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Consider a real but not symmetric matrix $A$. To test if the matrix has positive eigenvalues, I've learnt from this forum that a symmetric matrix will be given by $B=A+A^T$. If all the eigenvalues of $B$ are positive, then it follows that A also has all the eigenvalues positive. So this is a sufficient condition.

For example, consider

$$A=\begin{bmatrix}1&4\\0&1\end{bmatrix}$$

It so happens that $B$ has one negative eigenvalue. Whereas $A$ has both positive eigenvalues. So what is a necessary (and sufficient) condition that $A$ has all positive eigenvalues?

Looking at Gershgorin's theorem , further rises the possibility of complex eigenvalues.

References:

  1. Tests for positive definiteness for nonsymmetric matrices
  2. p.322,Linear Algebra and its Applications, Gilbert Strang.
  3. Necessary and sufficient condition for all the eigenvalues of a real matrix to be non-negative
Rodrigo de Azevedo
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Tilak Mallikarjun
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1 Answers1

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Your result is false: take $A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ then $A+ A^T=2I_2$ has his eigenvalues positive whereas eigenvalues of $A$ are $1 \pm i$ not even real...

GNUSupporter 8964民主女神 地下教會
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Robien1
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  • I agree. The proposition here, https://math.stackexchange.com/questions/387192/tests-for-positive-definiteness-for-nonsymmetric-matrices.%5C disregards the possibility of complex Eigen values. – Tilak Mallikarjun Dec 09 '17 at 11:42
  • Also it is falsely stated that positive Eigen values are possible if and only if A+A' is positive definite. It would be helpful in future if someone with a reputation of 50 or more can post a comment in that page https://math.stackexchange.com/questions/387192/tests-for-positive-definiteness-for-nonsymmetric-matrices.%5C by giving the counter examples provided by me and Robien1 – Tilak Mallikarjun Dec 09 '17 at 11:49