The Square Root of a Square Matrix

We can find the square root of a square matrix  
\[B\]
  if the matrix is similar to a matrix  
\[A^2\]
  having only diagonal entries ie if we can write  
\[A^2=P^{-1}BP\]
.
Having found the diagonal matrix  
\[A^2\]
,  
\[PAP^{-1}= \sqrt{B}\]
.
Example. Let  
\[B= \left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right) \]

The eigenvalues of  
\[B\]
  are the solutions to  
\[det(B- \lambda I)=0\]

\[\begin{equation} \begin{aligned} det( \left( \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right) -\left( \begin{array}{cc} \lambda & 0 \\ & \lambda \end{array} \right)) &= det (\left( \begin{array}{cc} 1- \lambda & 2 \\ 2 & 1- \lambda \end{array} \right) ) \\ &=(1- \lambda)^2 -4 \\ &= \lambda^2 -2 \lambda -3=(\lambda -3)(\lambda +1) =0 \end{aligned} \end{equation}\]

Hence  
\[\lambda = 3, \: -1\]
.
If  
\[\lambda =3, \: (B- \lambda I) \mathbf{v}_1 = \left( \begin{array}{cc} -2 & 2 \\ 2 & -2 \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}-2x+2y\\2x-2y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\]

Hence  
\[x=y=1, \: \rightarrow \mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix}\]

If  
\[\lambda =-1, \: (B- \lambda I) \mathbf{v}_2 = \left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right) \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2x+2y\\2x+2y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\]

Hence  
\[x=1, \: y=-1, \: \rightarrow \mathbf{v}_2 = \begin{pmatrix}1\\-1\end{pmatrix}\]

We can take  
\[P\]
  as the matrix with columns  
\[v_1 , v_2\]
.
\[P= \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \]

Then  
\[P^{-1}= - \frac{1}{2} \left( \begin{array}{cc} -1 & -1 \\ -1 & 1 \end{array} \right) \]

The matrix  
\[A^2\]
  will be a diagonal matrix with entries equal to the eigenvalues of  
\[B\]

\[A^2 = \left( \begin{array}{cc} -1 & 0 \\ 0 & 3 \end{array} \right) \rightarrow A = \left( \begin{array}{cc} i & 0 \\ 0 & \sqrt{3} \end{array} \right)\]

Then  
\[\sqrt{B} = \left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{cc} i & 0 \\ 0 & \sqrt{3} \end{array} \right) \frac{1}{2} \left( \begin{array}{cc} -1 & -1 \\ -1 & 1 \end{array} \right)= \frac{1}{2} \left( \begin{array}{cc} i + \sqrt{3} & i- \sqrt{3} \\ i- \sqrt{3} & i+ \sqrt{3} \end{array} \right) \]

Note that the matrix  
\[A\]
  is not unique. We can swap over the eigenvectors, hence the eigenvalues, take the negative square root of on or both eigenvalues. Either of these will result in a different square root.