let $T$ be a self adjoint operator, then set $A = [T]_B$ where $B$ is orthonormal basis, and we say that $T$ (resp $A$) is positive definite iff every eigenvalue of $T$ (resp $A$) is positive.

But when i am doing some exercises, there are matrix who is not self adjoint (not symmetric) but they are still positive definite. So is it necessary that the condition T being a self adjoint operator a must? I am a bit confused as in your definition the matrix A must be self adjoint. i.e the $2$ by $2$ matrix $$\begin {pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$$ is not symmetric but it is positive definite.

  • $\begingroup$ Why do you say this matrix is positive definite? $\endgroup$ – G Tony Jacobs Nov 26 '17 at 18:25
  • $\begingroup$ See en.wikipedia.org/wiki/…, which might be what you're looking for $\endgroup$ – G Tony Jacobs Nov 26 '17 at 18:27
  • $\begingroup$ There are certainly matrices which satisfy $\langle\mathbf{x},A\mathbf{x}\rangle>0$ but are not symmetric. So in this sense, being symmetric and being positive-definite are independent concepts. However, the vast majority of sources define positive-definite with symmetry as an additional requirement, because the theory of positive-definite symmetric matrices is a very rich field of study. Which definition your text/course chooses is another matter. $\endgroup$ – EuYu Nov 26 '17 at 18:29
  • $\begingroup$ While symmetric positive definite matrices enjoy the property of being diagonalizable (with positive diagonal entries), e.g., the non-symmetric positive definite matrix $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ is not diagonalizable. $\endgroup$ – amsmath Nov 26 '17 at 18:41

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