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?
Matrices – Transpose of Inverse vs Inverse of Transpose
inversematricestranspose
Related Solutions
Suppose $A$ an orthogonal matrix. Then we can write $A$ as $A = (a^1,a^2,...,a^n)$. With $a^i$ the column vectors of $A$.
Then we know that $a^1,a^2,...,a^n$ are pairwise orthogonal. E.g. $(a^i,a^j) = 0$ with $i≠j$ and $(a^i,a^j)= 1$ with $i=j$.
Now if you do the matrix multiplication $A^TA$ or $AA^T$ only the positions $a^{i,i}$ become ones, all others zeros.
There are great answers by fellow members. I would like to visualize just this particular problem. Lets say there are $4$ companies $A$,$B$,$C$ and $D$ and all of them sell three fruits Apples, Oranges and Pears. Because the numbers are less, I will assume that we want to see the daily sales in numbers of all companies.
Create the table for daily sales: $$\begin{bmatrix} &\text {Apples} & \text{Oranges}&\text{Pears} \\\text{Company 1}&10&2&5\\\text{Company 2} &5&3&10\\\text{Company 3} &4&3&2\\\text{Company 4} &5&10&5\\\end{bmatrix}$$
Just ignore the words and look at the numbers. The first row and first column are just for understanding. The numerical values of the table represent your matrix $A$. This table tells you the daily sales of each company for apples, oranges and pears.
$$A=\begin{bmatrix}10 & 2&5 \\5 & 3&10 \\4 & 3&2\\5 & 10&5 \end{bmatrix}$$ If we just write the table in another way, to see just the sales of a particular fruit from all the companies we will write, $$\begin{bmatrix} &\text {Company 1} & \text{Company 2}&\text{Company 3}&\text{Company 4} \\\text{Apples}&10&5&4&5\\\text{Oranges} &2&3&3&10\\\text{Pears} &5&10&2&5\\\end{bmatrix}$$ This can be written as: $$A^T=\begin{bmatrix}10 & 5&4&5 \\2 &3& 3&10 \\5 & 10&2&5\\ \end{bmatrix}$$ Now we keep both the tables together, $$\begin{bmatrix} &\text {Apples} & \text{Oranges}&\text{Pears} \\\text{Company 1}&10&2&5\\\text{Company 2} &5&3&10\\\text{Company 3} &4&3&2\\\text{Company 4} &5&10&5\\\end{bmatrix}\begin{bmatrix} &\text {Company 1} & \text{Company 2}&\text{Company 3}&\text{Company 4} \\\text{Apples}&10&5&4&5\\\text{Oranges} &2&3&3&10\\\text{Pears} &5&10&2&5\\\end{bmatrix}$$ If by some case there is a partnership between two companies say Company A and Company B, then what will be the total fruit sales? $$\text{Total fruit sales for the partnership} = \text{No of total apples + No of total oranges + No of total pears}$$
Total fruit sales for the partnership = Company 1 Apples X Company 2 Apples + Company 1 Oranges X Company 2 Oranges + Company 1 Pears X Company 2 Pears $$\text{Total fruit sales for the partnership} = 10X5 + 2X3 + 5X10=106$$ So the total sales of fruits for the partnership of Company A and Company B is $106$. This is nothing but the second element of the product $AA^T$.
$$AA^T=\begin{bmatrix}10 & 2&5 \\5 & 3&10 \\4 & 3&2\\5 & 10&5 \end{bmatrix}\begin{bmatrix}10 & 5&4&5 \\2 &3& 3&10 \\5 & 10&2&5\\ \end{bmatrix}=\begin{bmatrix}129&106&56&85 \\106&134&49&105 \\56&49&29&60\\ 85&105&60&150\end{bmatrix}$$
What does this product show? This product can be visualized as the total sales chart of each company as well as the total sales of mutual parnterships of companies. $$\begin{bmatrix} &\text {Company 1} & \text{Company 2}&\text{Company 3}&\text{Company 4} \\\text{Company 1}&129&106&56&85\\\text{Company 2} &106&134&49&105\\\text{Company 3} &56&49&29&60\\\text{Company 4} &85&105&60&150\\\end{bmatrix}$$
Crucial points to observe:
The diagonal elements of the matrix $AA^T$ are all just the squared sum of individual companies. For example the first element is the strength of sales of Company 1 and so on.
Each non diagonal element shows the total sales that would result due to the partnership between two companies. For example the second element of $AA^T$ is the total sales produced due to the partnership between Company 1 and Company 2.
The matrix $AA^T$ is symmetric, which can be visualized using the fact that the total sales due to the partnership of Company 1 and Company 2 is same as that of Company 2 and Company 1.
Useful insight from $AA^T$is that check the diagonal elements , whichever is the maximum, you can confirm that Company is stronger in sales. Another useful insight is you can check whether partnership with a particular company is beneficial or not. For example, Company 3 is having the lowest sales individually, so it is beneficial for Company 3 to form a partnership with Company 4 because the total sales would be 60 which is more than double of what Company 3 can have. So, we can check which partnerships would be most beneficial.
Diagonal elements: (A measure of) Individual strengths, Non Diagonal Elements: Partnership strengths.
Hope this helps...
Best Answer
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 $AB = BA = I$.