A is a real and non-symmetric matrix of order 3. prove that the rank of the matrix $A-A^t$ is 2. A is a real and non-symmetric matrix of order 3. prove that the rank of the matrix $A-A^t$ is 2.
We have $\rho(A)=\rho(A^t)$ where $\rho(A)$ is the rank of A.

Setting $A=\pmatrix{a&b&c\\d&e&f\\g&h&i} \ $, one has $\ B:=A-A^T=\pmatrix{0& b - d& c - g\\-b + d& 0& f - h\\-c + g& -f + h& 0}.$
The characteristic polynomial of $B$ is:
$$\tag{1}p(x)=- x\, (x^2 + \Delta)$$
where
$$\tag{2}\Delta:=b^2 + c^2 + f^2 + d^2 + g^2 + h^2 - 2\,b\,d  - 2\,c\,g  - 2\,f\,h$$
Letting $U=(b,c,f)$, $V=(d,g,h)$, we can write:
$$\tag{3}\Delta=\|U\|^2+\|V\|^2-2U.V=(U-V)^2 \ \ \text{in the sense of dot products}$$
But, as $A$ is assumed non symmetrical, $U\neq V$, thus $\Delta>0$.
Consequently, using (1), the spectrum of matrix $B$ is: 
$$\{0, -i\sqrt{\Delta}, +i\sqrt{\Delta}\}$$
Therefore, as $\Delta>0$, the rank of $B$ is always exactly 2.
Remark: I recognize that, at first, I thought that other cases could be possible.

Edit: there is an interesting geometrical interpretation of the eigenvalues.
Let us denote, in order to have simpler notations, 
$$B=\pmatrix{0& -\gamma & \beta \\ \gamma & 0& -\alpha \\-\beta& \alpha & 0} \ \ \text{and} \ \ X_0=\pmatrix{\alpha\\\beta\\\gamma}$$
Then, with :
$$X:=\pmatrix{x\\y\\z}, \ \ \text{on can check that} \ \ BX=X_0 \times X \ \ \text{(cross product)}$$
This explains that 


*

*$BX_0= X_0 \times X_0 = 0$, i.e., $BX_0=0X_0$, proving that $0$ is an eigenvalue of $B$ associated with eigenvector $X_0$.

*if $X_1$ is any vector in $X_0^{\perp}$ (the 2D plane orthogonal to $X_0$),
$BX_0= X_0 \times X_1$, which, in particular is an element of $X_0^{\perp}$ which is orthogonal to $X_0$ : it amounts to a $\pi/2$ rotation in plane $X_0^{\perp}$, this rotation being in full correspondance with multiplication by a purely imaginary number.
