What happens when a perfectly antisymmetric matrix develops a one-dimensional flaw — and why that flaw turns out to be unreasonably well-behaved.
Following P. N. Choudhury & K. C. Sivakumar, Linear Algebra and its Applications 482 (2015), 55–69.
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For any $A\in\mathbb{R}^{n\times n}$ define $$S(A)=\tfrac{1}{2}(A+A^t),\qquad K(A)=\tfrac{1}{2}(A-A^t).$$
$S(A)$ is symmetric, $K(A)$ is skew-symmetric, $A=S(A)+K(A)$, and this splitting is unique.
Existence: Check $(S(A))^t = S(A)$ and $(K(A))^t = -K(A)$ directly; sum gives $A$.
Uniqueness: If $A = S_1+K_1 = S_2+K_2$, then $M := S_1-S_2 = K_2-K_1$ is simultaneously symmetric and skew-symmetric, so $M = M^t = -M$, giving $M=0$. $\square$
$\operatorname{rk}(S(A))=0$ $\iff$ $A$ is skew-symmetric. The rank of $S(A)$ measures how far $A$ is from being skew-symmetric.
The quadratic form of $A$ at $x$ is the scalar $x^tAx$. The decomposition collapses it:
A $x^tKx$ is a $1\times1$ scalar, so it equals its own transpose:
$x^tKx = (x^tKx)^t = x^tK^tx = x^t(-K)x = -x^tKx.$
Hence $2x^tKx = 0$, so $x^tKx = 0$. Then $x^tAx = x^t(S(A)+K(A))x = x^tS(A)x + 0$. $\square$
Swapping $K(A)$ for any other skew-symmetric matrix leaves $x^tAx$ unchanged for every $x$. Two matrices with the same symmetric part define identical quadratic forms.
When $A$ is almost skew-symmetric we have $S(A)=ww^t$ (one nonzero eigenvalue $\lambda>0$, unit eigenvector $q$, $w=\sqrt\lambda\,q$). Substituting into slide 3:
$x^tAx = x^tS(A)x = (w^tx)^2 = \lambda(q^tx)^2$
An almost skew-symmetric matrix, however large its spin $K(A)$, is analytically no richer than a rank-one projection. One number $\lambda$ and one direction $q$ capture everything.
A $(w^tx)^2=0 \iff w^tx=0 \iff x\perp w$. The zero set is $w^\perp = N(S(A))$, a hyperplane of dimension $n-1$. The form is strictly positive only in the direction of $w$ (and its negative).
A No. The $(1,1)$ entry would need $v_1^2 = -1$, impossible over $\mathbb{R}$. $A$ is symmetric and rank-1 but has a negative eigenvalue $-1$. Positive semidefiniteness is the missing hypothesis.
$A = vv^t$ for some nonzero real $v$ $\iff$ $A$ is symmetric, PSD, and rank-one.
($\Rightarrow$) $(vv^t)^t=vv^t$ ✓; $(v^tx)^2\ge0$ ✓; columns are $v_j v$, so rank = 1 ✓.
($\Leftarrow$) Spectral theorem: $A=\sum\lambda_i q_iq_i^t$. Rank-1 forces exactly one nonzero eigenvalue $\lambda_1\ne0$, so $A=\lambda_1 q_1q_1^t$. PSD: $q_1^tAq_1=\lambda_1\ge0$, and $\lambda_1\ne0$ gives $\lambda_1>0$. Set $v=\sqrt{\lambda_1}\,q_1$; then $vv^t=A$. $\square$
Uniqueness: If also $A=uu^t$, then $u=\alpha v$ with $\alpha^2=1$, so $u=\pm v$.
$S(A)$ is always symmetric; it is PSD (as $x^tS(A)x=x^tAx=(w^tx)^2\ge0$); when rk$(S(A))=1$ this theorem guarantees a real $w$ with $S(A)=ww^t$.
$A\in\mathbb{R}^{n\times n}$, $n\ge2$, is almost skew-symmetric if $$\operatorname{rk}(S(A))=1.$$
$S(A)=\delta(A)\,vv^t$ for a unit $v$ and real $\delta(A)\ne0$. Since $-A$ has $S(-A)=-S(A)$, we always replace $A$ by $-A$ if needed so that $\delta(A)>0$. Then $S(A)=ww^t$ for a real $w=\sqrt{\delta(A)}\,v$.
