5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ look at $a$ and it transpose $a^\top$ $\endgroup$
    – abel
    Commented Jun 7, 2016 at 17:09

6 Answers 6

4
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Commented May 21, 2022 at 4:41
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $A=B+C$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Community Bot
    Commented May 21, 2022 at 4:09
0
$\begingroup$

"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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .