I am reading the Gilbert Strang's book "Introduction to Linear Algebra" and there is test for symmetric positive definite matrix which says that positive definite matrix's pivots are all positive. Let A be a positive definite symmetric matrix
A = [2 4; 4 10].
It's left upper determinants are all positive, it's eigenvalues are all positive. There is one thing I can not understand about finding pivots. Is there any strict rule for finding pivots, because for this matrix I have 2 ways and they contradict to the test that all pivots are positive.
- [2 4; 4 10] -> [2 4; 0 -2] (2 times row1 – row2), pivots are (2, -2)
- [2 4; 4 10] -> [2 4; 0 2] (row2 – 2 times row1), pivots are (2, 2)
I would appreciate if someone guide me in the right direction.
Best Answer
Yes, you have to start at the top left corner, that is positive. Reduce the first column. Now, look for nonzeroes in the second column. It turns out that you will have one on the second row, and it will be $>0$. Take that one as a pivot, and so on. Hence the rule is: take the first element available as a pivot, then all the pivots will be positive.