Eigenvalue of symmetric matrix with rational coefficients

Question

I'm looking for a 2×2 symmetric matrix with rational coefficients, such as √2 is eigenvalue. My idea is to find a 2×2 symmetric matrix such as 2 is eigenvalue, and then take the square root of this matrix. However, I get some troubles because the square root matrix is not with rational coefficients...

Bonus: Can you prove that √3 is never an eigenvalue for a 2×2 symmetric matrix with rational coefficients? :)

Answer 1

• Find a $2\times 2$ symmetric matrix of rational elements with eigenvalue $\sqrt2$.
• Prove that a $2\times 2$ symmetric matrix of rational elements can't have eigenvalue $\sqrt3$.

Consider a $2\times 2$ symmetric matrix $$\begin{pmatrix}a&c\\c&b\end{pmatrix}$$ with $a,b,c\in \mathbb{Q}$.

For it to have $\lambda$ as an eigenvalue, \begin{align*} (a-\lambda)(b-\lambda)-c^2&=0\\ \lambda^2-(a+b)\lambda+ab-c^2&=0\\ \lambda&=\frac{(a+b)\pm\sqrt{(a-b)^2+4c^2}}2\\ \text{Now}\qquad a+b&=0 &(\because \sqrt 2\text{ is one of the roots})\\ \text{or}\qquad a&=-b\\ \Rightarrow\sqrt{a^2+c^2}&=\sqrt2\end{align*} Choose $a=\pm1$, $b=\mp1$, $c=1$ so that the matrix becomes $$\begin{pmatrix}\pm1&1\\1&\mp1\end{pmatrix}$$ As for getting $\lambda=\sqrt3$, $\sqrt{a^2+c^2}=\sqrt3$, put $a\rightarrow \sqrt3\sin\theta$ and $b\rightarrow \sqrt3\cos\theta$. Both $a$ and $b$ can't be rational together since sine and cosine don't become $\pm\frac1{\sqrt3}$ together but do $\pm\frac1{\sqrt2}$. I believe that a rigorous reason can be got by drawing lines cutting graphs of sine and cosine. Try yourself!