Square root of a diagonal matrix $\lambda I$ Could you help me prove that if $M \in \mathcal{M}_{2 \times 2}(\mathbb{R})$ satisfies $X^2=\lambda I$, $ \ \ \ \lambda \in \mathbb{R}, \ \  \lambda <0$, then there exist $y,z \in \mathbb{R}, \ \ yz \le \lambda$ and $X=\left[\begin{array}{ccc}\sqrt{\lambda -yz}&y\\z&-\sqrt{\lambda - yz}\end{array}\right]$ or $X=\left[\begin{array}{ccc}-\sqrt{\lambda -yz}&y\\z&\sqrt{\lambda - yz}\end{array}\right]$?
I know that if $\lambda \ge 0 \ \ X=\lambda I$ or $X=-\lambda I$
Let $X=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$, $X^2=\left[\begin{array}{ccc}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{array}\right]$.
So it appears I need to solve
$\begin{cases} a^2+bc=\lambda & \\ab+bd=0 & \\ ac+cd=0 & \\ bc+d^2= \lambda & \end{cases}$.
I know that $det(X^2) = det(X)^2$, so $(ad-bc)^2= \lambda$. But I don't know how to use it.
 A: By Cayley-Hamilton, $X^2-tr(X)X+\det(X)I=0$. By assumption $0=tr(X)=a+d$.
Then we have  $a^2=d^2=\lambda-bc$.
This gives, with $d=-a$, the two solutions. Note that $X$ need not be $\pm \lambda I$.
A: By identifying the four entries of $X^2$ and $\lambda I$ we get the four equations shown in the Question involving $a,b,c,d$.
Subtract the fourth equation from the first, and we deduce:
$$ a^2 - d^2 = (a+d)(a-d) = 0 $$
For the product of these two factors to be zero, either $a+d$ or $a-d$ must be zero.  Let's first consider $a+d = 0$, i.e. that $a = -d$. [Note: The second and third equations state that multiples of $(a+d)$ are zero, so in this case they give no new information.]
In the first equation we have:
$$ a^2 + bc = \lambda $$
so $a = \pm \sqrt{\lambda - bc}$. If you identify $y=b$ and $z=c$, then you have the two matrices $X$ shown in the Question's opening paragraph.  The only detail to fill in is that we can pick real $y,z$ s.t. $yz \le \lambda$ if $\lambda < 0$.  Of course we can do this in a number of ways.  You can either pick an arbitrary $y \gt 0$, then choose sufficiently negative $z \lt \lambda / y$, or else pick $z \gt 0$ and consequently negative $y \lt \lambda / z$.  The point is that if $yz \lt \lambda$, then $\sqrt{\lambda - yz}$ is the square root of a positive number, hence real and we can set $a = \pm \sqrt{\lambda - yz}$ as outlined.
Before switching entirely to the cases where $a-d$ is zero, let's take special notice of the boundary cases where both $a+d$ and $a-d$ are zero, i.e. where $a=d=0$.  In these cases we see (either from the first or fourth equations) that $\lambda = bc$.  Such matrices have zero $a,d$ on the diagonal but we are free to choose $b,c$ on the anti-diagonal simply to make their product $\lambda$, and incidentally some matrices of this kind where $\lambda \gt 0$ were overlooked in the Question.
That said, now consider $a \neq -d$, so that necessarily $a = d \neq 0$.  Observe that the second and third equations:
$$ (a + d)b = 0 $$
$$ (a + d)c = 0 $$
require that $b=c=0$, since the factor $a+d$ is nonzero.  These matrices only produce diagonal squares but with $\lambda \ge 0$, at least when (as here) real numbers are used.
