# linear algebra: image, kernel and transpose

Given that $A$ is a $(n\times n)$ skew-symmetric matrix $(A^t = -A)$ and that $\mathrm{Img}(A) = \ker(A)$.

Note here $n$ must be even from the rank nullity theorem, $\mathrm{rank}(A) = \dim (\ker A )= n/2$.

I need to show that there exists a $n/2$-dim subspace $V\subset \mathbb{R}^n$ such that the bilinear form $\langle v, Aw\rangle$ is non-degenerate.

My question is that even on $\mathbb{R}^n$, since we know $$\mathrm{Img}(A) = \ker(A^T)^\perp =\ker(- A)^\perp = \ker(A)^\perp$$ together with the given condition $\mathrm{Img}(A) = \ker(A)$, we would have $\ker(A)^\perp = \ker(A)$ but this is impossible.

• Yeah, I would try to ask for clarification on this problem if it is for a class since there seems to be something rotten in the state of Denmark. – user357980 Jun 25 '17 at 17:29

Indeed, there is no nonzero skew-symmetric matrix $A$ such that ${\rm Im} A\subset {\rm ker} A$. Because this last condition is equivalent to $A^2=0$, and since $A$ is skew-symmetric this implies that $A^TA=0$ and consequently $\Vert Ax\Vert=0$ for every vector $x$, i.e $A=0$.