Show that a 2x2 scalar matrix $A$ over a field has a square root. Meaning a matrix say $B$ that satisfies $B^{2}$ = $A$.
Scalar Matrix over a field
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0@WillJagy: I was being dumb. – 2012-07-23
2 Answers
It is not the case that, in any field and given any field element $x,$ that either $x$ or $-x$ is a square. Consider $2 \in \mathbb Q.$
However, assuming ``scalar matrix'' means diagonal with all the same field elements on the diagonal, we do not need a field to write:
$ \left( \begin{array}{rr} 0 & x \\ 1 & 0 \end{array} \right) \cdot \left( \begin{array}{rr} 0 & x \\ 1 & 0 \end{array} \right) = \left( \begin{array}{rr} x & 0 \\ 0 & x \end{array} \right) $
There is nothing special about the number 2, as long as the dimension and the exponent match. Here is the 5 by 5 example: $ A \; = \; \left( \begin{array}{rrrrr} 0 & 0 & 0 & 0 & x \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right) $ $ A^2 \; = \; \left( \begin{array}{rrrrr} 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & x \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array} \right) $ $ A^3 \; = \; \left( \begin{array}{rrrrr} 0 & 0 & x & 0 & 0 \\ 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & x \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{array} \right) $ $ A^4 \; = \; \left( \begin{array}{rrrrr} 0 & x & 0 & 0 & 0 \\ 0 & 0 & x & 0 & 0 \\ 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & x \\ 1 & 0 & 0 & 0 & 0 \end{array} \right) $ $ A^5 \; = \; \left( \begin{array}{rrrrr} x & 0 & 0 & 0 & 0 \\ 0 & x & 0 & 0 & 0 \\ 0 & 0 & x & 0 & 0 \\ 0 & 0 & 0 & x & 0 \\ 0 & 0 & 0 & 0 & x \\ \end{array} \right) $
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0Yes, perhaps in that case it would also work in any unitary ring. =) Good point. +1 – 2012-07-23
This is not true in general over an arbitrary field $F$. Consider $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix} $. Asking for $A$ such that $A^2 = B$ is equivalent to being given $B$ and solve the system of equations $ a^2 + bc = e \\\ b(a+d) = f \\\ c(a+d) = g \\\ d^2 + bc = h. $ Just consider a simple case : $f = g = 0$. Then either $a+d \neq 0$, which means $b=c=0$ and you want to solve $a^2 = e$, $d^2 = h$ ; in an arbitrary field, it is possible to choose $e$ and $h$ such that the polynomial $x^2 - e$ has no roots (i.e. there exists fields such that you can find $e$ and $h$ such that...). Choosing such $e$'s and $h$'s would lead to $a+d = 0$, which means $a = -d$, hence $a^2 = d^2$ and a solution would give $e=h$, a contradiction if you choose them distinct to begin with. So a matrix with no square root would simply be $B = \begin{pmatrix} e & 0 \\ 0 & h \end{pmatrix}$ where $x^2 - e$ and $x^2 - h$ are distinct irreducible polynomials in $F[x]$.
Example : Think of $B = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix}$.
Hope that helps,
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0I never said I agreed with Arturo. Perhaps @Arturo Madigin should shed some light on his comment? – 2012-07-23