Hi I was wondering whether the following matrix is positive definite or not?
$$A=\begin{bmatrix} 9 & 3 &3\\3 & 10 & 7\\3 & 5 & 9\end{bmatrix}$$
Clearly A is not symmetric, but since
$$det\begin{bmatrix} 9\end{bmatrix}>0,\,det\begin{bmatrix} 9 & 3\\ 3 & 10\end{bmatrix}>0,\,\,\,det\begin{bmatrix} 9 & 3 &3\\3 & 10 & 7\\3 & 5 & 9\end{bmatrix}>0$$
A is positive definite, right?? (Symmetry is not necessary to be Positive Definite)
Am I right? Thanks in advance.
apply the test to $$H=\begin{bmatrix} 9 & 3 &3\\3 & 10 & 6\\3 & 6 & 9\end{bmatrix}$$ which is the symmetric part of your matrix, that is $H = (A + A^T)/2.$ The quadratic forms (for column vector $x$) $$ x^T Hx = x^T A x $$ because adding or subtracting a skew-symmetric matrix does not change the quadratic form; only the symmetric part matters.
$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 9 & 3 & 3 \\ 3 & 10 & 6 \\ 3 & 6 & 9 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 9 & 0 & 3 \\ 0 & 9 & 5 \\ 3 & 5 & 9 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & 9 & 5 \\ 0 & 5 & 8 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 5 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & - \frac{ 4 }{ 27 } \\ 0 & 1 & - \frac{ 5 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & \frac{ 5 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & \frac{ 47 }{ 9 } \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 3 } & 1 & 0 \\ - \frac{ 4 }{ 27 } & - \frac{ 5 }{ 9 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 9 & 3 & 3 \\ 3 & 10 & 6 \\ 3 & 6 & 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & - \frac{ 4 }{ 27 } \\ 0 & 1 & - \frac{ 5 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & \frac{ 47 }{ 9 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 5 }{ 9 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & \frac{ 47 }{ 9 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } \\ 0 & 1 & \frac{ 5 }{ 9 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 9 & 3 & 3 \\ 3 & 10 & 6 \\ 3 & 6 & 9 \\ \end{array} \right) $$