Hey need some help with this question:



I was thinking about doing a proof by diagonalizing A but I'm not sure if I can use that because I can't really show that A is diagonalizable?

Let's say A is an nxn matrix

we know if A if positive definite then x^TAx > 0 for all non zero vectors x ∈R^n

Then lets define e_i to be the vector whose ith element is 1 and the rest are 0. where i=1,2....n ie. e_1=(1,0,0,0,...0)

Then since e_i is a nonzero vector and A is positive definite

(e_i)^TA(e_i) > 0

but when you multiply this out you get

(e_i)^TA(e_i) = a_ii >0
where a_ii are the diagonal elements of A

therefore the diagonals of the matrix A are postive

This post was edited by cdexswzaq on Nov 16 2014 05:10pm

Wow that is so smart!! Never would have thought to use the standard basis vectors

Thank you so much

Hey sorry I have another question didn't want to start a new thread

I am working on this question:



And I found that the eigenvalues are 2,-4,-4 and the respective eigenvectors are (1,1,2),(-1,1,0),(-2,0,1). Since -4 has geometric multiplicity 2 I used gram schmidt on (-1,1,0) and (-2,0,1) to get (-1,1,0) and (-1,-1,1) to create an orthonormal basis so in the end here is my spectral decomposition:



But I get the wrong answer (I am not getting back the matrix A). I get very close to A though, I think I'm off by 2/3

Can someone help me find where I went wrong, been looking for a while now?

The mistake is where you normalized your (1,1,2) vector
should be √6 instead of √5
can check your answer here http://wims.unice.fr/wims/en_tool~linear~matrix.html

This post was edited by cdexswzaq on Nov 16 2014 06:51pm

ugh of course, thank you for finding that