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$$\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}\begin{bmatrix} 4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \\4x_1^2 + x_1x_2 + x_3x_1 + x_2x_1 + 2x_2^2 -x_2x_3 + x_1x_3-x_2x_3+3x_3^2$$

which I cannot put in a form that is $>0$. So how can this matrix be positive definite?

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    $\begingroup$ Orthogonalize (=find an orthogonal base for) the symmetric bilinear form $y^TAx$ and see if all the elements in the diagonal are strictly positive. $\endgroup$ – Saucy O'Path Apr 12 '19 at 21:26
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    $\begingroup$ See math.stackexchange.com/questions/1388421/… for a mechanical method of diagonalization of a symmetric matrix. $\endgroup$ – amd Apr 12 '19 at 21:41
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    $\begingroup$ or compute the eigenvalues of the matrix. $\endgroup$ – Luke Apr 12 '19 at 21:47
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Note that we can show that a matrix is positive definite by looking at its $n$ upper left determinants. Note that for the matrix

$$\begin{bmatrix} 4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix}$$

We have $$\begin{vmatrix} 4 \end{vmatrix} = 4$$

and $$\begin{vmatrix} 4 & 1\\ 1 & 2 \end{vmatrix} = 7$$

and lastly

$$\begin{vmatrix} 4 & 1 & 1\\ 1 & 2 & -1\\ 1 & -1 & 3 \end{vmatrix} = 13$$

Since $4,7,13 > 0$ the matrix is positive definite.

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I think you are asking about repeated completing the square: as matrices, one such $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 4 } & 1 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 5 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 13 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & \frac{ 1 }{ 4 } \\ 0 & 1 & - \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) $$

which can be revised (put all denominators into the diagonal matrix) to $$ \frac{1}{4} \left(4x+y+z \right)^2 + \frac{1}{28} \left(7y-5z \right)^2 + \frac{13}{7} \left(z \right)^2 $$

https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 4 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 4 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 4 & 0 & 1 \\ 0 & \frac{ 7 }{ 4 } & - \frac{ 5 }{ 4 } \\ 1 & - \frac{ 5 }{ 4 } & 3 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 1 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & - \frac{ 1 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & - \frac{ 5 }{ 4 } \\ 0 & - \frac{ 5 }{ 4 } & \frac{ 11 }{ 4 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & - \frac{ 3 }{ 7 } \\ 0 & 1 & \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & \frac{ 1 }{ 4 } \\ 0 & 1 & - \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 13 }{ 7 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 \\ - \frac{ 3 }{ 7 } & \frac{ 5 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & - \frac{ 3 }{ 7 } \\ 0 & 1 & \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 13 }{ 7 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 4 } & 1 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 5 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 13 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & \frac{ 1 }{ 4 } \\ 0 & 1 & - \frac{ 5 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 1 & 1 \\ 1 & 2 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) $$

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For the given matrix, there is no need for much computation. Multiplying the second column by $3/2$ and the second row by $2/3$ gives a matrix with the same (real) eigenvalues. It is $$\begin{bmatrix} 4 & 3/2 & 1\\2/3 & 2 & -2/3 \\ 1 & -3/2 & 3 \end{bmatrix}.$$

The Gershgorin circle theorem shows that all eigenvalues are contained in the union of the three intervals $|x-4|<\frac32+1$, $|x-2|<\frac23+\frac23$ and $|x-3|<\frac32+1$. Hence all eigenvalues are positive. Since the given matrix is symmetric, this implies that it is positive definite.

For matrices that are not essentially "diagonally dominated", the method of Wolfy's answer using the main subdeterminants seems best to me. For larger matrices, these subdeterminants can be computed by Gaussian elimination without swapping of rows - if the matrix is positive definite.

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