Square Root of a Matrix

To find the square root of the n X n matrix C, I use the eigenvalue
decomposition

[E L] = eig(C);

to obtain

sqrtC = E*sqrt(L)*E'.

Now, I would like an analytic expression for sqrtC(n,c) when C
is a correlation coefficient matrix with equal nondiagonal
components c:

For n = 2, C = [1 c; c 1]
For n = 3, C = [1 c c; c 1 c; c c 1]

etc.

Any ideas?

Greg

 

This isn't really the answer to your question, but just in case you
didn't come across this you can also use,

sqrtm

-Dan-

 

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Hello Greg,

The formula below should give you a square root of your matrix but not
always the one given by 'sqrtm'. I have a strong suspicion 'sqrtm' is not
consistent in its choice of square root, at least with respect to your
kind of problem. Your eigenvalues can easily be negative so their square
roots would then be imaginary. If your matrix is n x n, it has in general
2^n square roots, obtained by all possible combinations of signs on the
square roots of your n eigenvalues, but I don't know formulas like the one
below for them all, only that one. You can get one more by reversing all
the signs, but I'd have to put more work into the problem to find formulas
for them all. It took some sweat just to find this one!

x = repmat(sqrt((n-1)*c+1)/n-sqrt(c-1)/n*i,n,n);
x = x + diag(repmat(sqrt(c-1)*i,n,1));

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford

 

---------------------
Hello Greg,

I forgot that you stated c was a correlation coefficient, which
therefore must necessarily be less than or equal to 1, so the supposedly
imaginary part in the sqrt(c-1)*i term becomes real. The formula should
then read:

x = repmat(sqrt((n-1)*c+1)/n-sqrt(1-c)/n,n,n) + ...
diag(repmat(sqrt(1-c),n,1));

where x is a square root of your matrix. This agrees with 'sqrtm' as far
as I can observe. With c <= 1 both are choosing the positive square roots
of all the original eigenvalues. There is one eigenvalue in them equal to
+sqrt((n-1)*c+1) and n-1 eigenvalues all equal to +sqrt(1-c).

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford

 

Thanks.

How did you figure it out? I quit at the cubic eigenvalue equation
for n = 3. I would have tried MAPLE except I forgot how.

Greg

 

-----------------
Hello Greg,

I have to confess it was a case of blind luck. As I mentioned earlier,
I had forgotten that your c <= 1 and evaluated the square root matrices
for some particular integers c > 1 and for several values of n. I
discovered that the real and imaginary parts of their elements were each
square roots of something rational and thereby soon discovered the
formulas in terms of c and n (i.e., sqrt((n-1)*c+1) and sqrt(1-c).) These
formulas carry over to the case with c <= 1 except that there is no
imaginary part since sqrt(1-c) becomes real - the two square roots are
combined in real sums and differences. If I had tried things originally
with only c <= 1, it is doubtful I could have disentangled the two square
roots to find their formulas without a lot more work. There's fortuity
for you!

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford

 

I'll say. Did it ever occur to you to try MAPLE?

Greg

 

Hello Greg,

I thought I'd let you know that, thanks to your matrix example, I
managed to correct a mistaken notion I had previously held about the
number of possible square roots an n x n matrix can have. When an
eigenvalue is repeated, as occurs for your matrices for n greater than 2,
there are infinitely many possible sets of corresponding eigenvectors.
You can get them by performing arbitrary rotations (unitary
transformations) on the space of such eigenvectors. Hence, by assigning
both possible signs to the square roots of these repeated eigenvalues,
arbitrarily rotating the corresponding eigenvector space, and then working
your way back to a square root matrix, you can get infinitely many
different square roots! That was a surprise. Your matrices have
infinitely many square roots. I used TMW's Symbolic Math Toolbox (Maple)
to verify that this is true with your 3 x 3 matrix. I do think the 2^n
limit still applies to any matrix with non-repeating eigenvalues, however.

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford