# Thread: The square root of a 2x2 matrix

1. ## The square root of a 2x2 matrix

Given a symmetric 2x2 matrix, what would be the way to calculate the square root of it?
Here, JSTOR: An Error Occurred Setting Your User Cookie , it is announced that a relatively simple formula
could be applied for the resulting matrix entries, but I could not access it further.

Thanks

2. I think you'd also want your matrix to be positive semi-definite. Unless you're ok with complex numbers in your answer. What you'd normally do then is diagonalize your matrix thus: $A=PDP^{-1}$, where P is (obviously) invertible, and D is diagonal. Finally, you'd take the square root thus: $\sqrt{A}=P\sqrt{D}P^{-1}.$ However, this may not be the most general method available.

3. Ackbeets suggestion is the most "sophisticated" (and so simplest) method. Here's a more simple-minded method:

Any 2 by 2 matrix is of the form $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$.

Its square is $A^2= A*A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}a^2+ bc & ab+ bd \\ac+ cd & bc+ d^2\end{bmatrix}$. Set each of those equal to the corresponding member of the given matrix: if the given matrix is $\begin{bmatrix}x & y \\ y & z\end{bmatrix}$ you need to solve $a^2+ bc= x$, $ac+ bd= y$, $ac+ cd= y$, and $bc+ d^2= z$.

4. Reply to HallsofIvy: Not sure you did your matrix multiplication correctly. [EDIT]: Corrected now. I also imagine you meant to say that any 2 x 2 symmetric matrix is of the form ... You get quite the nonlinear system of equations there.

I'm assuming the matrix is over the reals. Otherwise, the OP would probably say "Hermitian" instead of "symmetric".

5. Thanks.
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?

6. He's not assuming $r_{1}$ and $r_{2}$ are the eigenvalues: he's constructing them to be the eigenvalues. I would assume that I is the identity matrix.

I can follow the steps he's taking, but I admit to being mystified how it is that $\sqrt{A}=mA+pI$. I've done some computations in Mathematica, and I just don't see it.

Diagonalizing the matrix is, in my opinion, the way to go. It's not that complicated.

7. Originally Posted by Ackbeet
Reply to HallsofIvy: Not sure you did your matrix multiplication correctly. I also imagine you meant to say that any 2 x 2 symmetric matrix is of the form ... You get quite the nonlinear system of equations there.

I'm assuming the matrix is over the reals. Otherwise, the OP would probably say "Hermitian" instead of "symmetric".
Hey, I didn't say it was easy! Thanks for the headsup- it wasn't my matrix multiplication that was wrong- I was trying not to assume that the root must also be symmetric, but then wrote the first part as if it were.

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