# Scalar Matrix over a field

Show that a 2x2 scalar matrix $A$ over a field has a square root. Meaning a matrix say $B$ that satisfies $B^{2}$ = $A$.

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When you say "show that a $2\times 2$ scalar matrix" do you mean every matrix? or just the existence of one such matrix, e.g. $\pmatrix{4 & 0 \\ 0 & 4}?$ –  user2468 Jul 23 '12 at 4:19
Hm. I think the meaning of "scalar matrix" tricked me. Perhaps OP should tell me if I misunderstood the meaning of that in my answer. –  Patrick Da Silva Jul 23 '12 at 4:25
@ArturoMagidin, we do not understand your hint. In $\mathbb Q,$ both $2,-2$ fail to be squares. –  Will Jagy Jul 23 '12 at 5:15
@WillJagy: I was being dumb. –  Arturo Magidin Jul 23 '12 at 15:08

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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Yes, perhaps in that case it would also work in any unitary ring. =) Good point. +1 –  Patrick Da Silva Jul 23 '12 at 4:51

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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I think "scalar matrix" means a field element times the identity matrix. I did not notice the word scalar either. –  Will Jagy Jul 23 '12 at 4:25
@Will Jagy : Well in that case, as you mentioned, if the field lacks square roots of some elements we can exhibit a counter example. It wouldn't be that hard. Let's wait and see if we answered OP's question. –  Patrick Da Silva Jul 23 '12 at 4:26
Arturo's comment settles the simplest case, but is not something I knew. Is it true that, in any field with any element $x,$ either $x$ or $-x$ is a square? No, false for the rationals. –  Will Jagy Jul 23 '12 at 4:31
@Will Jagy : That's not what Arturo meant. He meant that every element of a field $y$ is either of the form y=x^2$or$y=-x^2$. – Patrick Da Silva Jul 23 '12 at 4:41 Not to hijack this question, but I actually don't understand what Arturo meant. How is$2 \in \mathbb{Q}\$ of either form? –  WacDonald's Jul 23 '12 at 4:48