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Hint: If all steps were correct, then the answer would be yes. You already know ($A=4I, B=I$) that the answer is no. Therefore, there is a mistake in one of the two "they are similar" statements. The second statement looks pretty rock solid. The matrix $$A^{1/2}BA^{-1/2}$$ is clearly similar to $B$. Therefore, the mistake must be...

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You can say that such a matrix maps the first quadrant (octant, whatever) into itself via $x \mapsto Ax$. Either this map sends some nonzero vector $v$ in that quadrant to zero (in which case $0$ is an eigenvalue, and $v$'s a corresponding eigenvector), or... ...you can then consider the map, in the first quadrant/octant/whatever of the unit sphere, ...

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The matrix $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has two one-dimensional eigen spaces, one generated by the vector $[1,1]^T$ and one by $[1,-1]^T$. On the other hand, $A^HA=I$, so every vector in $\mathbb C^2$ is an eigenvector of $I$. Conclusion: No. The set of eigenvectors of $A^HA$ is not the same as the set of eigenvectors of $A$.

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Suppose $\left\Vert A\left(x^{\star}\right)\right\Vert=K<1$. For all $\epsilon>0$, there exists $r>0$ s.t. $0<\left\Vert x^{\star}-x\right\Vert<r$ implies $$\left|\left\Vert A\left(x\right)\right\Vert-\left\Vert A\left(x^{\star}\right)\right\Vert \right|\leq\left\Vert A\left(x\right)-A\left(x^{\star}\right)\right\Vert<\epsilon.$$ If ...

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Ok, after some fun roundabouts and a nice long learning experience, I have the following. it's long, but it answers things. The algebra I was attempting to perform was giving erroneous results because $e^{Ax}=(e^A)^x$ is only strictly true with matrix $A$ when $x$ is an integer. Otherwise calculations of $(e^A)^x$ will involve solving a set of ...

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