[Math] One and only one method to write A as sum of symmetric matrix and skew symmetric matrix

linear algebraproof-verificationsymmetric matrices

Question: Let $A$ be any square matrix

(a) Show that $A+A^T$ is symmetric and $A-A^T$ is skew-symmetric.

(b)Prove that there is one and only one way to write A as the sum of a symmetric matrix and a skew-symmetric matrix

For part(a), my ans:

since $(A+A^T)^T=A^T+(A^T)^T=(A^T+A)$ , thus $A+A^T$ is symmetric

since $(A-A^T)^T=A^T-(A^T)^T=A^T-A=-(A-A^T)$, thus $(A-A^T)$ is skew symmetric

For part (b),

Assume there is more than one way to write A as the sum of symmetric and skew_symmetric matrix

Let $B_1$ and $B_2$ be symmetric matrix , $C_1$ and $C_2$ be skew-symmetric matrix, such that

$$A=B_1+C_1$$

$$A=B_2+C_2$$

the difference of two-equation is $$C_2-C_1=B_1-B_2$$

Note that $C_2-C_1$ is a skew-symmetric matrix and $B_1-B_2$ is a symmetric.

The contradiction occurs since there does not exist matrix that are both skew-symmetric and symmetric [zero matrix is an exceptional case since it implies A=B+C is the one and only one solution. ]

Therefore, there is one and only one way to write A as sum of symmetric and skew-symmetric.

Is it a correct proof?

Best Answer

To satisfy the sticklers, you'll need to use or prove the fact that a square matrix is symmetric and skew-symmetric iff it's a zero matrix, and to show there's exactly one suitable decomposition you must provide it as $A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$.

Another strategy that doesn't require a proof by contradiction is to note symmetric $B$ and skew-symmetric $C$ satisfy $A=B+C$ iff $A=B+C$ and $A^T=B-C$, i.e. iff $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$. This has the advantage that you prove both parts (at most one, at least one) in a single step.

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