Proving $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$ using matrices I am trying to show that $\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2}+\sqrt{5})$. I have calculated the minimal polynomial $x^4-14x^2+9$ and I know that its basis as a $\mathbb{Q}$-vector space is $\{1, 
\sqrt{2}, \sqrt{5}, \sqrt{10}\}$. But I'm not sure how to construct an invertible matrix to express the basis as rational polynomials. I'm thinking of Fly by Night's comment here

As Mariano notes in the comments, there's a much quicker argument using degrees, but here's the matrix argument. (A nice bonus of this argument is that you can explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$.)
Let $\alpha = \sqrt{2} + \sqrt{5}$. Note that $\alpha \in \mathbb{Q}(\sqrt{2}, \sqrt{5}),$ so immediately we have $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{5})$. Suppose there exists an invertible rational $4 \times 4$ matrix $A$ such that
$$\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix} = A \begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix}.$$
Then, we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ which would prove that each of the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ can be expressed as a $\mathbb{Q}$-linear combination of powers of $\alpha$, and thus we must have $\mathbb{Q}(\sqrt{2}, \sqrt{5}) \subseteq \mathbb{Q}(\alpha)$. Since we already have inclusion the other way, we would then be done.
So, we just need to find such an $A$. Well, there's only one real candidate for what $A$ can be: the $i$th row of $A$ must be $\begin{pmatrix}a & b  & c & d \end{pmatrix}$, where $a + b\sqrt{2} + c\sqrt{5} + d\sqrt{10}$ is the unique way to write $\alpha^{i}$ as a $\mathbb{Q}$-linear combination of the basis elements $\{1, \sqrt{2}, \sqrt{5}, \sqrt{10}\}.$ (This is unique because $\alpha^{i} \in \mathbb{Q}(\sqrt{2}, \sqrt{5}).$)
Let's compute:
$$\alpha = 0 \cdot 1 + 1 \cdot \sqrt{2} + 1 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{2} = 7 + 2\sqrt{10} = 7 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 2 \cdot \sqrt{10},$$
$$\alpha^{3} = 17\sqrt{2} + 11\sqrt{5} = 0 \cdot 1 + 17 \cdot \sqrt{2} + 11 \cdot \sqrt{5} + 0 \cdot \sqrt{10},$$
$$\alpha^{4} = 89 + 28\sqrt{10} = 89 \cdot 1 + 0 \cdot \sqrt{2} + 0 \cdot \sqrt{5} + 28 \cdot \sqrt{10}.$$
So, our candidate for $A$ is
$$A = \begin{pmatrix} 0 & 1 & 1 &0 \\ 7 & 0 & 0 & 2 \\ 0 & 17 & 11 & 0\\ 89 & 0 & 0 &28 \end{pmatrix}.$$ A quick computation shows that this matrix has nonzero determinant, hence is invertible. At this point, we are already done, but we can actually invert $A$ to explicitly express the basis elements for $\mathbb{Q}(\sqrt{2}, \sqrt{5})$ in terms of powers of $\alpha$,
A more laborious computation (just kidding, I used a calculator) finds the inverse to be
$$A^{-1} = \begin{pmatrix} 0 & \frac{14}{9} & 0 & - \frac{1}{9} \\ -\frac{11}{6} & 0 & \frac{1}{6} & 0 \\ \frac{17}{6} & 0 & -\frac{1}{6} & 0\\ 0 & -\frac{89}{18} & 0 & \frac{7}{18} \end{pmatrix}.$$
Since we have
$$\begin{pmatrix}1 \\ \sqrt{2} \\ \sqrt{5} \\ \sqrt{10} \end{pmatrix} = A^{-1}\begin{pmatrix}\alpha \\ \alpha^{2} \\ \alpha^{3} \\ \alpha^{4}\end{pmatrix},$$ it follows that
$$1 =\frac{14}{9} \alpha^{2} - \frac{1}{9} \alpha^{4},$$
(Look! It's the minimal polynomial!)
$$\sqrt{2} = -\frac{11}{6}\alpha  + \frac{1}{6} \alpha^{3},$$
$$\sqrt{5} = \frac{17}{6}\alpha - \frac{1}{6} \alpha^{3},$$
$$\sqrt{10} = -\frac{89}{18}\alpha^{2} + \frac{7}{18}\alpha^{4}.$$
