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Suppose that $f:V\to V$ is a linear transformation such that the matrix representation of $f$ with respect to every basis is skew-symmetric; that is, if $B$ is a basis of $V$ then $[f]_B^B= - \bigl([f]_B^B \bigr)^T$. Does this imply that $f$ must be the zero linear transformation?.

Thanks in advance.

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  • $\begingroup$ That's true. You can prove by contradiction: for every basis $B$ you set which $f$'s matrix representation with respect to $B$ is skew symmetric, you can find a basis $B'$, that $f$'s matrix representation with respect to $B'$ will have a non-zero element in it's diagonal, which makes it not skew-symmetric. $\endgroup$ – Itay4 Apr 14 '17 at 17:53
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I am not sure over which field your vector spaces are, but if you assume real vector spaces then for every real square matrix $A$ there exists a nonsingular real matrix $X$, for which $B := X^{−1}AX$ is a tridiagonal matrix. If $B$ is also to be skew symmetric then it must be obviously zero which proves $f=0$.

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