Term between "orthogonal" and "orthonormal" If we have a bilinear form $(-, -)$ on a vector space $V$, a basis $\beta$ of $V$ is said to be orthogonal if $(v, w) = 0$ for every $v \neq w \in \beta$, and orthonormal if moreover $(v, w) = 1$ when $v = w$.
What if we relax the latter condition to $(v, v) \neq 0$ for $v \in \beta$?  Is there a name for that?  I've tried looking in the usual places and Googling but didn't come up with anything.
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To summarize, this is a property of the form, and not the basis. If there is one orthogonal basis satisfying this, then all of them do.

Your definition of orthogonal basis implies that if such a basis exists then $({-},{-})$ is symmetric, so I will assume out right that it is.
In characteristic 2, $({-},{-})$ need not have a orthogonal basis. We will ignore this case, as it is quite different from when the characteristic is not 2.
When the characteristic is not 2, then $({-},{-})$ always has an orthogonal basis $\beta$. Moreover, the number of basis elements $v \in \beta$ such that $(v,v) \not= 0$ is independent of $\beta$, and is called the rank of $({-},{-})$. So a symmetric bilinear form with $(v,v) \not= 0$ for all $v \in \beta$ has full rank, or more commonly is non-degenerate.
This is distinct from $({-},{-})$ being anisotropic. $({-},{-})$ is isotropic if there is $0\not= v \in V$ such that $(v, v) = 0$, and is anisotropic otherwise. But consider the form
$$
  ((x, y), (x', y')) = xx' - yy'
$$
over $\mathbb R^2$. $\{(1, 0), (0, 1)\}$ is an orthogonal basis for this form, and
$$
  ((1,0), (1, 0)) = 1,\quad ((0,1),(0,1)) = -1
$$
so the form is non-degenerate. But
$$
  ((1,1), (1,1)) = 1 - 1 = 0.
$$
Of course, any degenerate form is isotropic.

A good reference is Classical Groups and Geometric Algebra (2002) by Larry Grove.
