Find two square roots a 2 by 2 matrix filled with twos Given matrix $A = \begin{pmatrix} 2 && 2 \\  2 && 2\end{pmatrix}$, I want to find two square roots of A.
I have to go about this with only very introductory-type tools, those covered in an introductory matrix operations chapter.



*

*My Approach


*

*Since I know that the square root matrix is a 2x2 matrix, let the square root matrix be $B = \begin{pmatrix} a && b \\  c && d\end{pmatrix}$.

*Now, for B to be a square root matrix of A, the following must hold true: BB = A. Evaluating BB I get $BB = \begin{pmatrix} a^2 + bc && b(a + d) \\  c(a+d) && d^2 + bc\end{pmatrix}$

*This leaves me with the following equations:


*

*$a^2 + bc=2$

*$d^2 + bc=2$

*$b(a+d)=2$

*$c(a+d)=2$


*From here on I've tried solving the equations but none of my attempts yielded the correct solution. I get the feeling I'm overlooking something very basic.

Note that line 1 and 2 give $a^2 = 2-bc = d^2$ and line 3 and 4 give $b = 2/(a+d) = c$. Now since $b(a+d) =2\neq0$ we can't have $a=-d$, thus $a=d$. Therefore our four equations simplify to 


*

*$a^2+b^2=2$

*$2ab=2$


Rewriting the second we get $b = 1/a$ and substituting in the first gives $a^2 +a^{-2} = 2$ or $a^4-2a^2+1=0$. The two real solutions to this polynomial are $a=\pm 1$. We conclude that the two square roots of $A$ are
$$
\begin{pmatrix}
1&1\\1&1
\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}
-1&-1\\-1&-1
\end{pmatrix}
$$
