What Happens to an Inverse When We Transpose?
Transposing a matrix changes its rows into columns. This raises a natural question: if a square matrix can be undone by multiplying by its inverse, what matrix undoes its transpose? The answer is closely tied to the order of multiplication. Transposing a product reverses the order of its factors, and that reversal lets us check both defining equations for an inverse.
Throughout this tutorial, matrices have real entries. If \(\mathbf{A}\) is an \(m\times n\) matrix, its transpose \(\mathbf{A}^{\top}\) is the \(n\times m\) matrix whose entry in row \(i\), column \(j\) is the entry of \(\mathbf{A}\) in row \(j\), column \(i\). The transpose is defined for rectangular matrices as well as square ones. An inverse, however, is defined here for a square matrix.
The two equations are important: verifying only one product does not, by the definition alone, verify that a proposed matrix is an inverse. The key rule we need says what happens when a product is transposed. We first prove that rule from the entry formula for matrix multiplication.
The Transpose Product Rule
Proof. Both sides are \(p\times m\) matrices. For any row index \(i\) from \(1\) to \(p\) and column index \(j\) from \(1\) to \(m\), the \((i,j)\) entry of the left side is the \((j,i)\) entry of \(\mathbf{A}\mathbf{B}\). By the formula for matrix multiplication, this is \(\sum_{k=1}^{n}a_{jk}b_{ki}\). The \((i,j)\) entry of the right side is \(\sum_{k=1}^{n}( \mathbf{B}^{\top})_{ik}(\mathbf{A}^{\top})_{kj}=\sum_{k=1}^{n}b_{ki}a_{jk}\). Real numbers commute under multiplication, so \(a_{jk}b_{ki}=b_{ki}a_{jk}\) for each \(k\). The entries agree for every \(i,j\), and hence the matrices are equal. \(\square\)
The order reversal is part of the rule: the factor on the right before transposition becomes the factor on the left afterward. Also, \(\mathbf{I}_n^{\top}=\mathbf{I}_n\), since transposing the identity leaves its diagonal entries equal to \(1\) and its other entries equal to \(0\). These facts give the inverse result.
The Inverse of a Transpose
Proof. Set \(\mathbf{B}=\mathbf{A}^{-1}\). By the definition of inverse, \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\) and \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\). Transpose each equation and apply the transpose product rule. From the first equation we get \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}^{\top}\mathbf{A}^{\top}=\mathbf{I}_n^{\top}=\mathbf{I}_n\). From the second we get \((\mathbf{B}\mathbf{A})^{\top}=\mathbf{A}^{\top}\mathbf{B}^{\top}=\mathbf{I}_n^{\top}=\mathbf{I}_n\). Therefore \(\mathbf{B}^{\top}\) satisfies both defining inverse equations for \(\mathbf{A}^{\top}\). It is the inverse of \(\mathbf{A}^{\top}\), so \((\mathbf{A}^{\top})^{-1}=\mathbf{B}^{\top}=(\mathbf{A}^{-1})^{\top}\). \(\square\)
In particular, the theorem identifies a candidate inverse, but its proof does more than suggest the candidate: it verifies both products required by the definition. The dimensions also fit. Since \(\mathbf{A}^{-1}\) is \(n\times n\), its transpose is \(n\times n\), the correct size to be an inverse of \(\mathbf{A}^{\top}\).
Worked Example: A Two-by-Two Matrix
Let \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}7&-2\\-3&1\end{bmatrix}\). First check that \(\mathbf{B}\) is the inverse of \(\mathbf{A}\), using both products:
Thus \(\mathbf{B}=\mathbf{A}^{-1}\). The proposed inverse of the transpose is \(\mathbf{B}^{\top}\), so
Check the two products directly:
Both defining inverse equations hold, confirming that \((\mathbf{A}^{\top})^{-1}=\mathbf{B}^{\top}=(\mathbf{A}^{-1})^{\top}\).