$S(A^t)=\tfrac12(A^t+(A^t)^t)=S(A)$. So $A$ is almost skew-symmetric exactly when $A^t$ is. The property is shared by a matrix and its transpose.
Take $w=(1,2,3)^t$. Then $S(A)=ww^t=\begin{pmatrix}1&2&3\\2&4&6\\3&6&9\end{pmatrix}$, rank 1. Add any skew matrix $K$, e.g., with $(1,2)$-entry $+1$: $A=S(A)+K=\begin{pmatrix}1&3&3\\1&4&6\\3&6&9\end{pmatrix}$. Almost skew-symmetric ✓.
A From slides 3–4: $x^tAx = x^tS(A)x = (w^tx)^2 = \lambda(q^tx)^2\ge0$ for all $x$. $\square$
Three lines from the decomposition. No eigenvalue computation needed.
Every almost skew-symmetric matrix with $\delta(A)>0$ is PSD: $$x^tAx=(w^tx)^2=\lambda(q^tx)^2\ge0.$$
PSD matrices are always range-symmetric ($R(A)=R(A^t)$). For almost skew-symmetric $A$ this means the Moore–Penrose inverse $A^\dagger$ and the group inverse $A^\#$ coincide — two objects that are generally distinct collapse to one.
$A=(a_{ij})$ is almost skew-symmetric (with $\delta(A)>0$) $\iff$ all four hold:
(i) $a_{ii}\ge0$ for all $i$; (ii) at least one $a_{ii}>0$;
(iii) $a_{ii}=0 \Rightarrow a_{ik}+a_{ki}=0\ \forall k$;
(iv) $a_{ii}\ne0$, $a_{jj}\ne0$ $\Rightarrow$ $\exists\, k$ with $k^2=a_{jj}/a_{ii}$ and $a_{ij}+a_{ji}=2k\,a_{ii}$.
A $a_{11}=1$, $a_{22}=4$. Then $k^2=4/1=4$, so $k=2$. Check: $a_{12}+a_{21}=3+1=4=2k\cdot a_{11}=2(2)(1)=4$. ✓
Similarly for all other pairs — this is the $w=(1,2,3)^t$ example, so $k_{ij}=w_j/w_i$.
Whenever both $a_{ii},a_{jj}\ne0$: $(a_{ij}+a_{ji})^2 = 4\,a_{ii}\,a_{jj}$. The diagonal determines the symmetric off-diagonal pairs up to a sign choice.
Pick any $w\in\mathbb{R}^n$, $w\ne0$. Set $S(A)=ww^t$. Add any skew-symmetric $K$. Result: $A=ww^t+K$ is almost skew-symmetric.
Diagonal: $(1,4,9)$. Sign choices: $k_1=2$, $k_2=3$. Any skew $K$ works.
The set $\mathcal{A}_n$ is not closed and not closed under addition.
$A_m=\operatorname{diag}(1/m,0)\in\mathcal{A}_2$ but $A_m\to0\notin\mathcal{A}_2$.
$\operatorname{diag}(1,0),\operatorname{diag}(0,1)\in\mathcal{A}_2$ but their sum is $I$ with $\operatorname{rk}(S(I))=2\ne1$.
A For $I$, all $a_{ii}=1>0$, so condition (iv) requires $a_{ij}+a_{ji}=2k\cdot1$ for some $k$ with $k^2=1$. But the off-diagonal entries of $I$ are zero: $0+0=0\ne\pm2$. Condition (iv) is violated. Alternatively: $S(I)=I$ has rank $n\ne1$.
Let $A,B\in\mathcal{A}_n$ with $S(A)=uu^t$, $S(B)=ww^t$.
A $S(A+B)=uu^t+ww^t$. This has rank 1 iff $u$ and $w$ are linearly dependent. If $w=\alpha u$: $S(A+B)=(1+\alpha^2)uu^t$, rank 1 since $1+\alpha^2>0$ always — sums cannot cancel the defect. $\square$
Sum: $A+B\in\mathcal{A}_n$ $\iff$ $u\parallel w$ $\iff$ $\operatorname{diag}(A)$ and $\operatorname{diag}(B)$ are proportional.
