Can we say that there is $P$ such that $P^T AP=B?$

Question

Consider the quadratic form $Q(v)=v^{t}Av,v=(x,y,z,w)$ where matrix $A$ is given by \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ \end{bmatrix}

Then which of the following is true?

1. $Q$ has rank 3.

2. $xy+z^{2}=Q(Pv)$ for some invertible real matrix $P.$

3. $xy+y^{2}+z^{2}=Q(Pv)$ for some real invertible matrix $P.$

4. $x^{2}+y^{2}-zw=Q(Pv)$ for some some real invertible matrix $P.$

From $A$, $Q$ must be having the Rank $4$. So, (1) Must not be true.

if we do $R_3→-1/2R_3$ and $R_4→-1/2R_4$ we get, $$B=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1/2 \\ 0 & 0 & -1/2 & 0\\ \end{bmatrix}$$ Can we say that there is $P$ such that $P^T AP=B?$

Answer 1

** One possibility:**

If $A$ is a symmetric matrix it can always be diagonalized by an orthogonal matrix $P$ such that $P^{-1}=P^T$ and $P^{-1} A P= D= P^T A P$. For example:

Let $A=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}$ its eigen values are $-1,1,1,1$ with eigenvectors as the columns of P $P=\begin{pmatrix} 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\end{pmatrix}$ and $D=\begin{pmatrix} -1 & 0 & 0 &0 \\ 0 & 1 & 0 &0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} =B$