Square root of a squared block matrix

[GoodSpirit]

Hi everybody,

I’m trying to compute the square root of the following squared block matrix:

M=\(\displaystyle
\begin{bmatrix}
A &B\\
C &D\\
\end{bmatrix}
\)

(that is M^(1/2))as function of A,B,C, D.

Can you help me?

I sincerely thank you!

All the best

GoodSpirit

 

[Deleted member 4993]

Hi everybody,

I’m trying to compute the square root of the following squared block matrix:

M=\(\displaystyle
\begin{bmatrix}
A &B\\
C &D\\
\end{bmatrix}
\)

(that is M^(1/2))as function of A,B,C, D.

Can you help me?

I sincerely thank you!

All the best

GoodSpirit

Let

= [M]^1/2

and

S=\(\displaystyle
\begin{bmatrix}
E &F\\
G &H\\
\end{bmatrix}
\)

Find ^-1

And use

= [M]

and

= [M]^-1

You'll have 4 equations and 4 unknowns.

 

[lookagain]

Could this be a realistic alternative:


Using S and M from above:


$\displaystyle S^2 \ = \ M$


\(\displaystyle
\begin{bmatrix}
E &F\\
G &H\\
\end{bmatrix}
\)\(\displaystyle
\begin{bmatrix}
E &F\\
G &H\\
\end{bmatrix}
\) $\displaystyle \ = \$\(\displaystyle
\begin{bmatrix}
A & B\\
C & D\\
\end{bmatrix}
\)$\displaystyle \ \implies$


\(\displaystyle
\begin{bmatrix}
E^2 + FG &EF + FH\\
GE + HG &GF + H^2\\
\end{bmatrix}
\) $\displaystyle \ = \$\(\displaystyle
\begin{bmatrix}
A & B\\
C & D\\
\end{bmatrix}
\)


Now you have four equations and four unknowns.


Question: Is solving for the unknowns doable (as far as the degrees
of the resulting four equations not being too high)?

 

[daon2]

I believe it is a large assumption to assume S has an inverse. For example if S is the diagonal matrix diag(1,0) then it has two obvious square roots and neither are invertible. We must also have that A, D are square matrices. This looks like a fun problem, if doable.

 

[daon2]

I believe the best we can do is find the jordan form of the matrix but I don't see how to retain the need "in terms of A,B,C,D". we find a particular change of basis matrix P such that

$\displaystyle M = PSP^{-1}$

where $\displaystyle S$ is a block diagonal matrix whose eigenvalues lie on the diagonal and 1s on the super-diagonals of each block. This is annoying in general, but doable. Note if M is diagonalizable, S is just a diagonal matrix and has a trivial square root (maybe more than one).

Otherwise you are left with finding the square root of an arbitrary jordan block. This is not possible

Example: there is no matrix $\displaystyle T$ such that

$\displaystyle T^2=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}=M$

So I think your matrix needs to be either a) diagonalizable or if not: b) invertible (0 is not an eigenvalue).

 

[GoodSpirit]

Thank you all for your suggestions,

Unfortunately, the matrix M is not invertible and neither S. However, they are diagonalizable.
But there is more, M is symmetric and positive semi-definite. Well it is a covariance matrix…
A and D are symmetric and positive semi-definite(covariance matrices too) and $\displaystyle B=C^T$ and B is the cross covariance matrix of A and D.
My approach has been the approach that daon2 suggested.

I really thank you all for your answers!

All the best

GoodSpirit