Since $AA^{T}<I$, the eigenvalues of $A$ is real and between (-1, 1), and the eigenvalues of symmetric matrix $I+A$ is positive.
Can I say $I+A$ is positive definite? Is there anything wrong about my descriptions?
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1What does $AA^T<I$ even mean? The set of matrices is not ordered... – 5xum Aug 13 '15 at 07:28
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It means that $I-AA^{T}$ is positive definite. Sorry, my expression is not clear. – Y.Lee Aug 13 '15 at 07:34
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I supposed he mean $\text{}^txA\text{}^tAx < \langle x,x \rangle$. – InsideOut Aug 13 '15 at 07:35
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Your argument is correct: if $\lambda$ is an eigenvalue of $A'$ and $v$ is an associated eigenvector, then $$ 0<v'(I-AA')v=|v|^2-|A'v|^2=(1-\lambda^2)|v|^2\implies-1<\lambda<1. $$ But $A$ and $A'$ have the same set of eigenvalues (because $A^T-\alpha I$ is the transpose of $A-\alpha I$ and so has the same determinant as $A-\alpha I$), so the eigenvalues of $A$ are also in $(-1,1)$. From this, we infer that the eigenvalues of $I+A$ are all in $(0,2)$; in particular, these eigenvalues are strictly positive. This implies that $I+A$ is positive definite.
yurnero
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