Signature of $q(x,y)=5x^2-6xy+y^2$ a quadratic form Let $q:R^2\rightarrow R$ a quadratic form such that  $q(x,y)=5x^2-6xy+y^2$
find the signature of $q$
I know that $q_A=$ $\begin{pmatrix}
5 & -3\\
-3 & 1
\end{pmatrix}$
and how $q(1,0)=5\neq0$ then $(1,0)$ is an element of a orthogonal basis $\beta=\{(1,0),v\}$
and then how $\beta$ is orthogonal then the matrix associated to $q$ respect to $\beta$
should be
$\begin{pmatrix}
5 & 0\\
0 & 1
\end{pmatrix}$ is right?

$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rr} 
1 & 0 \\ 
 -  \frac{ 3 }{ 5 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
5 & 0 \\ 
0 &  -  \frac{ 4 }{ 5 }  \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
1 &  -  \frac{ 3 }{ 5 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rr} 
5 &  - 3 \\ 
 - 3 & 1 \\ 
\end{array}
\right) 
$$
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$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rr} 
1 & 0 \\ 
 \frac{ 3 }{ 5 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
5 &  - 3 \\ 
 - 3 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
1 &  \frac{ 3 }{ 5 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rr} 
5 & 0 \\ 
0 &  -  \frac{ 4 }{ 5 }  \\ 
\end{array}
\right) 
$$
$$  $$
