What is a necessary and sufficient condition for all the eigenvalues of a non-symmetric matrix to be positive?
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...
Tilak Mallikarjun
Updated on August 01, 2022Comments
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Tilak Mallikarjun over 1 year
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:
- Tests for positive definiteness for nonsymmetric matrices
- p.322,Linear Algebra and its Applications, Gilbert Strang.
- Necessary and sufficient condition for all the eigenvalues of a real matrix to be non-negative
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Tilak Mallikarjun almost 6 yearsI agree. The proposition here, math.stackexchange.com/questions/387192/… disregards the possibility of complex Eigen values.
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Tilak Mallikarjun almost 6 yearsAlso 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 math.stackexchange.com/questions/387192/… by giving the counter examples provided by me and Robien1