Intution about definiteness of a bilinear form, and relation to choice of basis We define for a bilinear symmetric Form on a vector space $V$ with finite dimension $n$ over the field $\mathbb{K}$ \begin{equation}   b:V \times V \to \mathbb{K}\\
\text{The to $b$ belonging quadratic form:}\\ q: V \to \mathbb{K}\\ v\mapsto b(v,v)  \end{equation}
And we say the bilinear form is:
\begin{align*}
\text{Positive definite:} & \Leftrightarrow q(v)>0 \:\forall v\in V-\{0\} \\
\text{Positive semi definite:} & \Leftrightarrow q(v)\geq0 \:\forall v\in V \\
\text{negative definite:} & \Leftrightarrow q(v)<0 \:\forall v\in V-\{0\} \\
\text{negative semi definite:} & \Leftrightarrow q(v)\leq0 \:\forall v\in V\\ \text{indefinite:} & \Leftrightarrow \exists v_1,v_2 \in V: q(v_1)>0>q(v_2)   \end{align*}
My first question is, what is the intution behind this definition? What do we understand for a bilinear form, which fullfills one of these attributes? are there perhaps helpful geometrical interpretations, or Examples which makes it "intutive, and useful" to come upon this definition.
Secondly, we define the Gram matrix of a bilinear form in relation to a base of $V$, lets say $B$  as the matrix \begin{equation} G_B(b)_{i,j} := b(v_i,v_j)\;\text{for}\;B= (v_1, \dots , v_n) \;\text{Thus, the diagonal entries are $q(v_i)$}  \end{equation}
Furthermore it is said that the bilinear form, has one of the said attributes, if the corresponding Gram-Matrix has the attribute.
Second question, how does this correspond to the change of basis for the matrix? is the attribute invariant.
I have previously asked in the Chat, about a small problem i had, where the Gram-Matrix given in a diagonal Form \begin{equation} \begin{bmatrix}1&0\\0&-1\end{bmatrix} \end{equation} Upon $\mathbb{R^2}$ It was said, that we can find basis, such that the corresponding bilinear form, positvely definite, indefinite, or negatively definit is. Someone tried to help me, and argumented that it is not possible to check definitness of a bilinear Form from diagonalised Gram matrix (IE, you need the whole matrix). But is not diagonalising equivellent to changing the basis into an Eigenbasis? So how does this work out for the above? If it is not invariant under base change, why do we bother with this term? (IE from different point of views we observe different attributes)
Thank you.
 A: If you look at these bilinear forms as matrices, then it's quite clear what this means. You just look at its eigenvalues ($\lambda_i$). You don't need the diagonalizing basis.

*

*$\lambda_i>0$: positive definite form

*$\lambda_i\ge0$: positive semi-definite form

*$\lambda_i<0$: negative definite form

*$\lambda_i\le0$: negative semi-definite form

*Other cases: indefinite form

In this situation, let's say that the diagonal matrix $\Lambda$ contains all the eigenvalues of your bilinear form $T$. The bilinear form $T$ is symmetrical and can then be written as $T=P\Lambda P^T$, where $P$ is an orthonormal basis (and full rank). Let's define $y=P^Tv$, where $v\in V$. So any vector $v$ from your vector field $V$ maps to a $y$. And by consequence, any bilinear application $v^TTv=y^T\Lambda y$. As $\Lambda$ is diagonal, then this can we simplified: $\sum_i\lambda_i y_i^2$. So as $y_i^2\ge0$, you see that the sign of the bilinear form will depend exclusively on the sign of the eigenvalues.
All of this do not depend on the choice of the basis, it depends only on the eigenvalues.
