# prove that for any nonsingular matrix $A$ there exist $X$ such that $X^2=A$

Prove that given any matrix A, where $$\det(A)\neq0$$ $$A\in M_{n,n}(\mathbb C)$$ the following equation $$X^2=A$$ always has a solution. Should I do something with Jordan Normal form? Any help will be appreciated

• This is basically asking to prove that an LDL or a cholesky factorization exists for nonsingular matrices. (You need to define what you mean by $X^2$) – user144410 Mar 16 '18 at 0:14
• If $A$ is diagonalizable, this is easy (let us know if you want a hint). For the general case, I would go for the Jordan normal form and show that the complex square root of any Jordan matrix (i.e. almost diagonal) exists. Maybe reasoning by blocks – Tal-Botvinnik Mar 16 '18 at 0:29
• This has been answered before – YAlexandrov Mar 16 '18 at 1:07

Swear i have done this one fairly recently...

$$\left( \begin{array}{rr} t & \frac{1}{2t} \\ 0 & t \\ \end{array} \right)^2 = \left( \begin{array}{rr} t^2 & 1 \\ 0 & t^2 \end{array} \right)$$



$$\left( \begin{array}{rrr} t & \frac{1}{2t} & \frac{-1}{8 t^3} \\ 0 & t & \frac{1}{2t} \\ 0 & 0 & t \end{array} \right)^2 = \left( \begin{array}{rrr} t^2 & 1 & 0\\ 0 & t^2 & 1 \\ 0 & 0 & t^2 \end{array} \right)$$



$$\left( \begin{array}{rrrr} t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5} \\ 0 & t & \frac{1}{2t} & \frac{-1}{8 t^3}\\ 0 & 0 & t & \frac{1}{2t}\\ 0 & 0 & 0 & t \end{array} \right)^2 = \left( \begin{array}{rrrr} t^2 & 1 & 0 & 0\\ 0 & t^2 & 1 & 0 \\ 0 & 0 & t^2 & 1 \\ 0 & 0 & 0 & t^2 \end{array} \right)$$



$$\left( \begin{array}{rrrrr} t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5}& \frac{-5}{128 t^7} \\ 0 & t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5}\\ 0 & 0 & t & \frac{1}{2t} & \frac{-1}{8 t^3}\\ 0 & 0 & 0 & t & \frac{1}{2t} \\ 0 & 0 & 0 & 0 & t \\ \end{array} \right)^2 = \left( \begin{array}{rrrrr} t^2 & 1 & 0 & 0 & 0\\ 0 & t^2 & 1 & 0 & 0 \\ 0 & 0 & t^2 & 1 & 0\\ 0 & 0 & 0 & t^2 & 1 \\ 0 & 0 & 0 & 0 & t^2 \end{array} \right)$$

 And $$\sqrt{t^2 + 1} \; \; = \; \; t \; \; \sqrt{1 + \frac{1}{t^2}} \; \; = \; \; t + \frac{1}{2t} - \frac{1}{8 t^3} + \frac{1}{16 t^5} -\frac{5}{128 t^7} + \frac{7}{256 t^9} -\frac{21}{1024 t^{11}} \cdots$$

The resemblance is not cosmetic or accidental. We have, in a Jordan block of size $n,$ an identity matrix $I$ and a nilpotent matrix $N$ with $N^n=0.$ We are asking for $\sqrt{t^2 I + N}.$ As with other real analytic functions, we can use the facts that $IN=NI$ commute to give a power series for the resulting matrix, and the series is finite because $N$ is nilpotent.

• I see, this proves that any Jordan block has a square root right? – Tal-Botvinnik Mar 16 '18 at 1:15
• Very cool, thanks – Tal-Botvinnik Mar 16 '18 at 1:19
• @Tal-Botvinnik yes, and gives an explicit construction. I watched Spassky-Fischer on television when I was a kid. I guess from Iceland, it was just a few hours earlier in the suburbs of New York – Will Jagy Mar 16 '18 at 1:19
• Wow, did you remember the infamous game 1 and that trapped bishop? – Tal-Botvinnik Mar 16 '18 at 1:21
• It might be interesting to prove that if $t=0$ is a possibility, then not every matrix has a square root. For example, prove that $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ is not the square of any $2 \times 2$ matrix. – Daniel Schepler Mar 16 '18 at 1:26

I would like to present here a little different approach to the problem without expanding the function into series (in reference to very compact Will's answer). Calculation of the square root can be made with the use of matrices which I'll call shifted scalar matrices and consequently only basic operations on matrices i.e. addition and multiplication are needed.

