       • To: mathgroup at smc.vnet.net
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 30 Oct 2007 03:17:21 -0500 (EST)
• References: <200710291054.FAA06915@smc.vnet.net>

```Your matrix M has the property:

MatrixPower[M, 2]
{{0, 0}, {0, 0}}

It is easy to prove using elementary linear algebra that such a
matrix has no square root. In fact one can prove more. Suppose than M
is an n by n matrix such that M^n=0 but M^(n-1) !=0 (in other words M
is nilpotent of order n). Then M has no square root.

The proof is easy so I won't bother to give it here.

Andrzej Kozlowski

On 29 Oct 2007, at 19:54, Roger Bagula wrote:

> M = {{-1, I}, {I, 1}}
> MatrixPower[M, 1/2]
> gives
> {{0, 0}, {0, 0}}
>
> So try it as {{a,b},{c,d}} squared:
> c = b; a = I*Sqrt[1 + b^2]; d = Sqrt[1 - b^2
> FullSimplify[{a^2 + b c + 1 == 0, a b + b d - I == 0, a c +
>          c d - I == 0, b c + d^2 - 1 == 0}]
> {True, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0, True}
> Solve[-I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0,b]
> {}
> which says there is no solution.
>
> This problem comes from the graph of SU(2) and of U(1)*SU(2)
> as a two vertex 3 directed connections (i,j,k) and a 2 vertex with 4
> directed connections (1,i,j,k}.
> Basically there is either a solution or there is none.
> Mathematica gives zero and the null set from two different approaches.
> {{0,i+k},{j,0}} and {{0,i+k},{j+IdentityMatrix,0}}
>
> It is pretty much a break down of mathematical definitions.
> The matrix M does appear to have not one, but four solutions
> the way I do it:
>
> M2={{+/-I*Sqrt[1 + b^2], b}, {b, +/-Sqrt[1 + b^2]}}
> I really may be doing it all wrong.
> b=+/-Sqrt[+/-1/2+I/2]
> which gives the stange answers from this code:
> Clear[b]
> M2 = {{I*Sqrt[1 + b^2], b}, {b, Sqrt[1 + b^2]}}
> Det[M2]
> Solve[Det[M2] == 0, b]
> b0 = b /. Solve[Det[M2] == 0, b][]
> M20 = {{-I*Sqrt[1 + b0^2], b0}, {b0, Sqrt[1 + b0^2]}}
> FullSimplify[M20]
> FullSimplify[M20.M20]
>
> All this leaves me really puzzled.
> Usually Mathematica takes away my doubts,
> but here it isn't any help at all.
> Maybe it is a paradox?
> Roger Bagula
>

```

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