A symmetric matrix is a square matrix that is equal to its transpose.

A skew-symmetric (or antisymmetric) matrix is a square matrix whose transpose is also its negative.

  1. Assume that $A \in M_n(\mathbb C)$ is a skew symmetric matrix. Prove that $adj(A)$ is symmetric or skew symmetric depending on $n$ being even or odd.
  2. If $A \in M_n(\mathbb C)$ is skew symmetric and $n$ is odd, then $A$ is not invertible.

Note : I wrote $(adjA)^T=adj(A^T)=adj(-A)$ but i don't know what happens next.

  • $\begingroup$ What is $F$ here? And could you define $adj(A)$. $\endgroup$ – Jacky Chong Oct 18 '16 at 22:41
  • $\begingroup$ @JackyChong adj(A) is the adjucate matrix ... $F$ is $\mathbb C$ i forgot it ... $\endgroup$ – Arman Malekzadeh Oct 18 '16 at 22:42
  • $\begingroup$ $A \cdot adj(A)=|A|, (-A)\cdot adj(-A)=|-A|=(-1)^n|A|.$ $\endgroup$ – mfl Oct 18 '16 at 22:45
  • $\begingroup$ @mfl i don't get the connection between $A.adj(A)$ and $adj(A)^T$ ... would you please write an answer ? $\endgroup$ – Arman Malekzadeh Oct 18 '16 at 22:50
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    $\begingroup$ You have that $A\cdot adj(A)=|A| I$ and $A\cdot adj(-A)=(-1)^{n+1}|A|I.$ Thus, if $A$ is invertible, we have $adj(-A)=(-1)^{n+1}adj(A),$ which is the identity you are looking for. Of course, it works also if $A$ is not invertible. $\endgroup$ – mfl Oct 18 '16 at 22:54

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