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Hint Consider the eigenvalues of $A \in SO(3, \Bbb R)$. In particular, (1) their product is $\det A = 1$, (2) they all have modulus $1$, and (3) any nonreal eigenvalues come in complex conjugate pairs. Now, if $A \neq I$, what can you say about the $1$-eigenspace of $A$?


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If $A$ is nonsingular, then $AA^{-1} = I$, so $$ 1 = ||I|| = ||AA^{-1}|| \leqslant ||A||\cdot||A^{-1}||. $$ In general, then $1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||$. Equality is thus not necessarily guaranteed for arbitrary nonsingular $A$; however, the inequality above implies that equality may occur. Consider an ...


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A direct argument may be obtained by arguing by induction. You should try the small cases $n = 2, 3, 4$ to see what this means in practice. Fix $k$. Say that a matrix $B$ satisfies condition null-$k$ if $b_{ij} = 0$ if $i \ge j - k$. (Clearly $B = A$ satisfies null-$0$.) Suppose $B$ satisfies null-$k$. Consider the product $C = A B$, and let $c_{ij}$ be ...


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$$ \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) $$


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The characteristic polynomial of a linear map is defined without any additional structure beyond that of a finite dimensional vector space while the notion of unitarity or normality requires the structure of an inner product so you should be suspicious if someone tells you that the characteristic polynomial tells you something about unitarity or normality. ...


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This is obvious using Hamilton-Cayley's theorem: the only eigenvalue of such a matrix is $0$, since its characteristic polynomial is $\chi_A(x)=(-1)^nx^n$. By Hamilton-Cayley, $$\chi_A(A)=0=(-1)^nA^n.$$


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The rank is 2 (the first two columns span the image) and trace is zero. If there is one non-zero eigenvalue $a$, there is another non-zero eigenvalue $-a$. Hence the image would be spanned by the two eigenvectors of the non-zero eigenvalues and the same would automatically hold for all powers of $A$. But the first column is contained in the kernel and in the ...



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