Keeping Transposes and Inverses in Order
Transpose and inverse are two different operations, but they interact in a predictable way. A transpose reverses the order of factors in a product, and an inverse of a product also reverses that order. In addition, taking the inverse of a transpose gives the transpose of the inverse. These rules let us simplify complicated-looking expressions without computing each inverse from scratch. The main discipline is to keep track of both the order of the factors and the conditions under which each inverse exists.
All matrices in the identities below are real. A matrix is invertible if it is square and has a matrix \(\mathbf{A}^{-1}\) such that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}\). The transpose \(\mathbf{A}^{\top}\) is formed by interchanging rows and columns. It exists for any matrix, but an inverse is defined only for a square invertible matrix. In particular, any formula that takes the inverse of a transpose or a product must satisfy the corresponding square-matrix and invertibility conditions.
This identity is useful because it tells us how to move an inverse and a transpose past one another. It does not mean that we may ignore their order inside a product. For example, if \(\mathbf{A}\) and \(\mathbf{B}\) are invertible \(n\times n\) matrices, the familiar product rules are \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}^{\top}\mathbf{A}^{\top}\) and \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}\). Each rule reverses the order of the factors.
Proof. Since \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\) and \(\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\), transpose both equalities. Using the product rule \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{Y}^{\top}\mathbf{X}^{\top}\) and \(\mathbf{I}_n^{\top}=\mathbf{I}_n\), we obtain \((\mathbf{A}^{-1})^{\top}\mathbf{A}^{\top}=\mathbf{I}_n\) and \(\mathbf{A}^{\top}(\mathbf{A}^{-1})^{\top}=\mathbf{I}_n\). Thus \((\mathbf{A}^{-1})^{\top}\) is a two-sided inverse of \(\mathbf{A}^{\top}\), proving the identity. \(\square\)
Applying the Identity to One Matrix
Worked Example: Finding the Inverse of a Transpose
Let \(\mathbf{A}=\begin{bmatrix}1&2\\3&5\end{bmatrix}\). Find \((\mathbf{A}^{\top})^{-1}\) using the transpose-inverse identity, then verify the answer by multiplication.
First, check the proposed inverse \(\mathbf{D}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}\) of \(\mathbf{A}\) by multiplying in both orders:
Therefore \(\mathbf{A}^{-1}=\mathbf{D}\). Transposing gives the candidate for the inverse of \(\mathbf{A}^{\top}\):
Check both products. The first row of each product is \((1,0)\), and the second row is \((0,1)\):
Both checks confirm that \((\mathbf{A}^{\top})^{-1}=(\mathbf{A}^{-1})^{\top}\).
The multiplication check is more than a numerical formality: a matrix is an inverse only if it works on both sides. The identity supplies a candidate, and the two products verify it directly. When working symbolically, the same two-sided condition is what justifies calling the candidate an inverse.
Combining Transposes and Product Inverses
Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are invertible \(n\times n\) matrices. To simplify an expression, first identify its outermost operation, then apply the relevant rule and continue inward. The product rules recalled above are standard identities; for instance, the candidate \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) for \((\mathbf{A}\mathbf{B})^{-1}\) works because \((\mathbf{A}\mathbf{B})(\mathbf{B}^{-1}\mathbf{A}^{-1})=\mathbf{A}\mathbf{I}_n\mathbf{A}^{-1}=\mathbf{I}_n\), and the reverse product is also \(\mathbf{I}_n\).
Proof. Set \(\mathbf{X}=\mathbf{A}^{\top}\mathbf{B}^{-1}\). The matrix \(\mathbf{B}(\mathbf{A}^{\top})^{-1}\) is its inverse: multiplying in the two orders gives \(\mathbf{X}\mathbf{B}(\mathbf{A}^{\top})^{-1}=\mathbf{A}^{\top}\mathbf{B}^{-1}\mathbf{B}(\mathbf{A}^{\top})^{-1}=\mathbf{I}_n\) and \(\mathbf{B}(\mathbf{A}^{\top})^{-1}\mathbf{X}=\mathbf{B}(\mathbf{A}^{\top})^{-1}\mathbf{A}^{\top}\mathbf{B}^{-1}=\mathbf{I}_n\). Hence \(\mathbf{X}^{-1}=\mathbf{B}(\mathbf{A}^{\top})^{-1}\). Transposing and reversing the product order gives \((\mathbf{X}^{-1})^{\top}=((\mathbf{A}^{\top})^{-1})^{\top}\mathbf{B}^{\top}\). By the transpose-inverse identity applied to \(\mathbf{A}\), \(((\mathbf{A}^{\top})^{-1})^{\top}=\mathbf{A}^{-1}\). Therefore \((\mathbf{X}^{-1})^{\top}=\mathbf{A}^{-1}\mathbf{B}^{\top}\), as required. \(\square\)
Worked Example: Inverting a Transposed Product
Let \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&2\\0&1\end{bmatrix}\). Simplify and compute \(((\mathbf{A}\mathbf{B})^{\top})^{-1}\), then verify the proposed inverse.
