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I'm stuck in a problem on positive definite matrices. I should show that "if all eigenvalues of A are positive, then A >= 0."

EDIT: assuming A is diagnosable.

I know the definition for a positive definite matrix is that x'Ax >= 0 for all x (but does "x" herein refer to the eigenvectors?). I've spent many hours on trying to figure this out, but I don't really grasp the idea of how to get started here..

Would greatly appreciate help!!

Best, Matthias

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The statement is incorrect. For instance, consider the matrix $$A = \begin{bmatrix} 1 & -100\\ 0 & 2\end{bmatrix}$$ The eigenvalues are $1$ and $2$. However, the matrix is not positive definite. For instance, $$\begin{bmatrix} 1 & 1\end{bmatrix} \begin{bmatrix} 1 & -100\\ 0 & 2\end{bmatrix} \begin{bmatrix} 1 \\ 1\end{bmatrix} = \begin{bmatrix} 1 & 1\end{bmatrix} \begin{bmatrix} -99 \\ 2\end{bmatrix} = -97 < 0$$ The statement is true if the matrix $A$ is symmetric and diagonalizable. However, the following statement is always true: "If the matrix is positive definite and symmetric, then all its eigenvalues are positive."

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  • $\begingroup$ Thank you so much for responding this fast!! I believe it is incorrect on the problem set then… assuming that A is supposed to be diagnosable, which I suppose is det(A)≠0, how may I affirm the statement? $\endgroup$ – Matthias Sep 15 '13 at 1:40
  • $\begingroup$ @Matthias Diagonalizable means $A$ can be written as $$A = X \Lambda X^{-1}$$ where $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on its diagonal. $\endgroup$ – user17762 Sep 15 '13 at 1:49
  • $\begingroup$ thanks. does it suffice to proof based on this definition and argue that if x'Ax >= 0; and A=XDX^-1; x'XDX'x >= 0 because D will be positive definite (since it only has positive eigenvalues and is diagonal) and X and X^-1 as well as x and x' will have the same eigenvalues, thus in any case the multiplication x'XDX'x >= 0. $\endgroup$ – Matthias Sep 15 '13 at 2:05
  • $\begingroup$ @user17762 Isn't your matrix $A$ diagonalisable? $\endgroup$ – lcv Sep 15 '13 at 2:41
  • $\begingroup$ @lcv Yes, it is. I realize that what I might have written earlier might have caused some confusion. Have updated it now and have added the word symmetric. $\endgroup$ – user17762 Sep 15 '13 at 2:50
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Usually one speaks of positive definite matrices for the class of symmetric (or more generally Hermitian) matrices. The statement indeed follows easily if you assume that $A$ can be diagonalized by a orthogonal matrix $V$ ( I.e. such that $V^T V = V V^T = I$ ) which happens to be always possible if the matrix is symmetric. Then, assuming you have $A= V^T D V$ with $D$ diagonal with $D_{j,j}\geq 0$

$$ x^T A x = x^T V^T D V x $$

And the last expression is the norm square of the vector $\sqrt{D} V x $ and so is positive for any $x$. I denoted with $X^T$ the transpose of $X$.

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Positive (negative) definite matrices are defined for symmetric ones, so assuming this your matrix is positive definite iff all its eigenvalues are positive. No need to mention diagonalization as any (complex or real) symmetric matrix is diagonalizable (even orthogonally so).

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