Let shifted identity matrix $M_k$ be a matrix with $1's$ on one of its overdiagonal where $k$ is a shift of this overdiagonal to the upper right corner direction from the main diagonal. The number $k$ is positive here.

For example
the shifted identity matrix $M_2$ (dimension $4 \times 4$).

$\begin {bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$

At this case the shifted scalar matrix can be written as $x_2M_2=\begin {bmatrix} 0 & 0 & x_2 & 0 \\ 0 & 0 & 0 & x_2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$.

Identity matrix is written as $I=M_0$ in this notation.

Shifted scalar matrices are easy for multiplication (which is commutative for them) - we have formula for $n \times n$ matrices:

• if $i+j < n$ then $\ \ (x_iM_i)(x_jM_j)=x_ix_jM_{i+j}$
• if $i+j \geq n$ then $\ \ (x_iM_i)(x_jM_j)=0$

From this multiplication table for shifted scalar matrices stems

$$\begin{array}{c|c|c } & x_0M_o & x_1 M_1 & x_2 M_2 & \dots\\ \hline x_0 M_o & x_0^2 M_0 & x_0x_1 M_1 & x_0x_2 M_2 & \dots\\ \hline x_1M_1 & x_1x_0 M_1 & x_1^2 M_2 & x_1x_2 M_3 & \dots\\ \hline x_2 M_2 & x_2x_0 M_2 & x_2x_1 M_3 & x_2^2 M_4 & \dots\\ \hline \dots & \dots & \dots & \dots & \dots\\ \end{array}$$

Assume that for Jordan cell $\begin {bmatrix} t^2 & 1 & 0 & 0 \\ 0 & t^2 & 1& 0 \\ 0 & 0 & t^2 & 1 \\ 0 & 0 & 0 & t^2\\ \end{bmatrix}$ the matrix of the square root has the form $X=\begin {bmatrix} x_0 & x_1 & x_2 & x_3 \\ 0 & x_0 & x_1& x_2 \\ 0 & 0 & x_0 & x_1 \\ 0 & 0 & 0 & x_0\\ \end{bmatrix} = x_0M_0+x_1M_1+x_2M_2+x_3M_3$

and its square $X^2=(x_0M_0+x_1M_1+x_2M_2+x_3M_3)^2$
On the other hand $A_J=t^2M_0+1\cdot M_1+0\cdot M_2+0\cdot M_3$.

Then from multiplication table we have following values of $x_0,x_1,x_2,x_3$
which can be presented in the vector form

$\begin {bmatrix} x_0^2 \\ 2x_0x_1 \\ 2x_0x_2+x_1^2 \\ 2x_0x_3+2x_1x_2\\ \end{bmatrix} = \begin {bmatrix} t^2 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix}$

For higher dimensions we can extend these vectors taking coefficients from extended multiplication table ...

$\begin {bmatrix} \color{red}{x_0}^2 \\ 2x_0\color{red}{x_1} \\ 2x_0\color{red}{x_2}+x_1^2 \\ 2x_0\color{red}{x_3}+2x_1x_2\\ 2x_0\color{red}{x_4}+2x_1x_3+x_2^2\\ \dots \\ \end{bmatrix} = \begin {bmatrix} t^2 \\ 1 \\ 0 \\ 0 \\ 0 \\ \dots\\ \end{bmatrix}$ but initial part is unchanged.

As we see every $x_k$ can be calculated from components of these vectors if the values $x_0\dots x_{k-1}$ are known. The pattern of coefficients it's easy to identify.
In fact every $x_k$ satisfies the formula $a_kx_k+b_k=0$ where $a_k=2x_0$ and $b_k=f(x_1, \dots, x_{k-1})$, solution always exists $x_k=-\dfrac{b_k}{a_k}$ if $x_0 \neq 0$.

Calculations for the square root provide

• $x_0=t$ $\ \ \ \ (-t)$

• $2tx_1=1 \ \ \Rightarrow \ \ x_1= \dfrac{1}{2t}$

• $2tx_2+ \left(\dfrac{1}{2t}\right)^2=0 \ \ \Rightarrow \ \ x_2= -\dfrac{1}{8t^3}$

• $2tx_3+2\left(\dfrac{1}{2t}\right)\left(-\dfrac{1}{8t^3}\right)=0 \ \ \Rightarrow \ \ x_3= \dfrac{1}{16t^5}$

as in Will's answer. Btw comparing two methods it is interesting that we could find expansion of function into series alternatively by basic (recursive) operations on matrices.

Procedure can be continued for higher dimensions in similar fashion. Conclusion: always it is possible to calculate the square root of invertible $( \det (A) \neq 0)$ matrix in complex field.

It's worth to notice that the described above procedure can be extended for other types of equations, not only $X^2=A \ \$...