Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
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 $$ \begin{pmatrix} 1&1\\1&1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} -1&-1\\-1&-1 \end{pmatrix} $$

share|cite|improve this answer
    
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.

share|cite|improve this answer
1  
...Which often works...so 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
3  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.