Assume $A^T + B^T = A - B$, prove $A$ is symmetric and $B$ is antisymmetric

Question

Let $A, B$ be square matrices ($n \times n$). Assume $A^T + B^T = A - B$ prove $A$ is symmetric and $B$ is antisymmetric.

---

Having trouble with this question. Thought about using $A^T + B^T = (A + B)^T$ but got stuck.

Answer 1

$$A^T+B^T=A-B$$ Transpose the whole equation, using that transposition is linear and involutive:

$$A+B=A^T-B^T$$

Which is the same as

$$A^T-B^T=A+B$$

Now, add this equation to the original one:

$$2 A^T=2A$$

Or substract it from the original one:

$$2B^T=-2B$$

Answer 2

A good way to look at it is to contruct A and B as sums of $A + B$ an $A - B$. You can then calculate $A^T$ and $B^T$ easily.

Answer 3

$A^T + B^T = (A+B)^T$

Therefore, $(A-B)^T = A+B$

Hence, $A = (A - B)^T - B$ and $B = (A - B)^T - A$

Now, transpose $A$ and $B$ to find what you want.

Answer 4

We have $a_{ij}+b_{ij}=a_{ji}-b_{ji}$ and also $a_{ji}+b_{ji}=a_{ij}-b_{ij}$.

If $i\neq j$ then these equalities together imply that $b_{ij}+b_{ji}=0$ and $a_{ij}=a_{ji}$.

Answer 5

You use that $A = A^T - B^T - B$ and then $A^T = (A^T - B^T - B)^T = A - B - B^T = A$. In similar way you get $B^T = (A-A^T-B^T)^T = A^T - A - B = -B$.

You also use that transposing twice get you back, ie that $A^{TT} = A$ and $B^{TT} = B$.

Answer 6

All answers are necessarily similar, so the point is just arranging the computations in an appealing way. I like this: $$ \begin{aligned} A^\top+B^\top&=A-B & & \\ A+B&=A^\top-B^\top & \text{Transpose} \\ A^\top-B^\top &=A+B& \text{Rearrange} \\ 2A^\top&=2A & \text{Subtract } (1)-(3) \\ 2B^\top&=-2B & \text{Add } (1)+(3) \\ \end{aligned} $$

Answer 7

You may rearrange the equation to get $B+B^T=A-A^T$. The LHS is symmetric, while the RHS is anti-symmetric. The only matrix that is both symmetric and anti-symmetric is zero. Hence $B+B^T=0=A-A^T$.