Minimal polynomial for symmetric form and linear operator 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!
 A: 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.
