The square root of a 2x2 matrix

[onako]

Given a symmetric 2x2 matrix, what would be the way to calculate the square root of it?
Here, http://www.jstor.org/stable/2689616?seq=2 , it is announced that a relatively simple formula
could be applied for the resulting matrix entries, but I could not access it further.

Thanks

 

[disregardthat]

A symmetric matrix B can be diagonalized.

Let B = P^{-1}DP, where P is the matrix with the eigenvector columns, and D is the matrix with the eigenvalues as diagonal entries.
You want a solution to X^2 = B, so it's reasonable to expect that X is on the form P^{-1}D^{\prime}P for some diagonal matrix D^{\prime}.

Now, we try to solve the equation:

X^2 = (P^{-1}D^{\prime}P)^2 = P^{-1}D^{\prime 2}P = B = P^{-1}DP \Rightarrow D^{\prime 2 } = D.

Now you solve this easy equation for D^{\prime}, hence determining X.

 

[Landau]

In simple words:
* diagonalize the matrix (i.e. move to a basis of eigenvectors)
* take the square root (which is now simple because the square root of a diagonal matrix is the diagonal matrix obtained from taking the square root of the diagonal entries),
* convert back to the original basis (if desired)

The matrix has to be positive semidefinite, so has to have nonnegative eigenvalues. But actually, all of this is probably in your definition of "square root of a matrix"?

 

[onako]

Thanks.
What do you think about this:
http://www.mathkb.com/Uwe/Forum.aspx/math/33058/Sqrt-of-a-2x2-matrix
Is he assuming that r_1,r_2 are the eigenvalues of A?
In step 3, is I the identity matrix?

 

[Landau]

Is he assuming that r_1,r_2 are the eigenvalues of A?

Since he writes "Step 1. Find the roots r_1,r_2 ( eigenvalues)", the answer is yes.

In step 3, is I the identity matrix?

Yes.