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?
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?
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 $AB = BA = I$.
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: $$(A^{-1})^T = (A^{-1})^TA^T(A^T)^{-1} = (AA^{-1})^T(A^T)^{-1} = \textrm{id}^T(A^T)^{-1} = (A^T)^{-1}.$$
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 $A^T$.
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.
$(A^{-1})^{T}=(A^T)^{-1}$
Well, if the underlying ring is not commutative, $(AB)^T = B^{T}A^T$ does not even hold for $1 \times 1$-matrices.
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 $A$.
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.