On an Equivalence between two statements involving bilinear forms and simmetrical matrices. Let $V$ be a $K$-vector space with $\dim_K(V) = n \ge 1$.
The statement I would like to prove is the following
Let $b$ be a bilinear symmetrical form on $V$ then in $V$ there is an orthogonal basis for $b$  $\iff$ Every symmetrical matrix $A \in M_n(K)$
is congruent to a diagonal matrix.
A link to a proof is fine if the proof is long to write down.
 A: There are some assumptions, for instance about $K$, missing from the statement of the problem and I try to mention them below. 
The matrix $A$ of a bilinear form $b$ with respect to a basis, not necessarily orthogonal, is symmetric, where:
$$
b(x,y)=y^tAx
$$
and $A_{ij}$ is $b(u_i,u_j)$, namely the inner product of two basis vectors $u_i$ and $u_j$. If the basis of the bilinear form $b$ is orthonormal then its matrix is diagonal. 
If there is an orthogonal basis for each form $b$, then each symmetric matrix $A$ is congruent to a diagonal matrix. This follows simply by changing the basis through the matrix $P$ to the orthogonal basis. 
For the reverse direction, if every symmetric matrix is congruent to a diagonal matrix, i.e. $D=PAP^T$, then for each symmetric bilinear form with matrix $A$, the matrix $P$ changes the basis of the form to an orthogonal basis and hence there is an orthogonal basis.

To give a rapid proof of the L.H.S, which is not necessary for the OP, we sketch that there is an orthogonal basis for finite dimensional vector space with symmetric bilinear form on it defined over field $K$ under certain condition. For the field $K$ with the characteristic not 2 or for the field of characteristic 2 and a non-alternate from $b$, the from is either fully zero or there is a non-null vector $v_1$ in $V$. From direct summand theorem, $V$ is the direct sum of the space spanned by $v_1$ and its orthogonal complement. Now by induction over the dimension, an orthogonal basis can be found for $V$.
