Why an Inverse Cannot Be Ambiguous
A square matrix is invertible when there is a matrix that undoes its multiplication on both sides. Once such a matrix has been found, it is important to know whether another, different matrix could do the same job. The answer is no: a square matrix has at most one two-sided inverse. The proof is short, but its ingredients are fundamental. It uses the identity matrix and associativity of matrix multiplication; it does not assume that matrices commute.
Throughout, take matrices over \(\mathbb{R}\), and let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). The identity matrix \(\mathbf{I}_n\) satisfies \(\mathbf{I}_n\mathbf{X}=\mathbf{X}\mathbf{I}_n=\mathbf{X}\) for every \(n\times n\) matrix \(\mathbf{X}\). Matrix multiplication is associative: whenever the products are defined, \((\mathbf{X}\mathbf{Y})\mathbf{Z}=\mathbf{X}(\mathbf{Y}\mathbf{Z})\). In general, it is not commutative: \(\mathbf{X}\mathbf{Y}\) need not equal \(\mathbf{Y}\mathbf{X}\).
The definition has two equations. To establish that a particular matrix is an inverse, both products must be checked. The uniqueness argument starts by supposing that two matrices satisfy both equations, then shows that they must be equal. It proves “at most one”; it does not show that an inverse exists or provide a method for finding one.
The Uniqueness Proof
Proof. Assume that \(\mathbf{B}\) and \(\mathbf{C}\) are both two-sided inverses of \(\mathbf{A}\). In particular, \(\mathbf{A}\mathbf{C}=\mathbf{I}_n\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\). Using the identity property and then associativity gives
Therefore \(\mathbf{B}=\mathbf{C}\), as required. If \(\mathbf{A}\) is invertible, at least one two-sided inverse exists by definition, and the result just proved shows that there cannot be a second, different one. Hence it has exactly one. \(\square\)
Each equality uses only a stated inverse equation, the identity property, or associativity. In particular, the argument does not switch the order of factors: the expression \(\mathbf{B}(\mathbf{A}\mathbf{C})\) becomes \((\mathbf{B}\mathbf{A})\mathbf{C}\) by changing the parentheses, not by moving \(\mathbf{A}\) past \(\mathbf{B}\) or \(\mathbf{C}\).
Proof. Since \(\mathbf{A}\) is invertible, it has at least one two-sided inverse. If two matrices could be used as its inverse, the Uniqueness of the Matrix Inverse theorem would make them equal. Thus there is exactly one matrix the notation \(\mathbf{A}^{-1}\) can denote. \(\square\)
Worked Checks
Worked Example: Verify an Inverse in Both Orders
Let \(\mathbf{A}=\begin{bmatrix}1&2\\3&5\end{bmatrix}\) and propose \(\mathbf{B}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}\). Check whether \(\mathbf{B}\) is a two-sided inverse.
First multiply \(\mathbf{A}\mathbf{B}\). Each entry is the dot product of a row of \(\mathbf{A}\) and a column of \(\mathbf{B}\):
Now check the reverse order rather than assuming it gives the same result:
Both defining equations hold, so \(\mathbf{B}\) is an inverse of \(\mathbf{A}\). By the uniqueness theorem, no other \(2\times2\) matrix can also be a two-sided inverse of this \(\mathbf{A}\).
Worked Example: Reject a Different Candidate
Use the matrices from the previous calculation. The inverse found there is \(\mathbf{B}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}\). A proposed alternative is \(\mathbf{C}=\begin{bmatrix}-5&3\\2&-1\end{bmatrix}\), which is different from \(\mathbf{B}\). The theorem says that \(\mathbf{C}\) cannot also be an inverse. Check this directly by computing \(\mathbf{A}\mathbf{C}\):
This is not \(\mathbf{I}_2\): for example, its top-left entry is \(-1\), not \(1\). Therefore \(\mathbf{C}\) is not a two-sided inverse. The uniqueness proof gives the general reason two distinct matrices cannot both be inverses; this multiplication identifies the specific failed condition for this proposed alternative.
Worked Example: A Three-by-Three Inverse
Consider \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}\) and the candidate \(\mathbf{B}=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\). Verify both products, then state what uniqueness tells us.
In the first order, row-by-column multiplication gives
For the reverse order, the entries are
Thus \(\mathbf{B}\) is a two-sided inverse of \(\mathbf{A}\). The uniqueness theorem then guarantees that any \(3\times3\) matrix satisfying both inverse equations for this \(\mathbf{A}\) must equal this \(\mathbf{B}\), entry by entry.
What the Result Does—and Does Not—Say
There are two separate questions about an inverse. Existence asks whether at least one matrix satisfies both inverse equations. Uniqueness asks whether more than one matrix can satisfy them. The theorem answers the second question. It does not imply that every square matrix is invertible, and it cannot turn a candidate into an inverse if either product fails to equal the identity.
For example, in the second worked example, \(\mathbf{A}\mathbf{B}=\mathbf{I}_2\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_2\) establish that an inverse exists. The theorem then rules out any different inverse. By contrast, comparing \(\mathbf{C}\) with \(\mathbf{B}\) alone would not prove that \(\mathbf{C}\) fails the definition; the direct product calculation shows that it fails.
The proof also depends on the order of factors. The step from \(\mathbf{B}(\mathbf{A}\mathbf{C})\) to \((\mathbf{B}\mathbf{A})\mathbf{C}\) is permitted by associativity. Replacing it with \((\mathbf{A}\mathbf{B})\mathbf{C}\) would change the order, and associativity does not justify that change.
Common Pitfall: Reordering Factors
A tempting but invalid argument is to treat matrix multiplication as if it were ordinary multiplication of real numbers and reorder factors while proving uniqueness. The following calculation shows why that is unsafe. Let \(\mathbf{X}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{Y}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\). Then
These products differ, so the false step \(\mathbf{X}\mathbf{Y}=\mathbf{Y}\mathbf{X}\) cannot be used in a general proof. The uniqueness argument avoids this error: it keeps the factors in the order \(\mathbf{B}\), \(\mathbf{A}\), \(\mathbf{C}\), and changes only their grouping.
When applying the theorem, check its hypotheses carefully. The matrices \(\mathbf{B}\) and \(\mathbf{C}\) must each be two-sided inverses of the same square matrix \(\mathbf{A}\), with products of compatible sizes. A matrix that satisfies only one of the two equations has not yet met the definition used here. Once both equations hold, the chain in the proof forces the candidates to coincide.
Check Your Understanding
Use the definition and the uniqueness proof to answer each question.
- In the uniqueness proof, which inverse equation justifies replacing \(\mathbf{A}\mathbf{C}\) by \(\mathbf{I}_n\)?
- Which inverse equation justifies replacing \(\mathbf{B}\mathbf{A}\) by \(\mathbf{I}_n\)?
- Why does \(\mathbf{B}(\mathbf{A}\mathbf{C})=(\mathbf{B}\mathbf{A})\mathbf{C}\) not require matrix multiplication to be commutative?
- Suppose \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\). What can the uniqueness theorem say about a different proposed two-sided inverse \(\mathbf{C}\)?
- Does uniqueness prove that every square matrix has an inverse? Explain the difference between existence and uniqueness.
