Often in examples, eigenvalue decomposition $A=U\Lambda U^T$, $A$ is usually assumed to be a symmetric matrix. I am wondering what are the differences and implications when $A$ is a non-symmetric (still positive values if that helps).
What can we say about the eigenvalues and eigenvectors of such decomposition?
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$\begingroup$The first implication of symmetry is normality. All Matrix, that suffice $A^TA = AA^T$ are acalled normal and the eigenvectors are orthogonal to each other. This makes it possible, to write $A = U Λ U^T$ instead of $A = U Λ U^{-1}$, which is correct for diagonizable matrices.
In addition, there always is this kind of decomposition. This is not always the case with any matrix. Often the best one can do is a Jordan normal form, that has 1 in some places of the upper diagonal of Λ.
Also Symmetric matrices have real eigenvalues. If the matrix is not symmetric anymore, there are possibly complex conjugate pairs of eigenvalues.
This means, that either $Λ \in \mathbb{C}^{n \times n}$ or you get $2\times 2$ blocks, instead of a diagonal matrix.
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