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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.

Jendrik Stelzner
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Yassin Rany
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6 Answers6

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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 $$C={(A-A^T)\over 2}\\ $$

Ejik_v_tumane
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  • 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. – Marc van Leeuwen May 21 '22 at 04:41
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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.

Ashwin Ganesan
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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 $A=B+C$

marwalix
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0

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

Eunsung Lim
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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

Marc van Leeuwen
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Jala srihith
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community May 21 '22 at 04:09
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"if someone does not mind to expose another way"...

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.

Marc van Leeuwen
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