I understand the inverse of a $2\times 2$ matrix:
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \frac{1}{ad-bc}\,\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
But what is the operation, $\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1/2}$?
Asked
Active
Viewed 1,277 times
1
-
2Given $A$, you could write $B = A^{-1/2}$ if $B^2 = A^{-1}$. This may or may not exist, and may or may not be unique (hence an issue with the notation....). – Apr 25 '16 at 01:30
-
What would be the operations involved? I'm trying to figure out this answer. – Antoni Parellada Apr 25 '16 at 01:39
-
Wikipedia has a formula. – Apr 25 '16 at 01:41
-
So we are taking the square root of the inverse. So first we calculate the inverse, and then we use the horrifying formula in the Wikipedia link, is that correct? – Antoni Parellada Apr 25 '16 at 01:47
-
Or the other way around, whichever seems less painful. – Apr 25 '16 at 01:47
-
The question you are referring to concerns symmetric, positive definite matrices. In this case, there is a unique solution $B$ to the equation $B^2=A^{-1}$ if one in addition assumes that $B$ is symmetric and positive definite. This can be obtained by diagonalizing, taking one over the square roots of the eigenvalues and then undo the diagonalization. – Andreas Cap Apr 25 '16 at 10:05