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Let $A=\begin{pmatrix}2&3\\4&-2\end{pmatrix}$.

(i) Find an invertible matrix $P$ such that $P^{-1}AP$ is diagonal.

(ii) Find $A^n$ (for positive integers $n$).

(iii) Find four (complex) solutions to $X^2=A$. Show that there are no other solutions.

I'm having troubles with coming up with a solution to part (iii), an idea was to use:

$$ X^2 = A \iff P^{-1}X^2P = P^{-1}AP = D$$ and note that $P^{-1}X^2P = (P^{-1}XP)^2$, so we can solve $Y^2 = D$ to get 4 solutions $Y_1,Y_2,Y_3,Y_4$ i.e. $$\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 2i \end{pmatrix}$$ which then leads to four solutions for $X$: $$X = P\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 2i \end{pmatrix} P^{-1}.$$

I'm having trouble showing that they are no more solutions.

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3 Answers 3

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Hint : $X$ and $A$ must have the same 2 (1-dimensional) eigenspaces. Use the fact that the eigenvalues of $X$ are distinct.

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You just showed it.

You assumed that $X$ was a solution, denoted $Y=P^{-1}XP$ and proved that $Y$ must satisfy $Y^{2}=D$ hence got $4$ possibilities for $Y$.

Since $X=PYP^{-1}$ this proves that there are only $4$ choices for $X$

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I quite like the approach to finding $X$ which I learned from the paper Bernard W. Levinger: The Square Root of a $2\times2$ Matrix, Mathematics Magazine, Vol. 53, No. 4 (Sep., 1980), pp. 222-224, DOI: 10.2307/2689616. This paper is among references in the Wikipedia article Square root of a 2 by 2 matrix.

The basic idea is that we are able to find $\newcommand{\Tra}{\operatorname{Tr}}\Tra(X)$ and $\det(X)$. Thus we know the characteristic polynomial and from Cayley-Hamilton theorem we have $X^2-\Tra(X)X+\det(X)I=0$. If we plug $X^2=A$ into the previous equation, we might be able to express $X$ using $A$. (Although the computations will be somewhat different in the case $\lambda_1=\lambda_2$.) At least if you are trying to solve this by hand, not using some software which immediately gives you the Jordan normal form and the matrix $P$.

This works only for $2\times2$ matrices. But I think the calculations with this approach are easier then finding the Jordan form and the corresponding base-change matrix.


Let us denote by $\lambda_1$ and $\lambda_2$ the eigenvalues of the matrix $X$. We know that $A=X^2$ then has the eigenvalues $\lambda_1^2$ and $\lambda_2^2$.

Let us denote $d=\lambda_1+\lambda_2=\Tra(X)$ and $t=\lambda_1\lambda_2=\det(X)$. We know that $$\det(A)=\lambda_1^2\lambda_2^2 = d^2$$ and $$\Tra(A)=\lambda_1^2+\lambda_2^2 = (\lambda_1+\lambda_2)^2-2\lambda_1\lambda_2=t^2-2d.$$ Since the matrix $A$ is given, this is a system of equations with the unknowns $t$ and $d$.

For the given matrix $$A=\begin{pmatrix}2&3\\4&-2\end{pmatrix}$$
we have $\det(A)=-16$ and $\Tra(A)=0$. Hence \begin{align*} d^2&=-16\\ t^2-2d&=0 \end{align*} i.e. \begin{align*} d^2&=-16\\ t^2&=2d \end{align*} We have two possible values $d=\pm4i$. For each value of $d$ we can find two possibilities for $t$. $$ \begin{array}{|c|c|} \hline d & t \\\hline 4i & 2+2i \\\hline 4i &-2-2i \\\hline -4i & 2-2i \\\hline -4i &-2+2i \\\hline \end{array} $$

For any of the above numbers we have \begin{align*} X^2-tX+dI&=0\\ A-tX+dI&=0\\ X&=\frac{A+dI}t \end{align*}

This gives us four possible values for $X$. For example, for $t=2+2i$ and $d=4i$ we get \begin{align*} X &=\frac1{2+2i}\begin{pmatrix}2+4i&3\\4&-2+4i\end{pmatrix}\\ &=\frac{1-i}4\begin{pmatrix}2+4i&3\\4&-2+4i\end{pmatrix}\\ &=\frac14\begin{pmatrix}6+2i&3-3i\\4-4i&2+6i\end{pmatrix}\\ \end{align*} We can check whether this is indeed a solution of the given equation simply by squaring it. Here is computation in WolframAlpha.

The remaining three solutions can be found in a similar way, simply by plugging the remaining values of $d$ and $t$. It is also clear that if $X$ is a solution, then so is $-X$. So we can simply compute one matrix for $d=4i$ and one matrix for $d=-4i$ and then add the opposite matrices.

We have also seen that the matrices of this form are the only possible candidates for $X$. So we know that there are no other solutions.

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