Using complete the square to determine positive definite matrices

Question

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

Answer 1

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.

Answer 2

Since $16>0$, but $16\cdot 9-12^2=0$, the matrix is only positive semidefinite, but not positive definite.

Answer 3

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 $$ 16 (x + \frac{3}{4}y)^2 = 16x^2 + 24 xy + 9 y^2 $$

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$$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rr} 16 & 12 \\ 12 & 9 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rr} 1 & - \frac{ 3 }{ 4 } \\ 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rr} 1 & - \frac{ 3 }{ 4 } \\ 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rr} 1 & \frac{ 3 }{ 4 } \\ 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rr} 16 & 0 \\ 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rr} 1 & 0 \\ - \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 16 & 12 \\ 12 & 9 \\ \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{ 3 }{ 4 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 16 & 0 \\ 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rr} 1 & 0 \\ \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 16 & 0 \\ 0 & 0 \\ \end{array} \right) \left( \begin{array}{rr} 1 & \frac{ 3 }{ 4 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 16 & 12 \\ 12 & 9 \\ \end{array} \right) $$