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I can't seem to find the answer to this using Google.

Is the transpose of the inverse of a square matrix the same as the inverse of the transpose of that same matrix?

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    $\begingroup$ This holds if the underlying ring is commutative as the answers show. If the underlying ring is not commutative, it might fail. $\endgroup$
    – user44400
    Jan 2, 2017 at 16:52
  • $\begingroup$ By which you mean, if the entries in the matrices do not have the commutative property, then this does not necessarily hold? Interesting. Maybe you should add that, with a little more detail, to the existing accepted answer. $\endgroup$
    – Void Star
    Jan 3, 2017 at 16:53
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    $\begingroup$ @user44400: great observation! I noticed now for $2\times 2$ matrices over quaternions that we can have invertible matrices whose transpose is not invertible ( of course the conjugate transpose is...) $\endgroup$
    – orangeskid
    Mar 25, 2018 at 9:26
  • $\begingroup$ It should also be said that the transpose just switches the indices of every element. Writing out an explicit formula for the $ij$ entry of $A^{-1}$ and swapping all indices would probably do the trick. Maybe a pain to write out but easy to believe. $\endgroup$ Dec 29, 2020 at 23:16
  • $\begingroup$ Orangeskid: please could you give an example? $\endgroup$
    – Andy Baker
    Jan 10 at 19:33

7 Answers 7

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Is $(A^{-1})^T = (A^T)^{-1}$ you ask.

Well $$\begin{align} A^T(A^{-1})^T = (A^{-1}A)^{T} = I^T = I \\ (A^{-1})^TA^T = (AA^{-1})^{T} = I^T = I \end{align} $$

This proves that the inverse of $A^T$ is $(A^{-1})^T$. So the answer to your question is yes.

Here I have used that $$ A^TB^T = (BA)^T. $$ And we have used that the inverse of a matrix $A$ is exactly (by definition) the matrix $B$ such that $AB = BA = I$.

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  • $\begingroup$ So while below @haslersn equates (A−1)T=(AT)-1, you don't AT-1 in your proof at all. How does your conclusion follow? $\endgroup$
    – atlex2
    Dec 29, 2018 at 19:31
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    $\begingroup$ @atlex2: I think you have misunderstood something. We are proving that $(A^T)^{-1} = (A^{-1})^T$. $\endgroup$
    – Thomas
    Dec 30, 2018 at 23:35
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Given that $A\in \mathbb{R}^{n\times n}$ is invertible, $(A^{-1})^T = (A^T)^{-1}$ holds.

Proof:

$A$ is invertible and $\textrm{rank }A = n = \textrm{rank }A^T,$ so $A^T$ is invertible as well. Conclusion: $$(A^{-1})^T = (A^{-1})^TA^T(A^T)^{-1} = (AA^{-1})^T(A^T)^{-1} = \textrm{id}^T(A^T)^{-1} = (A^T)^{-1}.$$

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    $\begingroup$ Strong proof that connects (A−1)T=(AT)-1 in a one-liner. Nice. $\endgroup$
    – atlex2
    Dec 29, 2018 at 19:31
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Suppose $A$ is an invertible square matrix. Then $A$ has a left-inverse; let us just call it $C$ to avoid confusing notation at first. So, $A$ being (left-)invertible means existence of some $C$ so that $$CA=I$$ Taking transposes on both sides and using the rule for transposing a product yields $$A^T C^T = I$$ The last equation shows that $A^T$ has a right-inverse (which is actually $C^T$), and so $A^T$ is (right-)invertible.

Do the analogous thing for the inverses from the other side, or appeal to the fact that for finite-dimensional matrices having an inverse from one side is enough to ensure that the same matrix is a two-sided inverse.

What we have then shown, is that $\left( A^T \right) ^{-1}$ exists and is given by the formula: $$\left( A^T \right) ^{-1} = \left( A^{-1} \right) ^T$$

This also shows, since the transpose of a transpose is the original matrix, that if $A$ is not invertible, then neither is $A^T$.

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I would derive the formula step by step this way.

Lets have invertible matrix A, so you can write following equation (definition of inverse matrix):

$AA^{-1} = I$

Lets transpose both sides of equation. (using $I^{T} = I$ , $(XY)^T = Y^TX^T$)

$(AA^{-1})^T = I^T$

$(A^{-1})^{T}A^T = I$

From the last equation we can say (based on the definition of inverse matrix) that $A^T$ is inverse of $(A^{-1})^{T}$. So we can write following.

$(A^{-1})^{T})^{-1}=A^T$

By inverting both sides of equation we obtain the desired formula.

$(A^{-1})^{T}=(A^T)^{-1}$

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Well, if the underlying ring is not commutative, $(AB)^T = B^{T}A^T$ does not even hold for $1 \times 1$-matrices.

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    $\begingroup$ Someone had pointed out the same thing in a comment seven months ago. By the way, while what you say is technically true, what the OP asked for wasn't a proof of $(AB)^T=B^TA^T$, but $(A^{-1}))^T=(A^T)^{-1}$, and the latter equality does hold in the $1\times1$ case even when the ring is not commutative. $\endgroup$
    – user1551
    Aug 30, 2017 at 0:01
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Let the inverse of $A$ be $B^T$ so that $B$ is the transpose of the inverse of$A$. Then by definition of the inverse

$$AB^{T}=I=B^{T}A.$$

Now taking the transpose

$$BA^T=I=A^TB,$$

which proves that $B$ is also the inverse of the transpose of $A$.

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The result is true. See part (4) of Theorem 3.5.3 in $\textit{A Modern Introduction to Linear Algebra}$ by Henry Ricardo (CRC Press, 2010), for example.

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