Proof that if all pivots of A are positive then A is a positive definite matrix

Hinted from Math 2270 – Lecture 33 : Positive Definite Matrices, by Dylan Zwick, foot note of page 4.

If A is symmetric then A is always diagonalizable: A = Q \Lambda Q^{T}, \text{rank}(Q) = \text{n}. Set x = Qz (x \neq 0), we have:

    \[ x^{T}Ax = z^{T}Q^{T}AQz = z^{T}Q^{T}Q \Lambda Q^{T}Qz = z^{T} \Lambda z = \sum_{i=1}^{n} \lambda_{i}z_{i}^{2} > 0 \]

Hinted from Introduction to Linear Algebra – Gilbert Strang [WORKING AREA]

We have:

    \[ x^{T}Ax = x^{T}\begin{bmatrix} r_{1}^{T}\\ r_{2}^{T}\\ ... \end{bmatrix}x = (x_{1}r_{1}^{T} + x_{2}r_{2}^{T} + ... + x_{i}r_{i}^{T} + ...)x \]

Now consider the expression x_{i}r_{i}^{T}x = \sum_{j=1}^{n}a_{i,j}x_{i}x_{j} , with a_{i,j} is the entry at position (i, j) of the matrix A. Now also consider the expression:  x_{j}r_{j}^{T}x = \sum_{i=1}^{n}a_{j,i}x_{j}x_{i}. We have a_{i,j} = a_{j,i} because A is a symmetric matrix. So yes,  x^{T}Ax can also be interpreted like this:

    \[ x^{T}Ax = \sum_{i=1}^{i=n}a_{i,i}x_{i}^2 + \sum_{i=1}^{n}\sum_{j=1}^{i-1} 2a_{i,j}x_{i}x_{j} \]

a_{1,1} is always a pivot. For some j, we can group an expression like this:

    \[ a_{1,1}\left (x_{1}^{2}+2\frac{a_{1,j}}{a_{1,1}}x_{1}x_{j}+\left ( \frac{a_{1,j}}{a_{1,1}} \right )^{2}x_{j}^{2} \right ) = a_{1,1}\left ( x_{1} + \frac{a_{1,j}}{a_{1,1}}x_{j} \right )^{2} \]

So:

    \[ x^{T}Ax = \sum_{j=2}^{n} a_{1,1}\left ( x_{1} + \frac{a_{1,j}}{a_{1,1}}x_{j} \right )^{2} + \sum_{j=2}^{n}\left ( a_{j,j} -  \frac{a_{j,1}}{a_{1,1}}a_{1,j}\right )x_{j}^{2} + \sum_{j=2}^{n}\sum_{k=1}^{j-1} 2a_{j,k}x_{j}x_{k} \]

Note that  a_{1,j} = a_{j,1} because A is symmetric, and \left ( a_{2,2} -  \frac{a_{2,1}}{a_{1,1}}a_{1,2}\right ) yields the pivot at position (2,2). \forall j = 2..n, \frac{a_{j,1}}{a_{1,1}} are in fact multipliers from elimination.


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