Worked example

A contradiction proof for non-invertibility

Suppose $A$ and $B$ are distinct $6 \times 6$ matrices, and

A^3=B^3, \qquad A^2B=B^2A.

We prove that $A^2+B^2$ is not invertible. Let

C=A^2+B^2.

First compare CB and CA:

CB=(A^2+B^2)B=A^2B+B^3,

while

CA=(A^2+B^2)A=A^3+B^2A.

Using the assumptions $A^2B=B^2A$ and $B^3=A^3$ , we obtain

CB=CA.

If $C$ were invertible, multiplying by $C^{-1}$ on the left would give $B=A$ , contradicting the assumption that $A$ and $B$ are distinct. Therefore $C=A^2+B^2$ is not invertible.

Common mistakes

Common mistake

Do not confuse one-sided inverses in the rectangular case

For a non-square matrix, having a left inverse does not automatically mean it has a right inverse. The square case is special: once an inverse exists, it is both a left inverse and a right inverse, and it is unique.

Common mistake

Do not guess invertibility from appearance

A matrix can look simple and still fail to be invertible. The correct test is to row-reduce it, or to use one of the equivalent conditions above.

Common mistake

Use the square one-sided identity theorem.

Solution

Answer

Exercise

Quick check

Suppose $A$ is invertible and $AB = I_p$ . Prove that $B = A^{-1}$ .

Use the fact that the inverse of $A$ is unique.

Solution

Guided solution

Quick check

Suppose $(A-B)A=I$ for square matrices $A$ and $B$ . Prove that $A$ and $B$ commute.

First use the square one-sided identity theorem, then expand $AB-BA$ .

Solution

Guided solution

Read this first

This page depends especially on 2.3 Gaussian elimination and RREF, 3.1 Matrix multiplication and identity matrices, and 3.2 Transpose and special matrices.

5.1 Invertible matrices

MATH1030: Linear algebra I

Before you start

Left and right inverses

Left inverse and right inverse

Do not treat one-sided inverse statements as symmetric

Invertible matrix

The inverse is unique

Why the inverse is unique

What invertibility means

A diagonal matrix is easy to invert

Row reduction and the inverse

Row-operation matrices are invertible

Invertibility and row reduction

Follow one inverse-by-row-reduction example

Trace one full row-reduction path

Equivalent formulations

Equivalent ways to recognize invertibility

How to use the invertibility dictionary

A nonzero null-space vector proves non-invertibility

Use an inverse to solve a system

Row-equivalence through invertible matrices

Row-equivalence is left multiplication by an invertible matrix

Reading a row-equivalence as one matrix equality

Row operations preserve linear relations among corresponding columns

Why the reduced row-echelon form is unique

A row-equivalence class has exactly one reduced row-echelon form

Rank

Column independence and linear combinations

Invertibility and the columns of a square matrix

Why the transpose also matters

Transpose and powers

Products that equal the identity

A one-sided identity is enough in the square case

Cyclic identities from ABCD=IABCD=IABCD=I

How to approach invertibility problems

Worked example

Find an inverse by row reduction

Practice-style inverse computations

A parameterized inverse from row reduction

Do not treat every parameter as a restriction

Reading supplied row-reduction tables

Recovering AAA and solving a transposed system from a reduction table

Four inverse computations read by pivot columns

A final identity block controls the whole chain

Proof practice without determinants

A vector identity that proves invertibility

A contradiction proof for non-invertibility

Common mistakes

Do not confuse one-sided inverses in the rectangular case

Do not guess invertibility from appearance

Do not permute factors without justification

Quick checks

If AAA is invertible, what is A−1AA^{-1}AA−1A?

Answer

If AAA is invertible, can the homogeneous system Ax=0Ax = 0Ax=0 have a nonzero solution?

Answer

If AAA is invertible, is AtA^tAt invertible?

Answer

If B=GAB = GAB=GA with GGG invertible and the columns of AAA satisfy a3=2a1−a2a_3 = 2a_1 - a_2a3​=2a1​−a2​, what relation must hold among the columns of BBB?

Answer

Why does uniqueness of RREF matter when defining rank?

Answer

Suppose v≠0v \ne 0v=0 and Av=0Av = 0Av=0. Can AAA be invertible?

Answer

Suppose A,B,C,D are square matrices and ABCD=IABCD=IABCD=I. Which cyclic identity is forced: DABC=IDABC=IDABC=I or DCBA=IDCBA=IDCBA=I?

Answer

In the parameterized inverse above, what is the (1,4) entry of Aα−1A_\alpha^{-1}Aα−1​ when α=3\alpha=3α=3?

Answer

If A2x=Ax+xA^2x=Ax+xA2x=Ax+x for every vector x, which matrix acts as the inverse of AAA?

Answer

If row operations transform [A∣Ip][A \mid I_p][A∣Ip​] into [Ip∣D][I_p \mid D][Ip​∣D], what is DDD?

Answer

If a row-operation product HHH satisfies HA=IpHA=I_pHA=Ip​ and HIp=EHI_p=EHIp​=E, what is HHH?

Answer

Exercise

Suppose AAA is invertible and AB=IpAB = I_pAB=Ip​. Prove that B=A−1B = A^{-1}B=A−1.

Guided solution

Suppose (A−B)A=I(A-B)A=I(A−B)A=I for square matrices AAA and BBB. Prove that AAA and BBB commute.

Guided solution

Cyclic identities from $ABCD=I$

Recovering $A$ and solving a transposed system from a reduction table

If $A$ is invertible, what is $A^{-1}A$ ?

If $A$ is invertible, can the homogeneous system $Ax = 0$ have a nonzero solution?

If $A$ is invertible, is $A^t$ invertible?

If $B = GA$ with $G$ invertible and the columns of $A$ satisfy $a_3 = 2a_1 - a_2$ , what relation must hold among the columns of $B$ ?

Suppose $v \ne 0$ and $Av = 0$ . Can $A$ be invertible?

Suppose `A,B,C,D` are square matrices and $ABCD=I$ . Which cyclic identity is forced: $DABC=I$ or $DCBA=I$ ?

In the parameterized inverse above, what is the `(1,4)` entry of $A_\alpha^{-1}$ when $\alpha=3$ ?

If $A^2x=Ax+x$ for every vector `x`, which matrix acts as the inverse of $A$ ?

If row operations transform $[A \mid I_p]$ into $[I_p \mid D]$ , what is $D$ ?

If a row-operation product $H$ satisfies $HA=I_p$ and $HI_p=E$ , what is $H$ ?

Suppose $A$ is invertible and $AB = I_p$ . Prove that $B = A^{-1}$ .

Suppose $(A-B)A=I$ for square matrices $A$ and $B$ . Prove that $A$ and $B$ commute.