I'm hoping to get help with the following problem.
Let $V$ be a vector space over $\mathbf{R}$ with $\mathrm{dim}~V = n < \infty$. Let $(\cdot, \cdot): V \times V \to \mathbf{R}$ be a positive definite symmetric form, and suppose the linear map $f: V \to V$ satisfies $(f(u), v) = (u, f(v))$ for each $u, v \in V$. Show that the minimal polynomial of $f$ splits.
I am stuck on this. I know that the minimal polynomial of $f$ splits with distinct roots if and only if the matrix of $f$ is diagonalizable. So I tried to use this fact. I have shown that $(\cdot, \cdot)$ can be represented as the identity matrix in a basis $e_1, \dots, e_n$ with $(e_i, e_j) = \delta_{ij}$, but now I'm stuck.
Edit: I've updated the comment above to clarify the equivalence holds iff the roots are distinct. This is an additional assumption not given in the problem, so I may not be going down the right track. Any hints would be helpful!
First answer, without tensor products.
Suppose the minimal polynomial of the linear transformation did not split. For my comfort, I'm going to use $T$ for the linear transformation and $m_T(t)$ for it's minimal polynomial. Then $m_T(t)$ factors into some number of linear factors and at least one irreducible quadratic factor. Let $t^2 + 2at+b$ be this quadratic factor. Since the quadratic factor is irreducible, we have that its discriminant, $d=4a^2-4b$, is negative. Let $\delta = -d/4$, so that $\delta = b-a^2$ is positive. Now we can complete the square to rewrite $m_T(t)$ as $(t+a)^2 +(b-a^2)=(t+a)^2+\delta$. Then we have that for any $u\ne 0\in V$, $$ (((T+a)^2 + \delta) u,u) = ((T+a)^2 u,u) + \delta(u,u)$$ $$ = ((T+a)u,(T+a)u) + \delta(u,u)=\|(T+a)u\|^2+\delta\|u\|^2>0.$$
However, since $t^2+at+b$ is a factor of the minimal polynomial of $T$, $T^2+aT+b$ must not be invertible (otherwise we could find a smaller polynomial which is 0 when evaluated at $T$ by removing this factor), so there is some $u\ne 0 \in V$ for which $(T^2+aT+b)u=0$, but then $((T^2+aT+b)u,u)=0$, which is a contradiction of the above inequality. Hence the minimal polynomial of $T$ cannot contain any irreducible quadratic factors.
Second answer, with tensor products.
We can make this into a question about complex vector spaces and Hermitian operators. Let $\newcommand{\tens}{\otimes}\newcommand{\CC}{\mathbb{C}}W=\CC\tens V$. Now we need to extend our original bilinear form, which I will call $B$, to a Hermitian form, $\phi$, defined on $W\times W$. This can be done by defining $\phi(\alpha \tens v, \beta \tens w)= \alpha\overline{\beta} B(v,w)$ and extending by linearity. Then if $A : V\to V$ was our old linear operator, we can extend $A$ to $W$ by tensoring with $1=\newcommand{\id}{\text{id}}\id_\CC$, to get $T=1\tens A : W\to W$. Now $T$ is Hermitian, since $$\phi(T(\alpha \tens v),\beta \tens w) = \phi(\alpha \tens Av,\beta \tens w) = \alpha\overline{\beta} B(Av,w) =\alpha\overline{\beta} B(v,Aw) = \phi(\alpha \tens v,\beta \tens Aw) =\phi(\alpha \tens v,T(\beta \tens w)).$$ Now let $\lambda\in\CC$ be an eigenvalue of $T$ with eigenvector $v$, then $\lambda\phi(v,v)=\phi(\lambda v,v)=\phi(Tv,v)=\phi(v,Tv)=\phi(v,\lambda v)=\overline{\lambda}\phi(v,v)$. Hence $\lambda = \overline{\lambda}$, so all of $T$'s eigenvalues are real. Hence the minimal polynomial of $T$ has real roots, and hence is a real polynomial.
Then $m_T(A)$ makes sense, and we want to show that it's zero so that we can say that $m_A(t)\mid m_T(t)$, and therefore say that all of the roots of $m_A(t)$ are real (hence $m_A(t)$ splits). Suppose $m_T(A)\ne 0$, then $m_T(A) u\ne 0$ for some $u\in V$. However then $0=m_T(T)(1\tens u) = m_T(1\tens A) (1\tens u) = (1\tens m_T(A))(1\tens u) = 1\tens (m_T(A)u) \ne 0$, which is a contradiction. Therefore $m_T(A)=0$, so $m_A(t)$ has all of its roots in the reals. Hence $m_A(t)$ splits.