Difference: $A-B\in\mathcal{A}_n$ $\iff$ $u\parallel w$ and $u\ne\pm w$. When $w=\pm u$, the symmetric parts cancel exactly and $A-B$ degenerates to skew-symmetric.
$A+B\in\mathcal{A}_n$ iff $\operatorname{diag}(A)$ and $\operatorname{diag}(B)$ are linearly dependent as vectors in $\mathbb{R}^n$ — a condition readable directly from the entries.
For $A\in\mathbb{R}^{m\times m}$ and $B\in\mathbb{R}^{n\times n}$, partitioned so $A_{22}$ and $B_{11}$ are both $k\times k$: $$A\oplus_k B = \begin{pmatrix}A_{11}&A_{12}&0\\A_{21}&A_{22}+B_{11}&B_{12}\\0&B_{21}&B_{22}\end{pmatrix}.$$ The blocks overlap and add at the seam.
A Theorem 3.5: $A\oplus_k B\in\mathcal{A}_{m+n-k}$ iff
(i) $\operatorname{diag}(A_{11})=0$ and $\operatorname{diag}(B_{22})=0$ — the "outer" diagonals vanish;
(ii) $\operatorname{diag}(A_{22})$ and $\operatorname{diag}(B_{11})$ are proportional — the "overlap" diagonals align.
Intuition: two independent nonzero diagonal blocks would force rk$(S)\ge2$.
Any almost skew-symmetric matrix with a block-tridiagonal zero-corner pattern decomposes as a subdirect sum of two almost skew-symmetric matrices (or one almost skew-symmetric + one skew-symmetric).
Declare $A\sim B$ if $\operatorname{rk}(K(A))=\operatorname{rk}(K(B))$. This partitions $\mathcal{A}_n$ into classes $\mathcal{A}_n^k$.
A No — real skew-symmetric matrices always have even rank. Nonzero eigenvalues of $K$ come in purely imaginary conjugate pairs $\pm i\lambda_j$, so rank $=2r$ always. The nonempty classes are $\mathcal{A}_n^0,\mathcal{A}_n^2,\mathcal{A}_n^4,\ldots,\mathcal{A}_n^{2\lfloor n/2\rfloor}$.
$K(A)=0$, so $A=S(A)=ww^t$. Exactly the rank-1 PSD symmetric matrices. As a manifold: $\mathbb{R}_{>0}\times\mathbb{R}P^{n-1}$, dimension $n$.
$A=ww^t+uv^t-vu^t$ for nonzero $w$ and linearly independent $u,v$. The simplest "mixed" class; first appears at $n=2$.
All use $w=(1,1,1,1,1,1)^t$, so $S(A)=J_6$. Only $K(A)$ changes. Write $K_{ij}=e_ie_j^t-e_je_i^t$.
| Class | $K(A)$ | Changed entries vs $J_6$ | rk$(K)$ | Nonempty? |
|---|---|---|---|---|
| $\mathcal{A}_6^0$ | $0$ | $A=J_6$ | 0 | Yes |
| $\mathcal{A}_6^1$ | — | — | 1 | Empty |
| $\mathcal{A}_6^2$ | $K_{12}$ | $(1,2)\to2,\;(2,1)\to0$ | 2 | Yes |
| $\mathcal{A}_6^3$ | — | — | 3 | Empty |
| $\mathcal{A}_6^4$ | $K_{12}+K_{34}$ | above + $(3,4)\to2,\;(4,3)\to0$ | 4 | Yes |
| $\mathcal{A}_6^5$ | — | — | 5 | Empty |
| $\mathcal{A}_6^6$ | $K_{12}+K_{34}+K_{56}$ | above + $(5,6)\to2,\;(6,5)\to0$ | 6 | Yes |
Each new spin plane adds one independent $2\times2$ rotation block $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Three such blocks fill all of $\mathbb{R}^6$, reaching the maximum possible skew rank of 6.
$K(A)_{ii}=(a_{ii}-a_{ii})/2=0$ always, so $a_{ii}=S(A)_{ii}=w_i^2\ge0$. The number of nonzero diagonal entries equals $|\operatorname{supp}(w)|\in\{1,\ldots,n\}$.
