Is $Q(u,u)$ (also $\langle u,Av\rangle$) considered a scalar product?

Question

Can you tell me if this is a good answer?: Let $u,v$ be vectors in $\Bbb R^n$, and let $A$ be an indefinite $n×n$ matrix (whose determinants of the main submatrices are both positive and negative). I thought that given that "$Av$" is actually a vector which a matrix has been applied to, we can consider $Av$ as $v$ in new coordinates imposed by the matrix $A$, also $\langle u,Av\rangle$ is a normal scalar product.

Is this a good point or I'm completely wrong?

Answer 1

I assume that scalar product means the same thing as inner product in this context.

The answer to your question is no. If $A$ is indefinite, then there necessarily exists a vector $v$ such that $Q(v,v) = v^TAv < 0$. In particular, plugging in an eigenvector of $A$ associated with a negative eigenvalue demonstrates this. Similarly, a positive semidefinite $A$ does not induce a scalar product because there exists a non-zero vector $v$ such that $v^TAv = 0$.

We could apply an argument similar to yours to deduce that for any invertible matrix $M$ (in fact, any matrix with linearly independent columns will do) is such that the function $Q(u,v) = \langle Mu, Mv \rangle$ defines a scalar product. Note that this can be rewritten as $Q(u,v) = u^T (M^TM)v$.

In fact, the following are equivalent:

• $A$ is positive definite,
• $Q(u,v) = u^TAv$ is a scalar product,
• There exists an invertible matrix $M$ such that $A = M^TM$.