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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.

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You have infinite many solutions... what do you think the correct solutions are? $B = \begin{pmatrix} 1 && 1 \\ 1 && 1\end{pmatrix}$ is a solution for your equations and for the problem, is this correct or you need a specific kind of solution? – N74 Feb 17 at 14:13
I agree that's a solution. However, I want to conclude that from the equations I obtained in my approach. Am I thinking about this in the right way? – Kermit the Hermit Feb 17 at 14:16
up vote 9 down vote accepted

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 $$ \begin{pmatrix} 1&1\\1&1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} -1&-1\\-1&-1 \end{pmatrix} $$

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I understand now. Your time and effort is much appreciated. – Kermit the Hermit Feb 17 at 14:38

Here is a trick which often works in cases like this:

If $B^2=A$ then $$AB=B^3=BA$$

So, if your $B$ is $$B=\begin{bmatrix} a & b \\c &d\end{bmatrix}$$

We have $$\begin{bmatrix} 2 & 2 \\2 &2\end{bmatrix}\begin{bmatrix} a & b \\c &d\end{bmatrix}=\begin{bmatrix} a & b \\c &d\end{bmatrix}\begin{bmatrix} 2 & 2 \\2 &2\end{bmatrix} \\ \begin{bmatrix} 2a+2c & 2b+2d \\2a+2c &2b+2d\end{bmatrix}=\begin{bmatrix} 2a+2b & 2a+2b \\2c+2d &2c+2d\end{bmatrix} $$ which gives $$a=d \\ b=c$$

Therefore you seek a matrix of the form $$B=\begin{bmatrix} a & b \\b &a\end{bmatrix}$$

Now solving $$\begin{bmatrix} a & b \\b &a\end{bmatrix}^2=\begin{bmatrix} 2 & 2 \\2 &2\end{bmatrix}$$ is easy.

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...Which often is there a situation where it doesn't work? – imranfat Feb 17 at 14:42
This is a really cool solution, far more elegant than my approach. – Kermit the Hermit Feb 17 at 14:49
@imranfat If $A=cI_n$ then the fact that $B$ commutes with $A$ given no information. Anyhow, if $A$ is $n\times n$ and the eigenvalues of $A$ are distinct (which is typically the case), then the space of matrices which commutes with $A$ is $n$ dimensional. THis means that $AB=BA$ reduces $B$ from $n^2$ to $n$ parameters. – N. S. Feb 17 at 15:24
Very interesting... – imranfat Feb 17 at 16:42

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