Question

If all eigenvalues of a nxn matrix are positive, does it imply that the matrix is positive definite?

Answer 1

yes. the determinant of a matrix is a product of its eigen values which is positive.

Answer 2

but positive determinant cannot judge if a matrix is positive definite isn't it? it could have been 2 negative eigenvalues that times together to get that positive determinant.

Answer 3

not quite. read it here. :D
http://en.wikipedia.org/wiki/Positive-definite_matrix
Cheers!

Answer 4

your concerns are right. Could've been, but that doesn't spoil the show. ;)

Answer 5

so i can safely say, as long as i found all the eigenvalues of a matrix to be positive, i don't have to go on to test for its other properties and I can just say that the matrix is positive definite?

Answer 6

yup. so it is.

Answer 7

also, what happens if it has double positive eigenvalues? say it has more than one eigenvalues that has the same value and are positive too. is it still positive definite?

Answer 8

\[\left[\begin{matrix}1 & -3 \\ 0 & 1\end{matrix}\right]\]

Answer 9

\[\left[\begin{matrix}1 & 1\end{matrix}\right]\left[\begin{matrix}1 & -3 \\ 0 & 1\end{matrix}\right]\left[\begin{matrix}1 \\ 1\end{matrix}\right]=-1\]

Answer 10

eigenvalues are 1 and 1

Answer 11

oh...so looks like repeated eigenvalues implies that the matrix is not positive definite because of \[x^{T}Ax<0\]..

Answer 12

this matrix is not symmetric...that is causing a problem

Answer 13

but a positive definite matrix may not neccessarily be symmetric, right?

Answer 14

has nothing to do with repeated eigenvalues

Answer 15

correct

Answer 16

a symmetric square matrix is PD iff the eigenvalues are positive

Answer 17

yea if that's the case, then why does it got nothing to do with repeated eigenvalues?

Answer 18

change the 1 in location 2,2 to 1.1

eigenvalues will be 1 and 1.1

the product that I have above will them be -.9

Answer 19

ohhh.... a symmetric matrix is only a positive definite if the eigenvalues are positive... if the matrix is not symmetric, having all its eigenvalues positive does not mean it is a positive definite anymore?

Answer 20

correct

Answer 21

ohhh I see....thanks a lot Zarkon! :D

Answer 22

as illustrated above ;)