I believe that pure imaginary eigenvalues can exist for $2n \times 2n$ matrices only when all the entries of the matrix are real. However, what if $n$ is odd? Is there still a matrix which has all its eigenvalues pure imaginary?
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4Absolutely. Consider the matrix $[i],$ for example. – Cameron Buie Sep 30 '23 at 21:05
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Does this answer your question? Why are all nonzero eigenvalues of the skew-symmetric real matrices pure imaginary? – cpiegore Sep 30 '23 at 21:09
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Are you talking about real matrices or complex matrices? If real, remember that the complex roots of a real polynomial must come in conjugate pairs. – Ted Shifrin Sep 30 '23 at 21:55
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For $n$ odd the characteristic polynomial is of odd degree Hence at least one root is a real number. – Ryszard Szwarc Sep 30 '23 at 22:24
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I am only considering a matrix having real entries. – MathCurious Oct 01 '23 at 02:27
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So, as several of us have explained, in that case $n$ must be even. – Ted Shifrin Oct 01 '23 at 03:03
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Thank you so much. It is clear. – MathCurious Oct 01 '23 at 03:51
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All eigenvalues of the zero matrix are purely imaginary. – user1551 Oct 01 '23 at 20:55