Looking for a symmetric matrix

Question

Do you know a method to find a particular $2 \times 2$ symmetric matrix $M$ with rational coefficients knowing that $\lambda = \sqrt 2$ is an eigenvalue of $M$?

Many thanks.

Answer 1

Let's go for a $2 \times 2$ matrix. If $A$ has rational coefficients it's minimal polynomial $\mu$ will have rational coefficients, and since $\sqrt{2}$ is an eigenvalue of $A$ so must $-\sqrt{2}$ be. That means $\mu = x^2-2$ from which we get $\mathrm{det}(A)=-2$ and $\mathrm{tr}(A)=0$. Moreover, A is symmetric so of the form $$\begin{pmatrix} a & b \\b & c\end{pmatrix}$$ and our conditions imply $c=-a$ and $-a^2-b^2=-2$ which gives us a possible solution $$\begin{pmatrix} 1 & 1 \\1 & -1\end{pmatrix}$$

Now if $n$ is larger than $2$ simply make it a block diagonal matrix with your favorite symmetric, rational matrix of size $(n-2) \times (n-2)$

Answer 2

You may take $M=\pmatrix{1&1\\ 1&-1}\oplus I_{n-2}$.