First multiply the matrices and compute their inverses, checking each inverse in both orders:
For example, \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_2\), since the off-diagonal entry in each product is \(1-1=0\). Likewise, both products for \(\mathbf{B}\) have off-diagonal entry \(2-2=0\). Apply the inverse-transpose identity and the product-inverse rule:
The factors on the right are \((\mathbf{A}^{-1})^{\top}=\begin{bmatrix}1&0\\-1&1\end{bmatrix}\) and \((\mathbf{B}^{-1})^{\top}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}\). Thus the result is
To verify, \((\mathbf{A}\mathbf{B})^{\top}=\begin{bmatrix}1&0\\3&1\end{bmatrix}\), and multiplying in both orders gives
The candidate is therefore the inverse of the transposed product. This computation also shows why the product order matters: the inverse and transpose rules each reverse order at their respective steps.
Reading a Nested Expression Carefully
For an expression with several operations, parentheses help identify what is being transposed or inverted. Consider \(\big((\mathbf{A}^{\top})^{-1}\mathbf{B}^{\top}\big)^{\top}\). The outer transpose reverses the two factors. The first factor then simplifies using the transpose-inverse identity, while the transpose of \(\mathbf{B}^{\top}\) is \(\mathbf{B}\). Thus the expression becomes \(\mathbf{B}\mathbf{A}^{-1}\). This simplification requires \(\mathbf{A}\) to be square and invertible; \(\mathbf{B}\) may be any matrix of compatible size for this expression.
Worked Example: Simplifying a Nested Expression
Let \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\). Compute \(\big((\mathbf{A}^{\top})^{-1}\mathbf{B}^{\top}\big)^{\top}\) in two ways and compare the results.
The inverse of \(\mathbf{A}\) is \(\begin{bmatrix}1&-2\\0&1\end{bmatrix}\), as multiplication on either side by \(\mathbf{A}\) gives \(\mathbf{I}_2\). Therefore \((\mathbf{A}^{\top})^{-1}=(\mathbf{A}^{-1})^{\top}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}\), and \(\mathbf{B}^{\top}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\). Compute the expression inside the outer transpose:
Transposing gives \(\begin{bmatrix}2&-3\\1&-1\end{bmatrix}\). Now simplify symbolically first: the outer transpose reverses the factors, so \(\big((\mathbf{A}^{\top})^{-1}\mathbf{B}^{\top}\big)^{\top} =(\mathbf{B}^{\top})^{\top}\big((\mathbf{A}^{\top})^{-1}\big)^{\top} =\mathbf{B}\mathbf{A}^{-1}\). Direct multiplication confirms the same result:
The two computations agree, entry by entry. The outer transpose reverses the factor order; it does not leave the original order unchanged.
Common Pitfalls and a Reliable Method
Pitfall: transposing a product without reversing its order. In the preceding example, the correct answer is \(\mathbf{B}\mathbf{A}^{-1}=\begin{bmatrix}2&-3\\1&-1\end{bmatrix}\). A tempting but incorrect step is to keep the factors in place and write \(\mathbf{A}^{-1}\mathbf{B}\). That product is instead \(\begin{bmatrix}1&-2\\0&1\end{bmatrix}\begin{bmatrix}2&1\\1&1\end{bmatrix}=\begin{bmatrix}0&-1\\1&1\end{bmatrix}\), which is not the answer. The correction is to use \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{Y}^{\top}\mathbf{X}^{\top}\) before simplifying either factor.
Pitfall: using an inverse identity when an inverse does not exist. The equality \((\mathbf{A}^{\top})^{-1}=(\mathbf{A}^{-1})^{\top}\) requires \(\mathbf{A}\) to be square and invertible. For a rectangular matrix, \(\mathbf{A}^{\top}\) is defined, but the ordinary matrix inverse of either matrix is not defined. For a square noninvertible matrix, neither \(\mathbf{A}^{-1}\) nor \((\mathbf{A}^{\top})^{-1}\) exists.
A dependable procedure for simplifying these expressions is:
- Check dimensions and confirm that every matrix being inverted is square and invertible.
- Apply one rule at a time, reversing factor order whenever a transpose or inverse is applied to a product.
- Use \((\mathbf{A}^{\top})^{-1}=(\mathbf{A}^{-1})^{\top}\) to exchange inverse and transpose.
- For a numerical answer, multiply the proposed inverse by the original matrix on both the left and the right.
These identities matter because they let us transform a matrix expression while preserving its value, often avoiding a fresh inverse computation. The rules also make clear that matrix operations are not interchangeable in arbitrary order. Writing each intermediate expression explicitly is a useful check: every factor should remain in the correct position, and every inverse should be justified by the stated hypotheses.
Check Your Understanding
For each question, state any required invertibility conditions and show enough work to justify the result.
- If \(\mathbf{C}\) is an invertible \(3\times3\) matrix, express \((\mathbf{C}^{\top})^{-1}\) using \(\mathbf{C}^{-1}\).
- For invertible \(2\times2\) matrices \(\mathbf{A}\) and \(\mathbf{B}\), simplify \(((\mathbf{A}\mathbf{B})^{\top})^{-1}\) into a product involving \((\mathbf{A}^{-1})^{\top}\) and \((\mathbf{B}^{-1})^{\top}\).
- Simplify \(\big((\mathbf{A}^{\top})^{-1}\mathbf{B}^{\top}\big)^{\top}\), assuming \(\mathbf{A}\) is invertible and the products are defined.
- For \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), verify directly that the transpose of \(\mathbf{A}^{-1}\) is the inverse of \(\mathbf{A}^{\top}\).
- Explain why \((\mathbf{X}\mathbf{Y})^{\top}=\mathbf{X}^{\top}\mathbf{Y}^{\top}\) is not the correct general product rule for transposes.
