If $A^T=-A$, then A is not invertible Let $n \in \mathbb{N}$ be odd and $A \in$Mat$(n,\mathbb{R})$ with $A^T=-A$. Show that $A$ is not invertible.
I have no idea how to start this...
 A: Hints:
1) What do you know about the determinant of an invertible matrix? What about the effect on the determinant of transposing a matrix and multiplying it by a scalar? 
2) Write $|A|=a$. Now use what you know, and the assumption that $A^T=-A$, in order to obtain equalities for $a$. Deduce that $a=0$.
A: Hint: $\det\left( A^{\mathrm T}\right ) = \det A$, and $\det(-A)=(-1)^n\det A$.
A: Let me give my own solution because it's a bit different:
If $A^T=-A$, then the eigenvalues of $A$ are purely imaginary (zero is of course a possibility), because if $Ax=\lambda x$, then:
$$
\lambda\, |x|^2=\lambda\langle x,x\rangle=\langle Ax,x\rangle=\langle x,A^Tx\rangle=
\langle x,-Ax\rangle=\langle x,-\lambda x\rangle=-\bar\lambda\,|x|^2.
$$
But if $n$ is odd, then $A$ should have a real eigenvalue, which is necessarily equal to zero.
A: Since $A$ is real and of odd size, its characteristic polynomial is of odd degree with real coefficients.  Such a polynomial has at least one real root, hence $A$ has at least one real eigenvalue.  BUT  any real eigenvalue of such a matrix must be zero, since if $Ax = \mu x$ for $\mu \in \Bbb R$ and $x \ne 0$, 
$\mu \langle x, x \rangle = \langle x, \mu x \rangle = \langle x, Ax \rangle = \langle A^Tx, x \rangle = -\langle Ax, x \rangle = -\langle \mu x, x \rangle = -\mu \langle x, x \rangle. \tag{1}$
whence, since $\langle x, x \rangle \ne 0$,
$\mu = -\mu \tag{2}$
or
$\mu = 0. \tag{3}$
The above considerations show that $0$ must be an eigenvalue of $A$, so $A$ is not invertible, having a non-trivial kernel.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Slightly more precisely, $A$ is invertible on a subspace of even dimension, because its eigenvalues come in pairs $\pm \lambda$, where the same pair can be repeated (eigenvalues have multiplicity).  When $n$ is odd this means $A$ is non-invertible on a subspace of odd dimension, in particular it is not an invertible matrix, since the space on which it is non-invertible cannot have dimension $0$.
The pairing follows from two general facts true for all matrices: $-A$ has eigenvalues the negative of $A$'s, and $A^T$ has the same eigenvalues as $A$. 
The same arguments show that a skew-Hermitian matrix, $A^T = -\overline{A}$, has an even number of eigenvalues with nonzero real part.   
