Prove that $ A = - A^{\top} $ and $ \text{rank}(A) \leq 1 $ imply $ A = \mathbf{0} $.

Question

Let $ A \in {\text{M}_{n}}(\Bbb{R}) $, and suppose that we have the following:

• $ A = - A^{\top} $.
• $ \text{rank}(A) \leq 1 $.

Why then is $ A = \mathbf{0} $?

Thanks!

Answer 1

Any at most rank 1 $n\times n$ real matrix $A$ can be written as $A=uv^T$ for some real $n$-vector $u$ and $v$. If $u$ or $v$ is zero, there is nothing to prove since then $A=0$ trivially.

Assume that $u$ and $v$ are nonzero. Using $A=-A^T$ and multiplying this equation with $v$ gives $$ Av=-A^Tv\implies (uv^T)v=(-vu^T)v\implies(v^Tv)u=-(u^Tv)v\implies u=-\frac{u^Tv}{v^Tv}v=:\alpha v, $$ since $v\neq 0$, that is, $u$ is a scalar multiple of $v$: $A=\alpha uu^T$. Hence $A$ is symmetric. A matrix, which is both symmetric ($A=A^T$) and skew-symmetric ($A=-A^T$) is a zero matrix: $$ A=\frac{1}{2}(A+A)=\frac{1}{2}(A^T-A^T)=0. $$

Answer 2

If a diagonal entry is $x$ then $A=-A^T$ $\Rightarrow$ $x=-x$, which implies $x=0$. Thus all diagonal entries are zero. Now if $a_{ij}$ is a non-diagonal entry, then $A=-A^T$ $\Rightarrow$ $a_{ij}=-a_{ji}$. Thus $a_{ij}\not=0$ $\Rightarrow$ $a_{ji}\not=0$. But $a_{ij}$ and $a_{ji}$ are in two different columns and two different rows (since $a_{ij}$ is off-diagnoal and therefore $i\not=j$), so if they are both non-zero the rank of the matrix must be at least two. Thus rank$(A)\leq1$ must imply $A=0$.

Answer 3

If $A$ has rank 1 then we have $A = w v^T $ for some vectors $v,w\in\mathbb{R}^n$. Since $A=-A^T$ all diagonal entries must be $0$. Try to show $w = \alpha v$. This gives $v_i^2=0$ so $v_i=0$ for all $1\leq i\leq n$.

Answer 4

Since $A$ is skew symmetric, it's diagonalizable with eigenvalues being purely imaginary and coming in conjugate pairs. Being of rank one forces all eigenvalues to be 0.

Edit: As $A$ is skew symmetric, it is normal. So $A=UDU^*$ for some diagonal matrix $D$ and unitary matrix $U$. We then have $-A^{\textsf{T}}=\overline{U}(-D)U^{\textsf{T}}$, which by assumption is equal to $UDU^*$. Since $A$ is real, $UDU^*=\overline{UDU^*}=\overline{U}\overline{D}U^{\textsf{T}}$. Comparing, we have $\overline{D}=-D$. So the eigenvalues must be purely imaginary. Since the eigenvalues are the roots of the characteristic polynomial which is real, they must appear in conjugate pairs.