While studying for linear algebra, I came across an assignment that I'm having trouble with. I don't know a bit which side to start from. Namely:
We say that two quadratic forms $a: V \to K$, $b: W \to K$ are equivalent if there exists a linear isomorphism $\alpha: V \to W$ so that $a = b \circ \alpha$. Prove that the quadratc forms $a$ and $b$ are equivalent iff they have the same signature.
Could someone give me some guidance or tips? Thank you in advance!
I assume $K = \mathbb R$, otherwise the question doesn't make sense. ($\mathbb C$ also works, but "signature" is slightly different in this case.)
If $a, b$ have the same signature $(p,q,r)$ then by definition there are isomorphisms $\alpha : \mathbb R^n \to V$ and $\beta : \mathbb R^n \to W$ such that $$ a\circ\alpha = c = b\circ\beta $$ where $c : \mathbb R^n \to \mathbb R$ is the quadratic form $$ c(x_1, \dotsc, x_p, y_1, \dotsc, y_p, z_1, \dotsc, z_r) = \sum_{i=1}^p x_i^2 - \sum_{i=0}^q y_i^2. $$ $\alpha$ and $\beta$ are isomorphisms, so what can you conclude from $a\circ\alpha = b\circ\beta$?
For the other direction, I think that thinking about bases is a bit clearer. If $a$ has signature $(p,q,r)$ then there is a basis $$ u_1, \dotsb, u_p, v_1, \dotsc, v_q, w_1, \dotsc, w_r \in V $$ such that $$ a\left(\sum_{i=1}^px_iu_i + \sum_{i=1}^qy_iv_i + \sum_{i=1}^rz_iw_i\right) = \sum_{i=1}^px_i^2 - \sum_{i=1}^qy_i^2. $$ Because $a$ and $b$ are equivalent there is a linear isomorphism $\alpha$ such that $a = b\circ\alpha$.
Now apply $\alpha$ to the above basis. What do you get? What does this say about $b$?