As above,
When A is anti-symmetric, prove that if $c\in\mathbb{R}$ is non-zero, $A+cI$ is always invertible.
I've tried manipulating eigenvalues but it doesn't seem to work out. What would be the best approach to this problem?
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best approach is to find a counterexample! – user Mar 11 '18 at 23:49
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Please confirm the object of your OP about A symmetric or anti-symmetric. – user Mar 11 '18 at 23:55
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My guess is that you meant anti-symmetric, not symmetric. If that's so, then all eigenvalues of $A$ are purely imaginary (I am assuming that $A$ is a real matrix). Therefore, if $c\in\mathbb R\setminus\{0\}$, then $-c$ is not an eigenvalue of $A$ and so $A+c\operatorname{Id}$ is invertible.
José Carlos Santos
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1How would you go about showing all eigenvalues are purely imaginary? – David Reak Mar 11 '18 at 23:58
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I've found an answer over here https://math.stackexchange.com/questions/57100/why-are-all-nonzero-eigenvalues-of-the-skew-symmetric-matrices-pure-imaginary. Thank you! – David Reak Mar 12 '18 at 00:02