Matrices

[Tolaso J Kos]

Let \( \displaystyle A_n =\begin{pmatrix}
n^2 &1 \\
-1& n^2
\end{pmatrix}, \; n \in \mathbb{N} \). Prove that \( x_n y_n >0 , \;\; \forall n \in \mathbb{N} \) where:

$$\prod_{i=1}^{n}A_i = \begin{pmatrix}
x_n &y_n \\
-y_n& x_n
\end{pmatrix}$$

[Papapetros Vaggelis]

Define \(\displaystyle{S=\left\{\begin{pmatrix}
a& b\\
-b& a
\end{pmatrix}\in\mathbb{M}_{2}(\mathbb{R}): a\,,b\in\mathbb{R}\right\}}\)

Obviously, \(\displaystyle{\mathbb{O}\,,I_{2}\in S}\) . Let

\(\displaystyle{A=\begin{pmatrix}
a_1&b_1 \\
-b_1& a_1
\end{pmatrix}\,,B=\begin{pmatrix}
a_2 &b_2 \\
-b_2& a_2
\end{pmatrix}\in S}\).

Then,

\(\displaystyle{A+B=\begin{pmatrix}
a_1+a_2&b_1+b_2 \\
-(b_1+b_2)& a_1+a_2
\end{pmatrix}\in S}\)

and

\(\displaystyle{A\cdot B=\begin{pmatrix}
a_1\,a_2-b_1\,b_2&a_1\,b_2+b_1\,a_2 \\
-(a_1\,b_2+b_1\,a_2)&a_1\,a_2-b_1\,b_2
\end{pmatrix}\in S}\)

Therefore, the set \(\displaystyle{S}\) is a subring of the ring \(\displaystyle{\left(\mathbb{M}_{2}(\mathbb{R}),+,\cdot\right)}\) .

We define \(\displaystyle{f:\mathbb{C}\longrightarrow S\,,a+i\,b\mapsto f(a+i\,b):=\begin{pmatrix}
a& b\\
-b& a
\end{pmatrix}}\)

This function is well defined. If \(\displaystyle{z_1=a_1+i\,b_1\,,z_2=a_2+i\,b_2\in\mathbb{C}}\) then

\(\displaystyle{\begin{aligned} f(z_1+z_2)&=f((a_1+a_2)+i\,(b_1+b_2))\\&=\begin{pmatrix}
a_1+a_2&b_1+b_2 \\
-b_1-b_2& a_1+a_2
\end{pmatrix}\\&=\begin{pmatrix}
a_1&b_1 \\
-b_1& a_1
\end{pmatrix}+\begin{pmatrix}
a_2 &b_2 \\
-b_2& a_2
\end{pmatrix}\\&=f(a_1+i\,b_1)+f(a_2+i\,b_2)\\&=f(z_1)+f(z_2)\end{aligned}}\)

and

\(\displaystyle{\begin{aligned} f(z_1\cdot z_2)&=f((a_1\,a_2-b_1\,b_2)+i\,(a_1\,b_2+a_2\,b_1))\\&=\begin{pmatrix}
a_1\,a_2-b_1\,b_2&a_1\,b_2+a_2\,b_1 \\
-(a_1\,b_2+a_2\,b_1)& a_1\,a_2-b_1\,b_2
\end{pmatrix}\\&=\begin{pmatrix}
a_1&b_1 \\
-b_1& a_1
\end{pmatrix}\cdot \begin{pmatrix}
a_2 &b_2 \\
-b_2& a_2
\end{pmatrix}\\&=f(a_1+i\,b_1)\cdot f(a_2+i\,b_2)\\&=f(z_1)\cdot f(z_2)\end{aligned}}\)

Finally, \(\displaystyle{f(1_{\mathbb{C}})=f(1+i\,0)=\begin{pmatrix}
1&0 \\
0& 1
\end{pmatrix}=I_2=1_{S}}\).

So, the function \(\displaystyle{f}\) is homomorphism and obviously

\(\displaystyle{f(\mathbb{C})=S\,,\rm{Ker}(f)=\left\{0\right\}}\) .

We have that \(\displaystyle{A_{n}\in S\,,\forall\,n\in\mathbb{N}}\) and thus \(\displaystyle{\prod_{i=1}^{n}A_{i}\in S\,,\forall\,n\in\mathbb{N}}\) .

Now,

\(\displaystyle{\forall\,n\in\mathbb{N}: \prod_{k=1}^{n}A_{i}=\prod_{k=1}^{n}f(k^2+i)=f\,\left(\prod_{k=1}^{n}(k^2+i)\right)}\)

and it is sufficient to prove that \(\displaystyle{Re\,\left(\prod_{k=1}^{n}(k^2+i)\right)\cdot Im\,\left(\prod_{k=1}^{n}(k^2+i)\right)>0}\) .

This is true because for every \(\displaystyle{m\,,t\in\mathbb{N}\,,m\neq t}\) holds

\(\displaystyle{(m^2+i)\cdot (t^2+i)=\left(m^2\,t^2-1\right)+i\,\left(m^2+t^2\right)}\) and

\(\displaystyle{\left(m^2\,t^2-1\right)\cdot \left(m^2+t^2\right)>0}\).

In conclusion, \(\displaystyle{x_{n}\cdot y_{n}>0\,,\forall\,n\in\mathbb{N}}\)