Show that A is a Positive-definite symmetric matrix

Question

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.

Answer 1

Hint: $$ a^2+ab+ac+ab+2b^2+2bc+ac+2bc+3c^2=(a^2+b^2+c^2+2ab+2ac+2bc) +(b^2+c^2+2bc)+c^2 =(a+b+c)^2+(b+c)^2+c^2 $$

Answer 2

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 $$ Q^T D Q = H. $$

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

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]

?

Answer 3

$ \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}$

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

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

Answer 4

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 $\det(A)>0$.