# Any $n \times n$ matrix $A$ can be written as $A = B + C$ with $B$ is symmetric and $C$ skew-symmetric.

How can I proof the following statement?

Any $n \times n$ matrix $A$ can be written as a sum $$A = B + C$$ where $B$ is symmetric and $C$ is skew-symmetric.

I tried to work out the properties of a matrix to be symmetric or skew-symmetric, but I could not prove this. Does someone know a way to prove it?

Thank you.

PS: The question Prove: Square Matrix Can Be Written As A Sum Of A Symmetric And Skew-Symmetric Matrices may be similiar, in fact gives a hint to a solution, but if someone does not mind in expose another way, our a track to reach to what is mentioned in the question of the aforementioned link.

• look at $a$ and it transpose $a^\top$
– abel
Commented Jun 7, 2016 at 17:09

Suppose $$A=B+C$$ If $$B^T=B,$$ $$C^T=-C,$$ then according to the known property of transposition of sum of matrices $$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$ Now we have $$A=B+C \tag 1\\$$ $$A^T=B-C\tag 2\\$$ Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\$$ Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\$$

• This is written to prove that if $A$ is written as $B+C$ with $B$ symmetric and $C$ anti-symmetric, then $B$ and $C$ must be given by the expressions you give. But the question was to prove that $A$ can be so written. Commented May 21, 2022 at 4:41

Given a matrix $A$, let $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$. Observe that $B^T=B$, so $B$ is symmetric. Also, $C^T = -C$, so $C$ is skew-symmetric.

Let $A$ be our matrix and

\begin{align}B&={A^T+A\over 2}\\C&={A-A^T\over 2}\end{align}

We have $B^T=B$ and $C^T=-C$ and $A=B+C$

For a matrix $A$,

$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$, where the first one is symmetric, and second one is skew-symmetric

Let $$A$$ be any square matrix. We can write $$A=\frac12(A+A^\top)+(A-A^\top)=P+Q$$, say, where $$P=\frac12(A+A^\top)$$ And $$Q=\frac12(A-A^\top)$$ we have $$P^\top=(\frac12(A+A^\top))^\top=\frac12(A+A^\top)^\top= \frac12{A^\top+(A^\top)^\top}=\frac12(A+A^\top)=P.$$ Therefore $$P$$ is a symmetric matrix. Now $$Q^\top=(\frac12(A-A^\top))^\top=\frac12(A-A^\top)^\top\\ =\frac12(A^\top-(A^\top)^\top) =\frac12(A^\top-A)=-\frac12(A-A^\top)=-Q.$$ Therefore $$Q$$ is a skew-symmetric matrix

Though nothing is mentioned in the statement about this, I will assume the matrices considered have entries in a field$$~F$$ that is not of characteristic$$~2$$ (in other words $$1$$ and $$-1$$ are distinct elements in $$F$$), since the statement is false for matrices with entries in a field of characteristic$$~2$$.
The set of $$n\times n$$ matrices with entries in $$F$$ is (with usual addition and scalar multiplication) a vector space $$V$$ over$$~F$$, and the operation of transposition defines a linear map $$T:V\to V$$. Since $$T\circ T=I$$, the polynomial $$X^2-1=(X-1)(X+1)\in F[X]$$ is an annihilating polynomial of $$T$$. Since this polynomial, which is split as indicated, has distinct roots $$1$$ and $$-1$$ in $$F$$, the operator $$T$$ is diagonalisable with eigenvalues in $$\{1,-1\}$$. This means that the sum of the two eigenspaces $$V_\lambda$$ for $$\lambda=1$$ and $$\lambda=-1$$, which is a direct sum, fills the whole space: $$V=V_1\oplus V_{-1}$$. But $$V_1$$ is the subspace of matrices $$M$$ satisfying $$T(M)=M$$, in other words of symmetric matrices, and similarly $$V_{-1}$$ is the subspace of anti-symmetric matrices.
The fact that $$V=V_1+V_{-1}$$ says that every matrix in $$V$$ can be written as sum of a symmetric and an anti-symmetric matrix (that was the question); moreover the fact that the sum is direct says that they can be so written in a unique manner.