Worked Example: A Three-by-Three Matrix
Consider the upper-triangular matrix \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}\), with proposed inverse \(\mathbf{B}=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\). Multiplication gives
So \(\mathbf{B}=\mathbf{A}^{-1}\). Transposing each matrix gives \(\mathbf{A}^{\top}=\begin{bmatrix}1&0&0\\1&1&0\\0&1&1\end{bmatrix}\) and \(\mathbf{B}^{\top}=\begin{bmatrix}1&0&0\\-1&1&0\\1&-1&1\end{bmatrix}\). Multiplying in the first order:
In the other order, the row-by-column products are
Thus both products are the identity, and the computed result agrees with the theorem.
Invertibility in Both Directions
The theorem proves that an invertible matrix has an invertible transpose. The converse follows as well. To justify it, note that \((\mathbf{A}^{\top})^{\top}=\mathbf{A}\): each entry returns to its original row and column when transposed twice.
Proof. If \(\mathbf{A}\) is invertible, the theorem shows that \(\mathbf{A}^{\top}\) is invertible. Conversely, suppose \(\mathbf{A}^{\top}\) is invertible. Apply the theorem to the square matrix \(\mathbf{A}^{\top}\). It follows that \((\mathbf{A}^{\top})^{\top}=\mathbf{A}\) is invertible. Thus each matrix is invertible exactly when the other is. \(\square\)
This equivalence concerns whether an inverse exists; the theorem also gives the inverse itself when it does exist. It does not say that the inverse of \(\mathbf{A}^{\top}\) is \(\mathbf{A}^{-1}\) without a transpose. In general, \((\mathbf{A}^{-1})^{\top}\) is the candidate, as the defining equations and examples show.
Worked Example: Checking the Order with a Product
The order reversal in the transpose product rule can be seen with two matrices that do not commute. Let \(\mathbf{C}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{D}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\). Their product and its transpose are
Now calculate the factors in reversed order after transposing:
This matches \((\mathbf{C}\mathbf{D})^{\top}\). By contrast, multiplying the transposes without reversing the order gives
which is not \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\). This computation shows why the product rule cannot be used with the factors left in their original order.
Common Mistake: Reversing the Order Incorrectly
A common incorrect calculation is to write \((\mathbf{C}\mathbf{D})^{\top}=\mathbf{C}^{\top}\mathbf{D}^{\top}\). The preceding example gives a concrete contradiction: the left side is \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\), whereas the claimed right side is \(\begin{bmatrix}1&1\\1&2\end{bmatrix}\). The correction is to reverse the factors: \((\mathbf{C}\mathbf{D})^{\top}=\mathbf{D}^{\top}\mathbf{C}^{\top}\).
The same order check matters in the inverse proof. Starting from \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\), transposition gives \((\mathbf{A}^{-1})^{\top}\mathbf{A}^{\top}=\mathbf{I}_n\), not \(\mathbf{A}^{\top}(\mathbf{A}^{-1})^{\top}=\mathbf{I}_n\). The second required equation comes from transposing the other inverse equation, \(\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\). Keeping these two steps separate ensures that both sides of the inverse definition have actually been checked.
The reliable method is therefore short but precise: begin with both inverse equations, transpose each equation, apply the product rule with its order reversal, and use \(\mathbf{I}_n^{\top}=\mathbf{I}_n\). The resulting two equations establish that the transpose of the inverse is the inverse of the transpose.
Check Your Understanding
For each question, use the definitions and the order of factors carefully.
- State the transpose product rule, including the matrix sizes that make the products valid.
- Why does proving only \((\mathbf{A}^{-1})^{\top}\mathbf{A}^{\top}=\mathbf{I}_n\) leave one inverse equation unchecked?
- If \(\mathbf{A}\) is invertible, write both equations that verify \((\mathbf{A}^{-1})^{\top}\) is the inverse of \(\mathbf{A}^{\top}\).
- For the matrices in the third worked example, calculate \(\mathbf{C}^{\top}\mathbf{D}^{\top}\) and compare it with \((\mathbf{C}\mathbf{D})^{\top}\).
- Prove that if \(\mathbf{A}^{\top}\) is invertible, then \(\mathbf{A}\) is invertible.
