Matrix with a parameter positive definite and Cholesky decomposition yield different conditions

Question

Find all $x$ such that matrix $A$ is positive definite and for all such $x$-es determine the Cholesky decomposition of $A$

$ \begin{bmatrix} 4 & 1 & 0 & 0 \\ 1 & 2 & x & 0 \\ 0 & x & 2 & 1 \\ 0 & 0 & 1 & 1 \\ \end{bmatrix} $

My attempt:

First, using the Sylvester's criterion I determined that $|x| < \sqrt{7}/2$ (the other condition being $|x|<\sqrt{14}/2$ but from making a Cholesky decomposition I found that $x$ should be such that it satisfies $|x|< \sqrt{3}/2$ and I'm not sure what to do right now. I double checked all my results and compared them to other peoples' results and they are all the same, but I'm not sure what to conclude.

Positive definiteness and Cholesky decomposition existence are equivalent, so why didn't I get the same conditions? Is the only option a mistake in calculations or what?

Answer 1

As far as deciding positive definite for a symmetric matrix, it is enough, and usually more accurate, to ignore the eigenvalues and just find $P^T A P = D,$ where $P$ is nothing special except $\det P \neq 0.$ There is an algorithm for this, I put information on how to construct $P$ a step at a time, see http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr

Given the outcome, I suggest checking when the variable matrix entry is set to $3/2 = 1.5 \; .$ Alright, did it myself, when the variable matrix entry is set to $3/2,$ the characteristic polynomial becomes $$ \frac{1}{4} \left( 2 \lambda^2 - 9 \lambda - 1 \right) \left( 2 \lambda^2 - 9 \lambda + 8 \right), $$ eigenvalues can be found with quadratic formula, anyway $\{ -0.108495, 1.21922, 3.2807764, 4.608495\} \; .$ The negative eigenvalue for this specific matrix is $(9 - \sqrt{89})/4$

I am using the variable $t$ instead of $x.$ take $$ P = \left( \begin{array}{cccc} 1&\frac{-1}{4}&\frac{t}{7}&\frac{t}{4 t^2 - 14} \\ 0&1&\frac{-4t}{7}&\frac{-4t}{4t^2-14} \\ 0&0&1&\frac{7}{4 t^2 - 14} \\ 0&0&0&1 \\ \end{array} \right) $$ $$ D = P^T AP = \left( \begin{array}{cccc} 4&0&0&0 \\ 0&\frac{7}{4}&0&0 \\ 0&0&\frac{14 - 4t^2}{7}&0 \\ 0&0&0&\frac{7 - 4 t^2}{14 - 4 t^2} \\ \end{array} \right) $$ For positive definite, by Sylvester's Law of Inertia, you need $$ 14 - 4 t^2 > 0 \; , $$ $$ 7 - 4 t^2 > 0 \; . $$ Both $$ |t| < \sqrt{ \frac{7}{2} } \approx 1.87\; , $$ $$ |t| < \sqrt{ \frac{7}{4} } \approx 1.32 \; , $$ The second implies both so we need $|t| < \sqrt{ \frac{7}{4} } \approx 1.32$

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Added: it occurred to me that I could double your matrix and make the variable entry $3$ instead of $3/2,$ to use my integer/rational software

$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 & 0 \\ \frac{ 3 }{ 14 } & - \frac{ 6 }{ 7 } & 1 & 0 \\ - \frac{ 3 }{ 10 } & \frac{ 6 }{ 5 } & - \frac{ 7 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 14 } & - \frac{ 3 }{ 10 } \\ 0 & 1 & - \frac{ 6 }{ 7 } & \frac{ 6 }{ 5 } \\ 0 & 0 & 1 & - \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 4 } & 1 & 0 & 0 \\ 0 & \frac{ 6 }{ 7 } & 1 & 0 \\ 0 & 0 & \frac{ 7 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$

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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
$$ H = \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$ $$ D_0 = H $$

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

$$ H = \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 14 } & 0 \\ 0 & 1 & - \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 14 } & - \frac{ 3 }{ 10 } \\ 0 & 1 & - \frac{ 6 }{ 7 } & \frac{ 6 }{ 5 } \\ 0 & 0 & 1 & - \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 & 0 \\ \frac{ 3 }{ 14 } & - \frac{ 6 }{ 7 } & 1 & 0 \\ - \frac{ 3 }{ 10 } & \frac{ 6 }{ 5 } & - \frac{ 7 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 14 } & - \frac{ 3 }{ 10 } \\ 0 & 1 & - \frac{ 6 }{ 7 } & \frac{ 6 }{ 5 } \\ 0 & 0 & 1 & - \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 4 } & 1 & 0 & 0 \\ 0 & \frac{ 6 }{ 7 } & 1 & 0 \\ 0 & 0 & \frac{ 7 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & \frac{ 7 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 10 }{ 7 } & 0 \\ 0 & 0 & 0 & - \frac{ 4 }{ 5 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 4 } & 0 & 0 \\ 0 & 1 & \frac{ 6 }{ 7 } & 0 \\ 0 & 0 & 1 & \frac{ 7 }{ 5 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 2 & 0 & 0 \\ 2 & 4 & 3 & 0 \\ 0 & 3 & 4 & 2 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) $$