10 replies to this topic

[roytheshort]

Normally, with a quadratic equation. There have to be two roots. For example, with this one.

x² - 4 = 0

The solution is
x = +2
x = -2

Simple right? But for some reason, I try this with the 2 * 2 identity matrix. And behold, 4 roots.

Let's start with this, A is an unkown 2*2 matrix.

A² =
[ 1 0 ]
[ 0 1 ]

Now, when I tried to find A, I found four solutions. (i is the Imaginary Unit)

A =
[1 0]
[0 1]

A =
[ -1 0 ]
[ 0 -1 ]

A =
[ (1/2)^(1/2) i(1/2)^(1/2)]
[ -i(1/2)^(1/2) -(1/2)^(1/2)]

A =
[ -(1/2)^(1/2) -i(1/2)^(1/2)]
[ i(1/2)^(1/2) (1/2)^(1/2)]

So, why four solutions for a quadratic in a 2*2 Matrix? Is it always four solutions?

[xshortguy]

The first thing to point out is that some matrices don't have square roots. But remember that the square root of a matrix is another matrix.

In general, if you want to compute the square root of a matrix A, you should diagonalize the matrix. This involves finding an eigenvector basis P and rewriting the matrix as A = PDP^{-1}. Then A^k = (PDP^{-1})^k = P(D^(k))P^{-1}.

[chance]

But for some reason, I try this with the 2 * 2 identity matrix. And behold, 4 roots.

The 2 x 2 identity matrix actually has an infinite number of square-root matrices.

[roytheshort]

But for some reason, I try this with the 2 * 2 identity matrix. And behold, 4 roots.

The 2 x 2 identity matrix actually has an infinite number of square-root matrices.

How would I find all of these solutions?

[chance]

But for some reason, I try this with the 2 * 2 identity matrix. And behold, 4 roots.

The 2 x 2 identity matrix actually has an infinite number of square-root matrices.

How would I find all of these solutions?

Remember that the Cayley-Hamilton theorem says a matrix satisfies its own characteristic equation. For matrix A:

A² = trace(A) A - det(A) I

If the trace of A is zero, and its determinant is -1, then A² = I. So you can construct an infinite number of sqrt(I) matrices of the form:

[x.............y]
[(1-x²)/y... -x]

To be clear, 1st column is
x
(1-x²)/y

and second column is
y
-x

I'm not sure this approach will generate all the possible solutions of sqrt(I), but it will generate an infinite number of them.

Edited by chance, 04 August 2012 - 09:39 PM.

[xshortguy]

I'm not sure this approach will generate all the possible solutions of sqrt(I), but it will generate an infinite number of them.

If we have a 2x2 matrix, then we know that the trace is the sum of the eigenvalues and the determinant is the product of the eigenvalues. It must be the case then that these eigenvalues are +1 and -1 (since there are at most two).

Thus any matrix A that satisfies A^2 = I must also satisfy (P D P^{-1})^2 = P D^2 P^{-1} = I. Given that , D^2 = I. So PP^{-1} = I. So all of the matrices that satisfy A^2 = I are strictly the ones that are similar to D. There's clearly an infinite number of these. (We can make them by taking D and picking any invertible matrix P to form such a matrix A.)

[chance]

I'm not sure this approach will generate all the possible solutions of sqrt(I), but it will generate an infinite number of them.

If we have a 2x2 matrix, then we know that the trace is the sum of the eigenvalues and the determinant is the product of the eigenvalues. It must be the case then that these eigenvalues are +1 and -1 (since there are at most two).

Thus any matrix A that satisfies A^2 = I must also satisfy (P D P^{-1})^2 = P D^2 P^{-1} = I. Given that , D^2 = I. So PP^{-1} = I. So all of the matrices that satisfy A^2 = I are strictly the ones that are similar to D. There's clearly an infinite number of these. (We can make them by taking D and picking any invertible matrix P to form such a matrix A.)

I don't understand the point you're making here.

My point was that the approach I outlined for computing sqrt(I) would produce an infinite number of solutions, but not necessarily every solution. Are you agreeing... or disagreeing... or something else? Or are you providing an alternative solution?

Edited by chance, 04 August 2012 - 09:35 PM.

[xshortguy]

I don't understand the point you're making here.

My point was that the approach I outlined for computing sqrt(I) would produce an infinite number of solutions, but not necessarily every solution. Are you agreeing... or disagreeing... or something else? Or are you providing an alternative solution?

I've shown that your matrices are indeed all of them for the n = 2 case.

[chance]

I don't understand the point you're making here.

My point was that the approach I outlined for computing sqrt(I) would produce an infinite number of solutions, but not necessarily every solution. Are you agreeing... or disagreeing... or something else? Or are you providing an alternative solution?

I've shown that your matrices are indeed all of them for the n = 2 case.

Thanks for clarifying. That wasn't clear from your explanation, so I'll re-read it.

On a related note, I've read references to a source of sqrt(I) matrices based on formulas for generating Pythagorean triples (a²+b²=c²). I always enjoy finding connections between various field of mathematics, so this seems very interesting. If anybody has any insight into how these are generated, I'd love to read it. All I can find is an (often quoted) reference to:

Mitchell, Douglas W. "Using Pythagorean triples to generate square roots of I2". The Mathematical Gazette 87, November 2003, 499-500.

Edited by chance, 04 August 2012 - 11:29 PM.

[roytheshort]

If I took the modulus of all of these matrices and plotted them, what shape would it look like?

[xshortguy]

If I took the modulus of all of these matrices and plotted them, what shape would it look like?

What is the "modulus" of a matrix?