# For every matrix $A\in M_{2}( \mathbb{C})$ there's $X\in M_{2}( \mathbb{C})$ such that $X^2=A$?

True\False?

For every matrix $A\in M_{2}( \mathbb{C})$ there's $X\in M_{2}( \mathbb{C})$ such that $X^2=A$.

I know that every complexed matrix has a Jordan form matrix $J$ such that $P^{-1}CP=J$, But it's not diagonalizable for sure.

Thanks

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Assume A^2=0 but A is not 0. Then if A=X^2, X^4=0, hence... –  Did Aug 13 '11 at 20:42
Hence all the eigenvalues of $X=0$, and then $X^2=0$. Very nice. –  Beni Bogosel Aug 13 '11 at 21:13

The $2$-by-$2$ matrices with no square roots are precisely those with Jordan form $\begin{bmatrix}0&1\\0&0\end{bmatrix}$. In general, every Jordan block corresponding to the eigenvalue $0$ has to have size $1$ in order for a square root to exist. @Did indicates an elegant way to see this in a comment.

The Wikipedia article on matrix functions indicates how functions can be applied to Jordan blocks provided sufficiently high order derivatives are defined at the eigenvalues, and this includes square root functions. For further details, a wonderful reference is Higham's Functions of matrices: theory and computation. In particular, a sufficient condition for the existence of square roots is that every Jordan block corresponding to the eigenvalue $0$ has size $1$.

Thanks to Marc for pointing out errors in the old version. See his answer.

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why do you consider only 0 to be an eigenvalue? –  user6163 Aug 13 '11 at 20:59
@Nir: I'm leaving some details for you to work out, but what I'm saying is that $0$ is the only eigenvalue that can possibly cause a problem. You can always find square roots of Jordan blocks with nonzero eigenvalues, and I recommend that you check this. The Wikipedia article should help suggest the form if you get stuck. –  Jonas Meyer Aug 13 '11 at 21:08
The second sentence "In general..." is false: $$\left(\begin{matrix}0&0&1\\0&0&0\\0&1&0\end{matrix}\right)^2=\left(\begin{matrix‌​}0&1&0\\0&0&0\\0&0&0\end{matrix}\right).$$ Damn, comments won't typeset matrices correctly. –  Marc van Leeuwen Apr 1 '13 at 15:31
@Marc: Thank you. $\begin{pmatrix}0&0&1\\0&0&0\\0&1&0\end{pmatrix}^2=\begin{pmatrix}0&1&0\\0&0&0\\0&0&‌​0\end{pmatrix}$ –  Jonas Meyer Apr 2 '13 at 4:47

As I was recommended , I figured out the answer, with the help of @Did and @JonasMeyer and here is my full explanation.

If A is diagonalizable, there's nothing to prove, cause we are under $\mathbb C$, so $X$ will be simply the roots of A's eigenvalues in the diagonal.

Let's assume that A is not diagonalizable, we know that every matrix under $\mathbb C$ has a Jordan form, so one of the jordan form that comes to the head is this nilpotent matrix : $A=\begin{pmatrix} 0 &1 \\ 0&0 \end{pmatrix}$ . Let's assume that there's $X$ so that $X^2=A$, and $A^2=0$, so $X^4=0$, so $X$ is nilpotent as well, but it can't be $0$ , so it has to implies $X^2=0$, But $X^2=0\neq A$, Contradiction.

So We find one matrix that does not have such $X$.

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This is excellent practice (for you and for the site) to show the proof you arrived at after the inputs from others on this page. Your proof is mostly correct, I will add two remarks. First, one cannot write that A=J with J=[0,1|0,0], but only A=P^{-1}JP for a given invertible P (and the remaining steps of the proof are easily adapted). Second, when you write that X nilpotent implies X^2=0, you might add the reason why. –  Did Aug 14 '11 at 16:29
I'll add a line to clarify the second remark of @Did Since $X$ is $2 \times 2$, its characteristic polynomial is of degree two, and characteristic polynomial of a matrix annihilates the matrix, Since $X$ is also nilpotent, $f(x)=x^2$ is the characteristic polynomial.Hence $X^2=0$ –  the8thone Aug 18 '14 at 17:54

This answer is just to indicate that in Jonas Meyer's answer the second sentence "In general..." is not correct: $$\left(\begin{matrix}0&0&1\\0&0&0\\0&1&0\end{matrix}\right)^2=\left(\begin{matrix}0&1&0\\0&0&0\\0&0&0\end{matrix}\right).$$ I mentioned this as comment, but the matrices won't typeset correctly in a comment.

Now that I'm writing an answer anyway, I'll add that for any other $2\times2$ non-diagonalisable matrix, which one can write as $\lambda(I_2+N)$ with $\lambda\neq0$ and $N^2=0$, one can explicitly get a square root $\sqrt\lambda(I_2+\frac12N)$, where $\sqrt\lambda$ is one of the two complex square roots of$~\lambda$. See this answer for the general case of square roots of complex square $n\times n$ matrices.

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The reason why the matrices were not shown correctly in your comment might be the problem explained in this post on meta: Too long LaTex formulae not shown in comments? –  Martin Sleziak Feb 10 '14 at 13:01