Is there any procedure to compute the root of a matrix equation??

Question

Actually I want to know the procedure, how could we calculate the value of $A$ where

$A^2$= $\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}$

Answer 1

Not every matrix has a square root. I assume that the given matrix is really the square of some $A$, and $A$ is real. The case that $A^2$ is diagonalizable over $\mathbb{R}$ should be easy and it is be omitted here.

If $A^2$ has a repeated real eigenvalue $\lambda^2$ with geometric multiplicity $1$, let $Q$ be an orthogonal matrix whose first column is a unit eigenvector of $A^2$. Then $Q^TA^2Q=\begin{pmatrix}\lambda^2&c\\0&\lambda^2\end{pmatrix}$ for some $c\in\mathbb{R}$. So you may take $A=Q\begin{pmatrix}\lambda&\frac{c}{2\lambda}\\0&\lambda\end{pmatrix}Q^T$.

If $A^2$ has a conjugate pair of nonreal eigenvalues, let $\lambda=re^{i\theta}$ be one of them and $u+iv$ be a corresponding eigenvector. Then we have $A^2u=r\cos(\theta)u-r\sin(\theta)v$ and $A^2v=r\cos(\theta)v+r\sin(\theta)u$. Hence $$ A^2 = rP\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}P^{-1}. $$ with $P=\begin{pmatrix}u&v\end{pmatrix}$ and we may take $$ A = \sqrt{r}P\begin{pmatrix}\cos\tfrac{\theta}2&\sin\tfrac{\theta}2\\-\sin\tfrac{\theta}2&\cos\tfrac{\theta}2\end{pmatrix}P^{-1}. $$ Note that the square root of a matrix is not unique in general. So the above only gives you a candidate square root.

Answer 2

$A=\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}$

You want to find $B=\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$ so that $B^2=A$

You can just calculate $B^2$ and solve for $a,b,c,d$ but a better way of doing it is to notice that $A$ commutes with $B$: $AB=B^2B=B^3=BB^2=BA$

From that, you can calculate $AB$ and $BA$ and since you know they are equal, it'll simplify $B$.

$AB=\begin{pmatrix} x & y\\ z & w\\ \end{pmatrix}\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}=\begin{pmatrix} xa+yc & xb+yd\\za+wc & zb+wd\\ \end{pmatrix}$

$BA=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{pmatrix} x & y\\ z & w\end{pmatrix}=\begin{pmatrix} xa+zb & ya+wb\\ xc+zd & yc+wd\end{pmatrix}$

$\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}=AB-BA=\begin{pmatrix} xa+yc & xb+yd\\za+wc & zb+wd \end{pmatrix}-\begin{pmatrix} xa+zb & ya+wb\\ xc+zd & yc+wd\end{pmatrix}=\begin{pmatrix} yc-zb & *\\* & zb-yc \end{pmatrix}$

So you get that $zb=yc$ which helps you simplify $B$:

• If $z=0=y$ then $A=\begin{pmatrix} x & 0\\ 0 & w \end{pmatrix}$ is diagonal so you can just take $B=\begin{pmatrix}a & 0\\ 0 & d \end{pmatrix}$ where $a^2=x$ and $d^2=2$

• If $y\not= 0$, you get $c=\cfrac{z}{y}b$ so $B=\begin{pmatrix} a & b\\ \cfrac{z}{y}b & d\\ \end{pmatrix}$

• If $z\not= 0$, you get $b=\cfrac{y}{z}c$ so $B=\begin{pmatrix} a & \cfrac{y}{z}c\\ c & d\\ \end{pmatrix}$

Then you just calculate $B^2$ and solve for $a,b,c,d$ (or the ones that remain in the expression of $B$).

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By the way there are some matrix that do no have a square root.

$A=\begin{pmatrix} 0 & 1\\ 0 & 0\\ \end{pmatrix}$

$y=1\not= 0$ so we have $B=\begin{pmatrix} a & b\\ \cfrac{z}{y}b & d\\ \end{pmatrix}=\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}$

Now $B^2=\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}\begin{pmatrix} a & b\\ 0 & d\\ \end{pmatrix}=\begin{pmatrix} a^2 & (a+d)b\\ 0 & d^2\\ \end{pmatrix}$

We want $B^2=A$, that is $\begin{pmatrix} a^2 & (a+d)b\\ 0 & d^2\\ \end{pmatrix}=\begin{pmatrix} 0 & 1\\ 0 & 0\\ \end{pmatrix}$

which implies $a=0=d$ and $1=(a+d)b=0b=0$ so you can not find a $B$ so that $B^2=A$