Using complete the square to determine positive definite matrices I realize this may be a basic question but I am having trouble following my notes.
I have the matrix $$\begin{bmatrix}16&12\\12&9\end{bmatrix} .$$ So I've got my equation from the matrix to be $(16x)^2 - 12xy - 12yx + 9y^2.$ Not sure where to go from here
 A: Since $16>0$, but $16\cdot 9-12^2=0$, the  matrix is only positive semidefinite, but not positive definite.
A: No. The bilinear form is $16x^2+24xy+9y^2$. And since this is equal to $\left(4x+3y\right)^2$, your bilinear form is not definite positive.
A: The direction that would usually be called completing the square is
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rr} 
1 & 0 \\ 
 \frac{ 3 }{ 4 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
16 & 0 \\ 
0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
1 &  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rr} 
16 & 12 \\ 
12 & 9 \\ 
\end{array}
\right) 
$$
which means semidefinite, not definite. In particular
$$ 16 (x + \frac{3}{4}y)^2 = 16x^2 + 24 xy + 9 y^2 $$
=================================
$$  D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j  $$
$$  P_{j-1} E_j = P_j  $$
$$ E_j^{-1} Q_{j-1}  = Q_j  $$
$$  P_j Q_j = I  $$
$$ P_j^T H P_j = D_j  $$
$$ Q_j^T D_j Q_j = H  $$
$$ H = \left( 
\begin{array}{rr} 
16 & 12 \\ 
12 & 9 \\ 
\end{array}
\right) 
$$
==============================================
$$ E_{1} = \left( 
\begin{array}{rr} 
1 &  -  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
$$
$$  P_{1} = \left( 
\begin{array}{rr} 
1 &  -  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
, \; \; \; Q_{1} = \left( 
\begin{array}{rr} 
1 &  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
, \; \; \; D_{1} = \left( 
\begin{array}{rr} 
16 & 0 \\ 
0 & 0 \\ 
\end{array}
\right) 
$$
==============================================
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rr} 
1 & 0 \\ 
 -  \frac{ 3 }{ 4 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
16 & 12 \\ 
12 & 9 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
1 &  -  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rr} 
16 & 0 \\ 
0 & 0 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rr} 
1 & 0 \\ 
 \frac{ 3 }{ 4 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
16 & 0 \\ 
0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rr} 
1 &  \frac{ 3 }{ 4 }  \\ 
0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rr} 
16 & 12 \\ 
12 & 9 \\ 
\end{array}
\right) 
$$