Declare $A\sim B$ if they have the same number of nonzero diagonal entries. This gives exactly $n$ nonempty classes $D_1,\ldots,D_n$.
Exactly one nonzero $a_{pp}$. Then $S(A)=a_{pp}e_pe_p^t$, so $a_{ij}+a_{ji}=0$ for all $i\ne j$. The matrix is antisymmetric in every off-diagonal pair — the nearest neighbour of skew-symmetric matrices in $\mathcal{A}_n$.
All diagonal entries nonzero. Generic case. The ratio constraint $(a_{ij}+a_{ji})^2=4a_{ii}a_{jj}$ holds for every pair; $S(A)$ is fully determined by the diagonal up to a sign per pair.
A Yes — for any valid $2r\le n$ and $k\in\{1,\ldots,n\}$, take $A=ww^t+K$ where $\operatorname{supp}(w)$ has size $k$ and $\operatorname{rk}(K)=2r$. The two parameters are independent, giving at most $\lfloor n/2\rfloor\cdot n$ nonempty cells. $\mathcal{A}_n^0\cap D_k$ contains all symmetric rank-1 matrices with exactly $k$ nonzero diagonal entries — parametrized by the $\binom{n}{k}$ support subsets.
$A$ is $n\times n$ with entries in $\{0,1\}$ and almost skew-symmetric $\iff$ there exists nonempty $U\subseteq[n]$ with $A=\mathbf{1}_U\mathbf{1}_U^t$ (the indicator matrix of $U\times U$). There are exactly $2^n-1$ such matrices — one per nonempty subset.
A $w_i^2=a_{ii}\in\{0,1\}$ gives $w_i\in\{0,\pm1\}$. Off-diagonal: $w_iw_j=S(A)_{ij}\in\{0,\frac12,1\}$. But $w_iw_j=-1$ if $w_i$ and $w_j$ have opposite signs — and $-1\notin\{0,\frac12,1\}$. Contradiction. So all nonzero $w_i$ share the same sign. Taking them positive gives $w=\mathbf{1}_U$.
Every 0,1 almost skew-symmetric matrix is symmetric ($A_{ij}=A_{ji}$), so $K(A)=0$ — all $2^n-1$ matrices lie in $\mathcal{A}_n^0$. The $\binom{n}{k}$ matrices with $|U|=k$ occupy $\mathcal{A}_n^0\cap D_k$.
Any matrix $M$ with entries in $\{a,b\}$ writes uniquely as $M=aJ+(b-a)A$ where $A$ is the 0,1 indicator matrix of the $b$-positions. Taking symmetric parts: $S(M)=a\mathbf{1}\mathbf{1}^t+(b-a)S(A)$.
A Exactly 2: the constant matrices $aJ$ and $bJ$ (provided $b\ne0$; if $b=0$ just $aJ$). No mixed matrix is almost skew-symmetric when $a\ne0$.
Proof sketch: rk$(S(A))=1$ requires $\mathbf{1}$ and $\mathbf{1}_U$ to be proportional (sum theorem), forcing $U=[n]$, giving $M=bJ$. Or $A=0$, giving $M=aJ$. All other cases have rk$(S(M))\ge2$.
| Alphabet | Count | Matrices |
|---|---|---|
| $\{0,b\}$ | $2^n-1$ | $b\cdot\mathbf{1}_U\mathbf{1}_U^t$, all nonempty $U$ |
| $\{a,b\}$, $a\ne0\ne b$ | $2$ | $aJ$ and $bJ$ only |
| $\{1,2\}$ ($n=3$) | $2$ | $J$ and $2J$ |
| $\{5,6\}$ ($n=3$) | $2$ | $5J$ and $6J$ |
The source paper goes further. Part II will cover:
Moore–Penrose inverse of almost skew-symmetric matrices — explicit formulas
Tucker's theorem analogue — sign patterns for almost skew-symmetric matrices
A new proof of Farkas' lemma via almost skew-symmetric matrices
P. N. Choudhury, K. C. Sivakumar — Tucker's theorem for almost skew-symmetric matrices and a proof of Farkas' lemma — Linear Algebra and its Applications 482 (2015), 55–69. doi:10.1016/j.laa.2015.05.019