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show that A is a Positive-definite symmetric matrix $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} $$

My try :

$A^T = A $, $A$ is symmetric

let $x \in \mathbb{R}^n$

$$x^TAx = \begin{bmatrix} a & b & c \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix} = a^2+ab+ac+ab+2b^2+2bc+ac+2bc+3c^2$$ $$=a^2+b^2+2ab+b^2+4c^2+4bc+2ac-c^2 = (a+b)^2+(b+2c)^2+c(2a-c)$$

I'm stuck here trying to find how can I prove it's positive.

Note : I know there's an easier method which consists of checking if all the eigenvalues are positive or checking if the leading principal minors are all positive but I have to show it this way using that definition.

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$ \begin{bmatrix} 1&1&1\\1&2&2\\1&2&3\end{bmatrix} = \begin{bmatrix} 1&0&0\\1&1&0\\1&1&1\end{bmatrix} \begin{bmatrix} 1&1&1\\0&1&1\\0&0&1\end{bmatrix}$

$\mathbf x^T \begin{bmatrix} 1&0&0\\1&1&0\\1&1&1\end{bmatrix} = \begin{bmatrix} (x+y+z)& (y+z) & z\end{bmatrix}$

$\mathbf x^TA\mathbf x =(x+y+z)^2 + (y+z)^2 + z^2$

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  • $\begingroup$ thank you !! if you don't mind I have a question : did you use a Cholesky decomposition or you found that decomposition just by intuition ? $\endgroup$ – rapidracim Mar 8 '17 at 20:37
  • 3
    $\begingroup$ I did it by eyeball. $\endgroup$ – Doug M Mar 8 '17 at 20:55
  • $\begingroup$ There is an algorithm for finding this sort of product, when the original matrix $H$ is all integers, one builds up a matrix $P$ of rational entries and $\det P = 1$ such that $P^T H P = D$ is diagonal. With $Q = P^{-1},$ this is $Q^T D Q = H.$ Usually $P$ and $Q$ are also upper triangular. Posted an answer. $\endgroup$ – Will Jagy Mar 8 '17 at 23:34
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Hint: $$ a^2+ab+ac+ab+2b^2+2bc+ac+2bc+3c^2=(a^2+b^2+c^2+2ab+2ac+2bc) +(b^2+c^2+2bc)+c^2 =(a+b+c)^2+(b+c)^2+c^2 $$

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There is a method, algorithm really, that deserves to be better known. Given a symmetric matrix $H$ of integers, it provides a matrix $P$ with rational (or integer) entries and $\det P = 1,$ along with a diagonal matrix $D,$ such that $$ P^T H P = D. $$ Since $\det P = 1 \;$ (and $P$ is usually upper triangular), it is not so hard to find $Q = P^{-1},$ after which $$ Q^T D Q = H. $$ See, for example, reference for linear algebra books that teach reverse Hermite method for symmetric matrices

Illustrated here, with notation change $D$ = h2.

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Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.

parisize = 4000000, primelimit = 500509
? h = [ 1,1,1; 1,2,2; 1,2,3]
%1 = 
[1 1 1]

[1 2 2]

[1 2 3]



? ht = mattranspose(h)
%2 = 
[1 1 1]

[1 2 2]

[1 2 3]

? ht - h
%3 = 
[0 0 0]

[0 0 0]

[0 0 0]

? p1 = [1,-1,-1; 0,1,0; 0,0,1]
%4 = 
[1 -1 -1]

[0 1 0]

[0 0 1]

? p1t = mattranspose(p1)
%5 = 
[1 0 0]

[-1 1 0]

[-1 0 1]

? h1 = p1t * h * p1
%6 = 
[1 0 0]

[0 1 1]

[0 1 2]

? p2 = [1,0,0; 0,1,-1; 0,0,1]
%7 = 
[1 0 0]

[0 1 -1]

[0 0 1]

? p2t = mattranspose(p2)
%8 = 
[1 0 0]

[0 1 0]

[0 -1 1]

? h2 = p2t * h1 * p2
%9 = 
[1 0 0]

[0 1 0]

[0 0 1]

? p = p1 * p2
%10 = 
[1 -1 0]

[0 1 -1]

[0 0 1]

? q = matadjoint(p)
%11 = 
[1 1 1]

[0 1 1]

[0 0 1]

? qt = mattranspose(q)
%12 = 
[1 0 0]

[1 1 0]

[1 1 1]

? qt * q
%13 = 
[1 1 1]

[1 2 2]

[1 2 3]

? h
%14 = 
[1 1 1]

[1 2 2]

[1 2 3]

? 
? h2
%15 = 
[1 0 0]

[0 1 0]

[0 0 1]

? qt * h2 * q
%16 = 
[1 1 1]

[1 2 2]

[1 2 3]

?
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0
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A theorem of Hurwitz says that $A$ is positive definite, if all leading principal minors have positive determinant. Since this is the case, $A$ is positive definite. We have $\det((1))=1$, $\det \begin{pmatrix} 1 & 1 \cr 1 &2 \end{pmatrix}=1$ and $\det(A)>0$.

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  • $\begingroup$ I know that. but I have to prove it using that specific definition as mentioned in my post. $\endgroup$ – rapidracim Mar 8 '17 at 19:19
  • $\begingroup$ The proof of Hurwitz theorem shows directly that $xTAx>0$ for this matrix. $\endgroup$ – Dietrich Burde Mar 8 '17 at 19:36

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